Introduction
Let X = ( X t ) t ≥ 0 X=(X_t)_{t\geq0} X = ( X t ) t ≥ 0 α \alpha α X 0 = 0 X_0=0 X 0 = 0
E [ e i ξ X t ] = e − t ∣ ξ ∣ α , 1 < α < 2. \mathbb{E}\left[e^{\mathrm{i}\xi X_t}\right]=e^{-t|\xi|^\alpha}, \qquad1<\alpha<2. E [ e i ξ X t ] = e − t ∣ ξ ∣ α , 1 < α < 2. For a nonzero real number a a a
T a = inf { t > 0 : X t = a } T_a=\inf\{t>0:X_t=a\} T a = inf { t > 0 : X t = a } be its first hitting time of a a a infinitely divisible if, for every integer n ≥ 1 n\geq1 n ≥ 1 n n n T a T_a T a
For background on stable processes and their potential theory we refer to [1 , 11 ] . Explicit distributional descriptions of stable hitting times are also developed in [3 ] .
The distribution of a point-hitting time of a stable process has attracted considerable attention. Yano, Yano and Yor [9 ] obtained several distributional identities, including a relation with the α \alpha α [4 ] proved unimodality for the hitting times of general strictly stable processes. In [5 Proposition 4.1(b)] , they also proved infinite divisibility of the logarithm of the hitting time, and conjectured, immediately after that proposition, that the hitting time itself is infinitely divisible.
For the symmetric process, this conjecture is related to a question about the random variable C α C_\alpha C α
α sin ( π / α ) 2 π ( 1 + ∣ x ∣ α ) , x ∈ R . \frac{\alpha\sin(\pi/\alpha)}{2\pi(1+|x|^\alpha)}, \qquad x\in\mathbb{R}. 2 π ( 1 + ∣ x ∣ α ) α sin ( π / α ) , x ∈ R . Indeed, if X ^ \widehat{X} X α \alpha α
X ^ T a = d ∣ a ∣ C α ; \widehat{X}_{T_a} \mathrel{\overset{d}{=}} |a|C_\alpha; X T a = d ∣ a ∣ C α ; see [9 Theorem 5.3, formula (5.12)] . Consequently, infinite divisibility of T a T_a T a C α C_\alpha C α [8 Theorem 1.1] proved that C α C_\alpha C α 1 < α ≤ 2 1<\alpha\le2 1 < α ≤ 2 T a T_a T a
Here is our main result. The function K α K_\alpha K α Lemma 3.2 by inversion of − L α ′ / L α -L'_\alpha/L_\alpha − L α ′ / L α L α ( q ) = E [ e − q T 1 ] L_\alpha(q)=\mathbb{E}[e^{-qT_1}] L α ( q ) = E [ e − q T 1 ]
Theorem 1.1 (Computer-assisted classification). There is a unique number α ∗ ∈ ( 1 , 2 ) \alpha_*\in(1,2) α ∗ ∈ ( 1 , 2 ) a ≠ 0 a\ne0 a = 0
T a is infinitely divisible ⟺ 1 < α ≤ α ∗ . (1) T_a\text{ is infinitely divisible}\qquad\Longleftrightarrow\qquad1<\alpha\le\alpha_*. \tag*{(1)} T a is infinitely divisible ⟺ 1 < α ≤ α ∗ . (1) More precisely,
1.6670659903 < α ∗ < 1.6670659915. (2) 1.6670659903<\alpha_*<1.6670659915. \tag*{(2)} 1.6670659903 < α ∗ < 1.6670659915. (2) It is the parameter coordinate of the unique solution of
K α ( t ) = 0 , ∂ t K α ( t ) = 0 (3) K_\alpha(t)=0,\qquad\partial_tK_\alpha(t)=0 \tag*{(3)} K α ( t ) = 0 , ∂ t K α ( t ) = 0 (3) in
[ 1.66706598 , 1.667066 ] × [ .162799 , .162801 ] . [1.66706598,1.667066]\times[.162799,.162801]. [ 1.66706598 , 1.667066 ] × [ .162799 , .162801 ] . For 1 < α < α ∗ 1<\alpha<\alpha_* 1 < α < α ∗ K α ( t ) > 0 K_\alpha(t)>0 K α ( t ) > 0 t > 0 t>0 t > 0 α = α ∗ \alpha=\alpha_* α = α ∗ ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) T 1 T_1 T 1 K α ( t ) d t / t K_\alpha(t)\,dt/t K α ( t ) d t / t
The endpoint α = 2 \alpha=2 α = 2 e − ∣ a ∣ q e^{-|a|\sqrt{q}} e − ∣ a ∣ q
The proof has three parts. First, a finite large-frequency expansion constructs K α K_\alpha K α K α ≥ 0 K_\alpha\ge0 K α ≥ 0 [ 15 / 8 , 2 ) [15/8,2) [ 15/8 , 2 )
The finite enclosures concern whole parameter–time rectangles or whole parameter intervals, rather than samples. Appendix B specifies the certificate format, the finite acceptance conditions and the accompanying reproducibility files. The computational proof uses the guarantees of Arb/FLINT ball arithmetic and the explicit analytic truncation bounds established below. It is not a formal proof-assistant verification. No assertion about the classification of nonsymmetric stable processes is made here.
Normalization and the Laplace transform
Let X = ( X t ) t ≥ 0 X=(X_t)_{t\geq0} X = ( X t ) t ≥ 0 α \alpha α
E [ e i ξ X t ] = e − t ∣ ξ ∣ α , ξ ∈ R , t ≥ 0 , (4) \mathbb{E}[e^{i\xi X_t}]=e^{-t|\xi|^\alpha},\qquad\xi\in\mathbb{R},\quad t\geq0, \tag*{(4)} E [ e i ξ X t ] = e − t ∣ ξ ∣ α , ξ ∈ R , t ≥ 0 , (4) where 1 < α < 2 1<\alpha<2 1 < α < 2 a ≠ 0 a\ne0 a = 0
T a = inf { t > 0 : X t = a } , T = T 1 , L α ( q ) = E [ e − q T ] . T_a=\inf\{t>0:X_t=a\},\qquad T=T_1,\qquad L_\alpha(q)=\mathbb{E}[e^{-qT}]. T a = inf { t > 0 : X t = a } , T = T 1 , L α ( q ) = E [ e − q T ] . Write β = 1 / α \beta=1/\alpha β = 1/ α θ = π α / 2 \theta=\pi\alpha/2 θ = π α /2
Lemma 2.1. The random variable T a T_a T a
T a = d ∣ a ∣ α T . (5) T_a\mathrel{\mathop=^{\mathrm{d}}}|a|^\alpha T. \tag*{(5)} T a = d ∣ a ∣ α T . (5) For q > 0 q>0 q > 0
u q ( x ) = 1 π ∫ 0 ∞ cos ( x ξ ) q + ξ α d ξ . u_q(x)=\frac{1}{\pi}\int_0^\infty\frac{\cos(x\xi)}{q+\xi^\alpha}\,\mathrm{d}\xi. u q ( x ) = π 1 ∫ 0 ∞ q + ξ α cos ( x ξ ) d ξ . Then
E [ e − q T a ] = u q ( a ) u q ( 0 ) , u q ( 0 ) = q β − 1 α sin ( π β ) . (6) \mathbb{E}[e^{-qT_a}]=\frac{u_q(a)}{u_q(0)},\qquad u_q(0)=\frac{q^{\beta-1}}{\alpha\sin(\pi\beta)}. \tag*{(6)} E [ e − q T a ] = u q ( 0 ) u q ( a ) , u q ( 0 ) = α sin ( π β ) q β − 1 . (6) Proof. The resolvent identity is the standard hitting-time formula; in this normalization it is stated in [9 ] , formula (5.10). The value of u q ( 0 ) u_q(0) u q ( 0 ) ξ = q β v \xi=q^\beta v ξ = q β v
∫ 0 ∞ ( 1 + v α ) − 1 d v = π α sin ( π / α ) . \int_0^\infty(1+v^\alpha)^{-1}\,\mathrm{d}v=\frac{\pi}{\alpha\sin(\pi/\alpha)}. ∫ 0 ∞ ( 1 + v α ) − 1 d v = α sin ( π / α ) π . Moreover,
0 ≤ u q ( 0 ) − u q ( a ) ≤ 1 π ∫ 0 ∞ 1 − cos ( a ξ ) ξ α d ξ < ∞ . 0\leq u_q(0)-u_q(a)\leq\frac{1}{\pi}\int_0^\infty\frac{1-\cos(a\xi)}{\xi^\alpha}\,\mathrm{d}\xi<\infty. 0 ≤ u q ( 0 ) − u q ( a ) ≤ π 1 ∫ 0 ∞ ξ α 1 − cos ( a ξ ) d ξ < ∞. Since u q ( 0 ) → ∞ u_q(0)\to\infty u q ( 0 ) → ∞ q ↓ 0 q\downarrow0 q ↓ 0 [9 ] , formula (5.16). □
The normalization entails no loss of generality. If the characteristic exponent is c ∣ ξ ∣ α c|\xi|^\alpha c ∣ ξ ∣ α c > 0 c>0 c > 0 ∣ a ∣ α T / c |a|^\alpha T/c ∣ a ∣ α T / c
Lemma 2.2. For Re q > 0 \operatorname{Re}q>0 Re q > 0
L α ( q ) = α sin θ sin ( π β ) π ∫ 0 ∞ e − q β r r α 1 + 2 cos θ r α + r 2 α d r . (7) L_\alpha(q)=\frac{\alpha\sin\theta\sin(\pi\beta)}{\pi}\int_0^\infty\frac{e^{-q^\beta r}r^\alpha}{1+2\cos\theta\,r^\alpha+r^{2\alpha}}\,\mathrm{d}r. \tag*{(7)} L α ( q ) = π α sin θ sin ( π β ) ∫ 0 ∞ 1 + 2 cos θ r α + r 2 α e − q β r r α d r . (7) Proof. First suppose that q > 0 q>0 q > 0
∫ 0 ∞ e i x q + x α d x = i ∫ 0 ∞ e − y q + e i θ y α d y . \int_0^\infty\frac{e^{ix}}{q+x^\alpha}\,\mathrm{d}x
=i\int_0^\infty\frac{e^{-y}}{q+e^{i\theta}y^\alpha}\,\mathrm{d}y. ∫ 0 ∞ q + x α e i x d x = i ∫ 0 ∞ q + e i θ y α e − y d y . There is no pole in this quadrant because θ < π \theta<\pi θ < π α > 1 \alpha>1 α > 1
u q ( 1 ) = sin θ π ∫ 0 ∞ e − y y α q 2 + 2 q cos θ y α + y 2 α d y . u_q(1)=\frac{\sin\theta}{\pi}\int_0^\infty\frac{e^{-y}y^\alpha}{q^2+2q\cos\theta\,y^\alpha+y^{2\alpha}}\,\mathrm{d}y. u q ( 1 ) = π sin θ ∫ 0 ∞ q 2 + 2 q cos θ y α + y 2 α e − y y α d y . Substitute y = q β r y=q^\beta r y = q β r u q ( 0 ) u_q(0) u q ( 0 ) Re q > 0 \operatorname{Re}q>0 Re q > 0
Formula (2.4) is also obtained from the independent-product representation in [4 ] , Section 2.4 and final remark (a). The contour argument above fixes all normalizing constants directly. The formula is for the transform L α L_\alpha L α
The canonical logarithmic-derivative kernel
Throughout the paper, powers and logarithms of complex numbers have their principal values. Put
β = α − 1 , δ = π ( 1 − α / 2 ) , m = α + 1 , D δ ( z ) = 1 − 2 cos ( δ ) z + z 2 , κ α = α sin δ sin ( π / α ) π . (8) \beta=\alpha^{-1},\qquad\delta=\pi(1-\alpha/2),\qquad m=\alpha+1,\qquad D_\delta(z)=1-2\cos(\delta)z+z^2,\qquad\kappa_\alpha=\frac{\alpha\sin\delta\sin(\pi/\alpha)}{\pi}. \tag*{(8)} β = α − 1 , δ = π ( 1 − α /2 ) , m = α + 1 , D δ ( z ) = 1 − 2 cos ( δ ) z + z 2 , κ α = π α sin δ sin ( π / α ) . (8) Thus 0 < δ < π / 2 0<\delta<\pi/2 0 < δ < π /2 Lemma 2.2 is
L α ( q ) = κ α ∫ 0 ∞ e − q β r r α D δ ( r α ) d r , ℜ q > 0. (9) L_\alpha(q)=\kappa_\alpha\int_0^\infty\frac{e^{-q\beta r}r^\alpha}{D_\delta(r^\alpha)}\,\mathrm{d}r,\qquad\Re q>0. \tag*{(9)} L α ( q ) = κ α ∫ 0 ∞ D δ ( r α ) e − q β r r α d r , ℜ q > 0. (9) The integral converges locally uniformly in this half-plane. In particular, it defines the analytic continuation of the real Laplace transform. Write F α = − L α ′ / L α F_\alpha=-L_\alpha'/L_\alpha F α = − L α ′ / L α L α L_\alpha L α
Lemma 3.1 (Finite expansion with an explicit error). Let N ≥ 2 N\ge2 N ≥ 2 Y > 0 Y>0 Y > 0
c n = sin ( ( n + 1 ) δ ) sin δ , A n = c n Γ ( m + n α ) Γ ( m ) , (10) c_n=\frac{\sin((n+1)\delta)}{\sin\delta},\qquad A_n=c_n\frac{\Gamma(m+n\alpha)}{\Gamma(m)}, \tag*{(10)} c n = sin δ sin (( n + 1 ) δ ) , A n = c n Γ ( m ) Γ ( m + n α ) , (10) d 0 = 1 + β , d n = n A n − ∑ j = 1 n − 1 A j d n − j ( 1 ≤ n < N ) . (11) d_0=1+\beta,\qquad d_n=nA_n-\sum_{j=1}^{n-1}A_jd_{n-j}\qquad(1\le n<N). \tag*{(11)} d 0 = 1 + β , d n = n A n − j = 1 ∑ n − 1 A j d n − j ( 1 ≤ n < N ) . (11) Put μ = 2 − 1 / 2 \mu=2^{-1/2} μ = 2 − 1/2 η = sin ( δ / 4 ) sin ( 3 δ / 4 ) \eta=\sin(\delta/4)\sin(3\delta/4) η = sin ( δ /4 ) sin ( 3 δ /4 )
C A = max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) η Γ ( m + N α ) Γ ( m ) μ m + N α , (12) C_A=\frac{\max(|c_N|,|c_{N-1}|)}{\eta}\frac{\Gamma(m+N\alpha)}{\Gamma(m)\mu^{m+N\alpha}}, \tag*{(12)} C A = η max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) Γ ( m ) μ m + N α Γ ( m + N α ) , (12) C H = max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) η Γ ( m + N α + 1 ) Γ ( m ) μ m + N α + 1 , (13) C_H=\frac{\max(|c_N|,|c_{N-1}|)}{\eta}\frac{\Gamma(m+N\alpha+1)}{\Gamma(m)\mu^{m+N\alpha+1}}, \tag*{(13)} C H = η max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) Γ ( m ) μ m + N α + 1 Γ ( m + N α + 1 ) , (13) λ Y = 1 − ∑ n = 1 N − 1 ∣ A n ∣ Y − n − C A Y − N . (14) \lambda_Y=1-\sum_{n=1}^{N-1}|A_n|Y^{-n}-C_AY^{-N}. \tag*{(14)} λ Y = 1 − n = 1 ∑ N − 1 ∣ A n ∣ Y − n − C A Y − N . (14) If λ Y > 0 \lambda_Y>0 λ Y > 0 L α ( q ) ≠ 0 L_\alpha(q)\ne0 L α ( q ) = 0 ℜ q ≥ 0 \Re q\ge0 ℜ q ≥ 0 ∣ q ∣ ≥ Y |q|\ge Y ∣ q ∣ ≥ Y
∣ F α ( q ) − ∑ n = 0 N − 1 d n q − n − 1 ∣ ≤ C F ∣ q ∣ − N − 1 , (15) \left|F_\alpha(q)-\sum_{n=0}^{N-1}d_nq^{-n-1}\right|\le C_F|q|^{-N-1}, \tag*{(15)} F α ( q ) − n = 0 ∑ N − 1 d n q − n − 1 ≤ C F ∣ q ∣ − N − 1 , (15) where a valid constant is
C F = 1 λ Y ( 2 ∑ n = N 2 N − 2 ∣ ∑ 0 ≤ j < N 0 ≤ n − j < N A j d n − j ∣ Y N − n + β C H + C A ∑ n = 0 N − 1 ∣ d n ∣ Y − n ) . (16) C_F=\frac{1}{\lambda_Y}\left(2\sum_{n=N}^{2N-2}\left|\sum_{\substack{0\le j<N\\0\le n-j<N}}A_jd_{n-j}\right|Y^{N-n}+\beta C_H+C_A\sum_{n=0}^{N-1}|d_n|Y^{-n}\right). \tag*{(16)} C F = λ Y 1 2 n = N ∑ 2 N − 2 0 ≤ j < N 0 ≤ n − j < N ∑ A j d n − j Y N − n + β C H + C A n = 0 ∑ N − 1 ∣ d n ∣ Y − n . (16) All these bounds are uniform when their right-hand sides are evaluated by interval arithmetic on a closed parameter interval in ( 1 , 2 ) (1,2) ( 1 , 2 )
Proof. The recurrence c 0 = 1 c_0=1 c 0 = 1 c 1 = 2 cos δ c_1=2\cos\delta c 1 = 2 cos δ c n = 2 cos δ c n − 1 − c n − 2 c_n=2\cos\delta\,c_{n-1}-c_{n-2} c n = 2 cos δ c n − 1 − c n − 2
1 D δ ( z ) = ∑ n = 0 N − 1 c n z n + z N ( c N − c N − 1 z ) D δ ( z ) . (17) \frac{1}{D_\delta(z)}=\sum_{n=0}^{N-1}c_nz^n+\frac{z^N(c_N-c_{N-1}z)}{D_\delta(z)}. \tag*{(17)} D δ ( z ) 1 = n = 0 ∑ N − 1 c n z n + D δ ( z ) z N ( c N − c N − 1 z ) . (17) In (9) , make the substitution w = q β r w=q^{\beta}r w = q β r w w w arg w = ( arg q ) / 2 \arg w=(\arg q)/2 arg w = ( arg q ) /2 w α / q w^{\alpha}/q w α / q ( α / 2 − 1 ) arg q (\alpha/2-1)\arg q ( α /2 − 1 ) arg q δ / 2 \delta/2 δ /2 ℜ w ≥ μ ∣ w ∣ \Re w\geq\mu|w| ℜ w ≥ μ ∣ w ∣ z = w α / q z=w^{\alpha}/q z = w α / q
∣ D δ ( z ) ∣ = ∣ 1 − e i δ z ∣ ∣ 1 − e − i δ z ∣ ≥ η ( 1 + ∣ z ∣ ) 2 . (18) |D_{\delta}(z)|=|1-e^{i\delta}z||1-e^{-i\delta}z|\geq\eta(1+|z|)^2. \tag*{(18)} ∣ D δ ( z ) ∣ = ∣1 − e i δ z ∣∣1 − e − i δ z ∣ ≥ η ( 1 + ∣ z ∣ ) 2 . (18) Indeed ∣ 1 − x e i θ ∣ ≥ ( 1 + x ) ∣ sin ( θ / 2 ) ∣ |1-xe^{i\theta}|\geq(1+x)|\sin(\theta/2)| ∣1 − x e i θ ∣ ≥ ( 1 + x ) ∣ sin ( θ /2 ) ∣ x ≥ 0 x\geq0 x ≥ 0 ∣ arg z ∣ = δ / 2 |\arg z|=\delta/2 ∣ arg z ∣ = δ /2 m > 0 m>0 m > 0 ℜ q = 0 \Re q=0 ℜ q = 0
Normalize the transform and its first moment as follows:
A ( q ) = L α ( q ) κ α Γ ( m ) q − 1 − β = 1 Γ ( m ) ∫ arg w = ( arg q ) / 2 e − w w α D δ ( w α / q ) d w , (19) A(q)=\frac{L_{\alpha}(q)}{\kappa_{\alpha}\Gamma(m)q^{-1-\beta}}=\frac{1}{\Gamma(m)}\int_{\arg w=(\arg q)/2}\frac{e^{-w}w^{\alpha}}{D_{\delta}(w^{\alpha}/q)}\,dw, \tag*{(19)} A ( q ) = κ α Γ ( m ) q − 1 − β L α ( q ) = Γ ( m ) 1 ∫ a r g w = ( a r g q ) /2 D δ ( w α / q ) e − w w α d w , (19) H ( q ) = 1 Γ ( m ) ∫ arg w = ( arg q ) / 2 e − w w α + 1 D δ ( w α / q ) d w . (20) H(q)=\frac{1}{\Gamma(m)}\int_{\arg w=(\arg q)/2}\frac{e^{-w}w^{\alpha+1}}{D_{\delta}(w^{\alpha}/q)}\,dw. \tag*{(20)} H ( q ) = Γ ( m ) 1 ∫ a r g w = ( a r g q ) /2 D δ ( w α / q ) e − w w α + 1 d w . (20) Then F α ( q ) = β H ( q ) / ( q A ( q ) ) F_{\alpha}(q)=\beta H(q)/(qA(q)) F α ( q ) = β H ( q ) / ( q A ( q )) (17) and (18) yield
A ( q ) = ∑ n = 0 N − 1 A n q − n + E A ( q ) , H ( q ) = ∑ n = 0 N − 1 ( m + n α ) A n q − n + E H ( q ) , (21) A(q)=\sum_{n=0}^{N-1}A_nq^{-n}+E_A(q),\qquad H(q)=\sum_{n=0}^{N-1}(m+n\alpha)A_nq^{-n}+E_H(q), \tag*{(21)} A ( q ) = n = 0 ∑ N − 1 A n q − n + E A ( q ) , H ( q ) = n = 0 ∑ N − 1 ( m + n α ) A n q − n + E H ( q ) , (21) with ∣ E A ( q ) ∣ ≤ C A ∣ q ∣ − N |E_A(q)|\leq C_A|q|^{-N} ∣ E A ( q ) ∣ ≤ C A ∣ q ∣ − N ∣ E H ( q ) ∣ ≤ C H ∣ q ∣ − N |E_H(q)|\leq C_H|q|^{-N} ∣ E H ( q ) ∣ ≤ C H ∣ q ∣ − N
max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) η Γ ( m ) e − μ u u α + N α ∣ q ∣ − N \frac{\max(|c_N|,|c_{N-1}|)}{\eta\Gamma(m)}e^{-\mu u}u^{\alpha+N\alpha}|q|^{-N} η Γ ( m ) max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) e − μu u α + N α ∣ q ∣ − N over u > 0 u>0 u > 0 ∣ A ( q ) ∣ ≥ λ Y > 0 |A(q)|\geq\lambda_Y>0 ∣ A ( q ) ∣ ≥ λ Y > 0 (11) by the polynomial for A A A q − 1 q^{-1} q − 1 N N N β H − A ∑ n < N d n q − n \beta H-A\sum_{n<N}d_nq^{-n} β H − A ∑ n < N d n q − n q A qA q A (16) . This proof uses only a finite algebraic identity. In particular, no convergence of the infinite asymptotic expansion is asserted. □
Lemma 3.2 (Existence and uniqueness of the canonical kernel). For each 1 < α < 2 1<\alpha<2 1 < α < 2 K α : [ 0 , ∞ ) → R K_{\alpha}:[0,\infty)\to\mathbb{R} K α : [ 0 , ∞ ) → R
F α ( q ) = ∫ 0 ∞ e − q t K α ( t ) d t (22) F_{\alpha}(q)=\int_{0}^{\infty}e^{-qt}K_{\alpha}(t)\,dt \tag*{(22)} F α ( q ) = ∫ 0 ∞ e − q t K α ( t ) d t (22) for all sufficiently large real $q. It is smooth on [ 0 , ∞ ) [0,\infty) [ 0 , ∞ )
K α ( n ) ( 0 ) = d n , n ≥ 0. (23) K_{\alpha}^{(n)}(0)=d_n,\qquad n\geq0. \tag*{(23)} K α ( n ) ( 0 ) = d n , n ≥ 0. (23) Its time derivatives and its first real parameter derivative are continuous locally in ( α , t ) ∈ ( 1 , 2 ) × [ 0 , ∞ ) (\alpha,t)\in(1,2)\times[0,\infty) ( α , t ) ∈ ( 1 , 2 ) × [ 0 , ∞ )
Proof. Fix a compact parameter interval I ⊂ ( 1 , 2 ) I\subset(1,2) I ⊂ ( 1 , 2 ) N = 2 N=2 N = 2 c c c λ c > 0 \lambda_c>0 λ c > 0 I I I F α F_{\alpha} F α ℜ q ≥ c \Re q\geq c ℜ q ≥ c I I I q = c ± i y q=c\pm iy q = c ± i y w = u e ± i π / 4 w=ue^{\pm i\pi/4} w = u e ± iπ /4 ∣ y ∣ |y| ∣ y ∣ ∣ arg ( w α / q ) ∣ < δ / 2 |\arg(w^{\alpha}/q)|<\delta/2 ∣ arg ( w α / q ) ∣ < δ /2 I I I (18) still applies. Differentiating the finite remainder introduces only log w \log w log w ∣ log w ∣ ≤ u + u − 1 + π / 4 |\log w|\leq u+u^{-1}+\pi/4 ∣ log w ∣ ≤ u + u − 1 + π /4 e − u / 2 e^{-u/\sqrt{2}} e − u / 2 I I I (9) is immediate.
Fix h > 0 h > 0 h > 0 n ≥ 1 n \ge1 n ≥ 1
R n ( q ) = ∑ k = 1 n b k ( q + h ) k , b k = ∑ j = 0 k − 1 ( k − 1 j ) h k − 1 − j d j , r n ( t ) = e − h t ∑ k = 1 n b k t k − 1 ( k − 1 ) ! . (24) R_n(q) = \sum_{k=1}^{n} \frac{b_k}{(q+h)^k}, \qquad b_k = \sum_{j=0}^{k-1} \binom{k-1}{j} h^{k-1-j} d_j, \qquad r_n(t) = e^{-ht} \sum_{k=1}^{n} \frac{b_k t^{k-1}}{(k-1)!}. \tag*{(24)} R n ( q ) = k = 1 ∑ n ( q + h ) k b k , b k = j = 0 ∑ k − 1 ( j k − 1 ) h k − 1 − j d j , r n ( t ) = e − h t k = 1 ∑ n ( k − 1 )! b k t k − 1 . (24) These coefficients give F α − R n = O ( ∣ q ∣ − n − 1 ) F_\alpha-R_n=O(|q|^{-n-1}) F α − R n = O ( ∣ q ∣ − n − 1 ) r n ( j ) ( 0 ) = d j r_n^{(j)}(0)=d_j r n ( j ) ( 0 ) = d j j < n j<n j < n
K α ( t ) = r n ( t ) + 1 2 π ∫ − ∞ ∞ e ( c + i y ) t ( F α ( c + i y ) − R n ( c + i y ) ) d y . (25) K_\alpha(t)=r_n(t)+\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{(c+iy)t}\left(F_\alpha(c+iy)-R_n(c+iy)\right)\,dy. \tag*{(25)} K α ( t ) = r n ( t ) + 2 π 1 ∫ − ∞ ∞ e ( c + i y ) t ( F α ( c + i y ) − R n ( c + i y ) ) d y . (25) Absolute convergence, the elementary inverse transform of R n − R 1 R_n-R_1 R n − R 1 n n n c c c n > j n>j n > j j j j t t t t = 0 t=0 t = 0 q j ( F α − R n ) q^j(F_\alpha-R_n) q j ( F α − R n ) K α ( j ) ( 0 ) = r n ( j ) ( 0 ) = d j K_\alpha^{(j)}(0)=r_n^{(j)}(0)=d_j K α ( j ) ( 0 ) = r n ( j ) ( 0 ) = d j
For real q > c q>c q > c (25) is F α ( q ) − R n ( q ) F_\alpha(q)-R_n(q) F α ( q ) − R n ( q ) r n r_n r n (22) . Uniqueness follows from uniqueness of Laplace transforms after exponential weighting. Finally, the differentiated remainder bounds above permit parameter differentiation of (25) and prove the asserted local continuity.
Lemma 3.3 (A sign criterion for additive infinite divisibility). The distribution of T 1 T_1 T 1 K α ( t ) ≥ 0 K_\alpha(t)\geq0 K α ( t ) ≥ 0 t > 0 t>0 t > 0
ν α ( d t ) = K α ( t ) t d t . (26) \nu_\alpha(dt)=\frac{K_\alpha(t)}{t}\,dt. \tag*{(26)} ν α ( d t ) = t K α ( t ) d t . (26) Proof. For a nonnegative infinitely divisible random variable, the Lévy–Khintchine representation (see [6 , 7 ] ) and its derivative are
− log L ( q ) = d q + ∫ 0 ∞ ( 1 − e − q t ) ν ( d t ) , − L ′ ( q ) L ( q ) = d + ∫ 0 ∞ t e − q t ν ( d t ) , (27) -\log L(q)=dq+\int_0^\infty(1-e^{-qt})\nu(dt), \qquad-\frac{L'(q)}{L(q)}=d+\int_0^\infty te^{-qt}\nu(dt), \tag*{(27)} − log L ( q ) = d q + ∫ 0 ∞ ( 1 − e − q t ) ν ( d t ) , − L ( q ) L ′ ( q ) = d + ∫ 0 ∞ t e − q t ν ( d t ) , (27) where d ≥ 0 d\geq0 d ≥ 0 ∫ ( 1 ∧ t ) ν ( d t ) < ∞ \int(1\wedge t)\nu(dt)<\infty ∫ ( 1 ∧ t ) ν ( d t ) < ∞ F α ( q ) → 0 F_\alpha(q)\to0 F α ( q ) → 0 q → ∞ q\to\infty q → ∞ d = 0 d=0 d = 0 t ν ( d t ) = K α ( t ) d t t\nu(dt)=K_\alpha(t)dt t ν ( d t ) = K α ( t ) d t K α ≥ 0 K_\alpha\geq0 K α ≥ 0
Conversely, suppose K α ≥ 0 K_\alpha\geq0 K α ≥ 0 c > 0 c>0 c > 0 (22) . Then ν c ( d t ) = e − c t K α ( t ) d t / t \nu_c(dt)=e^{-ct}K_\alpha(t)dt/t ν c ( d t ) = e − c t K α ( t ) d t / t (22) from c c c c + s c+s c + s
L α ( c + s ) L α ( c ) = exp { − ∫ 0 ∞ ( 1 − e − s t ) ν c ( d t ) } . (28) \frac{L_\alpha(c+s)}{L_\alpha(c)}=\exp\left\{-\int_0^\infty(1-e^{-st})\nu_c(dt)\right\}. \tag*{(28)} L α ( c ) L α ( c + s ) = exp { − ∫ 0 ∞ ( 1 − e − s t ) ν c ( d t ) } . (28) Thus the actual tilted probability law P c ( d t ) = e − c t P ( T 1 ∈ d t ) / L α ( c ) P_c(dt)=e^{-ct}\mathbb{P}(T_1\in dt)/L_\alpha(c) P c ( d t ) = e − c t P ( T 1 ∈ d t ) / L α ( c ) n ≥ 1 n\geq1 n ≥ 1 Q n , c Q_{n,c} Q n , c ν c / n \nu_c/n ν c / n Q n , c ∗ n = P c Q_{n,c}^{*n}=P_c Q n , c ∗ n = P c
( ∫ 0 ∞ e c t Q n , c ( d t ) ) n = ∫ 0 ∞ e c t P c ( d t ) = L α ( c ) − 1 < ∞ . (29) \left(\int_0^\infty e^{ct}Q_{n,c}(dt)\right)^n=\int_0^\infty e^{ct}P_c(dt)=L_\alpha(c)^{-1}<\infty. \tag*{(29)} ( ∫ 0 ∞ e c t Q n , c ( d t ) ) n = ∫ 0 ∞ e c t P c ( d t ) = L α ( c ) − 1 < ∞. (29) Consequently Q n ( d t ) = L α ( c ) 1 / n e c t Q n , c ( d t ) Q_n(dt)=L_\alpha(c)^{1/n}e^{ct}Q_{n,c}(dt) Q n ( d t ) = L α ( c ) 1/ n e c t Q n , c ( d t )
Q n ∗ n ( d t ) = L α ( c ) e c t P c ( d t ) = P ( T 1 ∈ d t ) . Q_n^{*n}(dt)=L_\alpha(c)e^{ct}P_c(dt)=\mathbb{P}(T_1\in dt). Q n ∗ n ( d t ) = L α ( c ) e c t P c ( d t ) = P ( T 1 ∈ d t ) . This proves infinite divisibility directly from its definition. The already established necessary direction then identifies its Lévy measure as (26) , including the required integrability.
Remark 3.4. The first two coefficients are
K α ( 0 ) = 1 + α − 1 , K α ′ ( 0 ) = 2 cos δ Γ ( 2 α + 1 ) Γ ( α + 1 ) > 0. K_{\alpha}(0)=1+\alpha^{-1}, \qquad K'_{\alpha}(0)=2\cos\delta\frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}>0. K α ( 0 ) = 1 + α − 1 , K α ′ ( 0 ) = 2 cos δ Γ ( α + 1 ) Γ ( 2 α + 1 ) > 0. A law is self-decomposable if, for every b ∈ ( 0 , 1 ) b\in(0,1) b ∈ ( 0 , 1 ) Z = d b Z 1 + Y b Z\overset{d}{=}bZ_1+Y_b Z = d b Z 1 + Y b Z 1 Z_1 Z 1 Z Z Z Y b Y_b Y b [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) k ( t ) / t k(t)/t k ( t ) / t k k k [6 Theorem 15.10] . Uniqueness in Lemma 3.3 rules out self-decomposability whenever the law is infinitely divisible; outside that range it is ruled out a fortiori. Thus this argument also recovers the result of Wang and Yin [10 Corollary 1.2] : non-self-decomposability for every 1 < α < 2 1<\alpha<2 1 < α < 2 α = 2 \alpha=2 α = 2 (9) degenerates.
Positivity up to the critical index
We retain the notation
β = α − 1 , δ = π ( 1 − α / 2 ) , D δ ( z ) = 1 − 2 cos ( δ ) z + z 2 , d 0 = 1 + β . \beta=\alpha^{-1}, \qquad\delta=\pi(1-\alpha/2), \qquad D_{\delta}(z)=1-2\cos(\delta)z+z^2, \qquad d_0=1+\beta. β = α − 1 , δ = π ( 1 − α /2 ) , D δ ( z ) = 1 − 2 cos ( δ ) z + z 2 , d 0 = 1 + β . In this section K α K_{\alpha} K α F α = − L α ′ / L α F_{\alpha}=-L'_{\alpha}/L_{\alpha} F α = − L α ′ / L α
K α ( t ) > 0 ( 1 < α < α ∗ , t > 0 ) , K α ∗ ( t ) ≥ 0 ( t > 0 ) , (30) K_{\alpha}(t)>0 \quad(1<\alpha<\alpha_*,\ t>0), \qquad K_{\alpha_*}(t)\ge0 \quad(t>0), \tag*{(30)} K α ( t ) > 0 ( 1 < α < α ∗ , t > 0 ) , K α ∗ ( t ) ≥ 0 ( t > 0 ) , (30) and show that the last kernel has precisely one zero. Every decimal endpoint of a parameter or time interval below is an exact rational number. The finite inequalities used in the proof, together with their exact interval records and verification programs, are indexed in Section B. The analytic reductions and the quantities tested by those programs are given explicitly here.
The cut and a common zero-free region
Put
κ α = α sin δ sin ( π β ) π , η α = α sin δ π , M α ( q ) = ∫ 0 ∞ e − q β r r α D δ ( r α ) d r . \kappa_{\alpha}=\frac{\alpha\sin\delta\sin(\pi\beta)}{\pi}, \qquad\eta_{\alpha}=\frac{\alpha\sin\delta}{\pi}, \qquad M_{\alpha}(q)=\int_0^{\infty}e^{-q\beta r}\frac{r^{\alpha}}{D_{\delta}(r^{\alpha})}\,dr. κ α = π α sin δ sin ( π β ) , η α = π α sin δ , M α ( q ) = ∫ 0 ∞ e − q β r D δ ( r α ) r α d r . For α > 1 \alpha>1 α > 1 L α = κ α M α L_{\alpha}=\kappa_{\alpha}M_{\alpha} L α = κ α M α M α M_{\alpha} M α α = 1 \alpha=1 α = 1
M 1 ( q ) = ∫ 0 ∞ e − q r r 1 + r 2 d r . (31) M_1(q)=\int_0^{\infty}e^{-qr}\frac{r}{1+r^2}\,dr. \tag*{(31)} M 1 ( q ) = ∫ 0 ∞ e − q r 1 + r 2 r d r . (31) The endpoint α = 1 \alpha=1 α = 1 K 1 K_1 K 1 Lemma 3.2 , with M 1 M_1 M 1 L α L_{\alpha} L α
Lemma 4.1 (Cut phase). For 1 ≤ α < 2 1\le\alpha<2 1 ≤ α < 2 upper-bank continuation of M α M_{\alpha} M α − 1 ≤ q < 0 -1\le q<0 − 1 ≤ q < 0 θ α ( v ) \theta_{\alpha}(v) θ α ( v ) θ α ( 0 + ) = 0 \theta_{\alpha}(0+)=0 θ α ( 0 + ) = 0
− π β < θ α ( v ) < 0 ( 0 < v ≤ 1 ) . -\pi\beta<\theta_{\alpha}(v)<0 \qquad(0<v\le1). − π β < θ α ( v ) < 0 ( 0 < v ≤ 1 ) . Consequently
w α ( v ) = − 1 π Im F α ( − v + i 0 ) > 0 , ∫ 0 b w α ( v ) d v = − θ α ( b ) π < β ( 0 < b ≤ 1 ) . w_{\alpha}(v)=-\frac{1}{\pi}\operatorname{Im}F_{\alpha}(-v+i0)>0, \qquad\int_0^b w_{\alpha}(v)\,dv=-\frac{\theta_{\alpha}(b)}{\pi}<\beta\qquad(0<b\le1). w α ( v ) = − π 1 Im F α ( − v + i 0 ) > 0 , ∫ 0 b w α ( v ) d v = − π θ α ( b ) < β ( 0 < b ≤ 1 ) . Proof. Let x = v β x=v^{\beta} x = v β
H α ( x ) = ∫ 0 ∞ e − x u u α 1 + 2 cos δ u α + u 2 α d u . H_{\alpha}(x)=\int_{0}^{\infty}e^{-xu}\frac{u^{\alpha}}{1+2\cos\delta\,u^{\alpha}+u^{2\alpha}}\,du. H α ( x ) = ∫ 0 ∞ e − xu 1 + 2 cos δ u α + u 2 α u α d u . Rotating the integration ray by − π β -\pi\beta − π β p = exp [ − i π ( β − 1 / 2 ) ] p=\exp[-i\pi(\beta-1/2)] p = exp [ − iπ ( β − 1/2 )]
M α ( − v + i 0 ) = η α − 1 e − i π β N α ( x ) , N α ( x ) = i e − i x − η α H α ( x ) . (32) M_{\alpha}(-v+i0)=\eta_{\alpha}^{-1}e^{-i\pi\beta}N_{\alpha}(x),\qquad N_{\alpha}(x)=ie^{-ix}-\eta_{\alpha}H_{\alpha}(x). \tag*{(32)} M α ( − v + i 0 ) = η α − 1 e − iπ β N α ( x ) , N α ( x ) = i e − i x − η α H α ( x ) . (32) In particular Im N α ( x ) = cos x > 0 \operatorname{Im}N_{\alpha}(x)=\cos x>0 Im N α ( x ) = cos x > 0 0 < x ≤ 1 0<x\leq1 0 < x ≤ 1 H α H_{\alpha} H α 1 / 2 1/2 1/2 cos δ ≥ 0 \cos\delta\geq0 cos δ ≥ 0 1 + u 2 α ≥ 2 u α 1+u^{2\alpha}\geq2u^{\alpha} 1 + u 2 α ≥ 2 u α x H α ( x ) ≤ 1 / 2 xH_{\alpha}(x)\leq1/2 x H α ( x ) ≤ 1/2 H α ′ ( x ) < 0 H_{\alpha}'(x)<0 H α ′ ( x ) < 0
w α ( v ) = β v β − 1 π 1 − η α { H α ( x ) sin x + H α ′ ( x ) cos x } ∣ N α ( x ) ∣ 2 > 0 , (33) w_{\alpha}(v)=\frac{\beta v^{\beta-1}}{\pi}\frac{1-\eta_{\alpha}\{H_{\alpha}(x)\sin x+H_{\alpha}'(x)\cos x\}}{\lvert N_{\alpha}(x)\rvert^{2}}>0, \tag*{(33)} w α ( v ) = π β v β − 1 ∣ N α ( x ) ∣ 2 1 − η α { H α ( x ) sin x + H α ′ ( x ) cos x } > 0 , (33) since η α < 1 \eta_{\alpha}<1 η α < 1 1 / 2 1/2 1/2
The phase arg N α ( v β ) − π β \arg N_{\alpha}(v^{\beta})-\pi\beta arg N α ( v β ) − π β ( − π β , π ( 1 − β ) ) (-\pi\beta,\pi(1-\beta)) ( − π β , π ( 1 − β )) − π w α -\pi w_{\alpha} − π w α α > 1 \alpha>1 α > 1 M α ( 0 ) = 1 / κ α > 0 M_{\alpha}(0)=1/\kappa_{\alpha}>0 M α ( 0 ) = 1/ κ α > 0 α = 1 \alpha=1 α = 1 M 1 ( q ) = − log q + O ( 1 ) M_{1}(q)=-\log q+O(1) M 1 ( q ) = − log q + O ( 1 )
Lemma 4.2 (Zero-free slit region). For 1 ≤ α ≤ 1.6671 1\leq\alpha\leq1.6671 1 ≤ α ≤ 1.6671 M α M_{\alpha} M α
{ q : Re q ≥ − 1 } ∖ ( − ∞ , 0 ] , \{q:\operatorname{Re}q\geq-1\}\setminus(-\infty,0], { q : Re q ≥ − 1 } ∖ ( − ∞ , 0 ] , and its boundary values on the cut segment [ − 1 , 0 ) [-1,0) [ − 1 , 0 )
Proof. We describe separately the origin, the exterior and the finite vertical boundary; their argument changes, rather than boundary nonvanishing alone, are needed to exclude interior zeros.
For 1 ≤ α ≤ 6 / 5 1\leq\alpha\leq6/5 1 ≤ α ≤ 6/5 ε = α − 1 \varepsilon=\alpha-1 ε = α − 1 s = q β s=q^{\beta} s = q β r = ∣ s ∣ r=\lvert s\rvert r = ∣ s ∣ u − α u^{-\alpha} u − α ε > 0 \varepsilon>0 ε > 0
M α ( q ) = M α ( 0 ) − Γ ( 1 − ε ) ε s ε + R α ( s ) , ∣ R α ( s ) ∣ ≤ 60 ∣ s ∣ , ∣ R α ′ ( s ) ∣ ≤ 60. M_{\alpha}(q)=M_{\alpha}(0)-\frac{\Gamma(1-\varepsilon)}{\varepsilon}s^{\varepsilon}+R_{\alpha}(s),\qquad\lvert R_{\alpha}(s)\rvert\leq60\lvert s\rvert,\quad\lvert R_{\alpha}'(s)\rvert\leq60. M α ( q ) = M α ( 0 ) − ε Γ ( 1 − ε ) s ε + R α ( s ) , ∣ R α ( s )∣ ≤ 60 ∣ s ∣ , ∣ R α ′ ( s )∣ ≤ 60. The two divergent terms have their removable logarithmic limit when ε = 0 \varepsilon=0 ε = 0 − π / 4 -\pi/4 − π /4 − 3 π / 4 -3\pi/4 − 3 π /4 ( 1 + u α ) 2 / 10 (1+u^{\alpha})^2/10 ( 1 + u α ) 2 /10 10 ( 5 / 4 + 1 / 3 + 1 + π / 2 ) < 42 10(5/4+1/3+1+\pi/2)<42 10 ( 5/4 + 1/3 + 1 + π /2 ) < 42 ε M α ( 0 ) ≥ 1 \varepsilon M_{\alpha}(0)\geq1 ε M α ( 0 ) ≥ 1 Γ ( 1 − ε ) ≤ e 2 ε \Gamma(1-\varepsilon)\leq e^{2\varepsilon} Γ ( 1 − ε ) ≤ e 2 ε ℓ = log ( 1 / r ) − 2 \ell=\log(1/r)-2 ℓ = log ( 1/ r ) − 2
Re M α ( q ) ≥ 1 − e − ε ℓ ε − 60 r > 2 ( ∣ q ∣ ≤ 10 − 4 ) . \operatorname{Re}M_{\alpha}(q)\geq\frac{1-e^{-\varepsilon\ell}}{\varepsilon}-60r>2\qquad(\lvert q\rvert\leq10^{-4}). Re M α ( q ) ≥ ε 1 − e − ε ℓ − 60 r > 2 (∣ q ∣ ≤ 1 0 − 4 ) . Here the quotient has its continuous value at ε = 0 \varepsilon=0 ε = 0 ℓ > 4 \ell>4 ℓ > 4 20 / 9 20/9 20/9 q F α ( q ) → 0 qF_{\alpha}(q)\to0 q F α ( q ) → 0 6 / 5 ≤ α ≤ 13 / 10 6/5\leq\alpha\leq13/10 6/5 ≤ α ≤ 13/10
∣ L α ( q ) − 1 ∣ < 20 ∣ q ∣ 1 / 6 < 1 ( ∣ q ∣ ≤ 10 − 8 ) . \lvert L_{\alpha}(q)-1\rvert<20\lvert q\rvert^{1/6}<1\qquad(\lvert q\rvert\leq10^{-8}). ∣ L α ( q ) − 1 ∣ < 20 ∣ q ∣ 1/6 < 1 (∣ q ∣ ≤ 1 0 − 8 ) . For 13 / 10 ≤ α ≤ 1.6671 13/10\leq\alpha\leq1.6671 13/10 ≤ α ≤ 1.6671 C α ( 0 ) ( 10 − 12 ) 1 − β < 1 / 2 C_{\alpha}^{(0)}(10^{-12})^{1-\beta}<1/2 C α ( 0 ) ( 1 0 − 12 ) 1 − β < 1/2 ∣ L α ( q ) − 1 ∣ ≤ C α ( 0 ) ∣ q ∣ 1 − β \lvert L_{\alpha}(q)-1\rvert\leq C_{\alpha}^{(0)}\lvert q\rvert^{1-\beta} ∣ L α ( q ) − 1 ∣ ≤ C α ( 0 ) ∣ q ∣ 1 − β
γ = 3 π β 4 , Δ = sin α γ − δ 2 sin α γ + δ 2 , V α = 2 2 − α ( 1 2 − α + 1 α − 1 ) , (34) \gamma=\frac{3\pi\beta}{4},\qquad\Delta=\sin\frac{\alpha\gamma-\delta}{2}\sin\frac{\alpha\gamma+\delta}{2},\qquad V_{\alpha}=2^{2-\alpha}\left(\frac{1}{2-\alpha}+\frac{1}{\alpha-1}\right), \tag*{(34)} γ = 4 3 π β , Δ = sin 2 α γ − δ sin 2 α γ + δ , V α = 2 2 − α ( 2 − α 1 + α − 1 1 ) , (34) one may take
C α ( 0 ) = max { sin ( π β ) + κ α V α Δ , κ α V α sin 2 ( δ / 2 ) } . C_{\alpha}^{(0)}=\max\left\{\sin(\pi\beta)+\frac{\kappa_{\alpha}V_{\alpha}}{\Delta},\frac{\kappa_{\alpha}V_{\alpha}}{\sin^2(\delta/2)}\right\}. C α ( 0 ) = max { sin ( π β ) + Δ κ α V α , sin 2 ( δ /2 ) κ α V α } . These last two estimates follow by integrating min ( ∣ s ∣ u , 2 ) u − α \min(|s|u,2)u^{-\alpha} min ( ∣ s ∣ u , 2 ) u − α
For ∣ q ∣ ≥ 400 |q|\ge400 ∣ q ∣ ≥ 400 ∣ arg q ∣ ≤ π / 2 + 1 / 20 |\arg q|\le\pi/2+1/20 ∣ arg q ∣ ≤ π /2 + 1/20 A α ( q ) = q d 0 M α ( q ) / Γ ( α + 1 ) A_{\alpha}(q)=q^{d_0}M_{\alpha}(q)/\Gamma(\alpha+1) A α ( q ) = q d 0 M α ( q ) /Γ ( α + 1 ) D δ − 1 D_{\delta}^{-1} D δ − 1
∣ A α ( q ) − 1 ∣ < 1 / 4 ( 1 ≤ α ≤ 13 / 10 ) , ∣ A α ( q ) − 1 ∣ < 1 / 2 ( 13 / 10 ≤ α ≤ 1.6671 ) , |A_{\alpha}(q)-1|<1/4\quad(1\le\alpha\le13/10),\qquad|A_{\alpha}(q)-1|<1/2\quad(13/10\le\alpha\le1.6671), ∣ A α ( q ) − 1∣ < 1/4 ( 1 ≤ α ≤ 13/10 ) , ∣ A α ( q ) − 1∣ < 1/2 ( 13/10 ≤ α ≤ 1.6671 ) , and F α ( q ) − d 0 / q = O ( ∣ q ∣ − 2 ) F_{\alpha}(q)-d_0/q=O(|q|^{-2}) F α ( q ) − d 0 / q = O ( ∣ q ∣ − 2 ) 1000 / ∣ q ∣ 2 1000/|q|^2 1000/∣ q ∣ 2 arg w = arg q / 2 \arg w=\arg q/2 arg w = arg q /2 w = q β r w=q^{\beta}r w = q β r 2 / 3 2/3 2/3 sin ( δ / 5 ) sin ( 3 δ / 5 ) \sin(\delta/5)\sin(3\delta/5) sin ( δ /5 ) sin ( 3 δ /5 )
On q = − 1 + i y q=-1+iy q = − 1 + i y 0 ≤ y ≤ 400 0\le y\le400 0 ≤ y ≤ 400
Table 1.
For clarity, these are conditions on entire closed parameter–y y y b j b_j b j y j y_j y j B j B_j B j ∣ L α ′ ∣ |L'_{\alpha}| ∣ L α ′ ∣ [ y j , y j + 1 ] [y_j,y_{j+1}] [ y j , y j + 1 ] b j − ( y j + 1 − y j ) B j > 0 b_j-(y_{j+1}-y_j)B_j>0 b j − ( y j + 1 − y j ) B j > 0 M M M B j B_j B j [ 0 , 400 ] [0,400] [ 0 , 400 ] r = ( 1 + y 2 ) 1 / 2 r=(1+y^2)^{1/2} r = ( 1 + y 2 ) 1/2 p = e − i δ / α p=e^{-i\delta/\alpha} p = e − i δ / α Δ \Delta Δ
B ( y ) = β sin ( π β ) r β − 1 e − R e ( p ( − 1 + i y ) β ) + β κ α Δ min { Γ ( 2 − α ) ( 3 / 4 ) 2 − α r − β , Γ ( α + 2 ) ( 3 / 4 ) α + 2 r − 2 − β } . B(y)=\beta\sin(\pi\beta)r^{\beta-1}e^{-\mathop{\rm Re}(p(-1+iy)^\beta)}
+\frac{\beta\kappa_{\alpha}}{\Delta}\min\left\{\frac{\Gamma(2-\alpha)}{(3/4)^{2-\alpha}}r^{-\beta},\frac{\Gamma(\alpha+2)}{(3/4)^{\alpha+2}}r^{-2-\beta}\right\}. B ( y ) = β sin ( π β ) r β − 1 e − Re ( p ( − 1 + i y ) β ) + Δ β κ α min { ( 3/4 ) 2 − α Γ ( 2 − α ) r − β , ( 3/4 ) α + 2 Γ ( α + 2 ) r − 2 − β } . This follows by differentiating the residue and the rotated integral, using R e ( s e − i γ ) ≥ ( 3 / 4 ) ∣ s ∣ \mathop{\rm Re}(se^{-i\gamma})\ge(3/4)|s| Re ( s e − iγ ) ≥ ( 3/4 ) ∣ s ∣ y y y r β − 1 r^{\beta-1} r β − 1
d d y R e ( p ( − 1 + i y ) β ) = β r β − 1 cos ( ( 1 − β ) ( π − arg ( − 1 + i y ) ) ) > 0. \frac{d}{dy}\mathop{\rm Re}\left(p(-1+iy)^\beta\right)=\beta r^{\beta-1}\cos\left((1-\beta)(\pi-\arg(-1+iy))\right)>0. d y d Re ( p ( − 1 + i y ) β ) = β r β − 1 cos ( ( 1 − β ) ( π − arg ( − 1 + i y )) ) > 0. Thus this formula supplies the required whole-segment B j B_j B j
All three rows place the vertical image in { z : R e z < 0 or I m z < 0 } \{z:\mathop{\rm Re}z<0\text{ or }\mathop{\rm Im}z<0\} { z : Re z < 0 or Im z < 0 } ( − 3 π / 2 , 0 ) (-3\pi/2,0) ( − 3 π /2 , 0 ) ( − π , 0 ) (-\pi,0) ( − π , 0 ) q = − 1 + 400 i q=-1+400i q = − 1 + 400 i − d 0 arg q + arg A α ( q ) -d_0\arg q+\arg A_{\alpha}(q) − d 0 arg q + arg A α ( q ) 3 / 2 < d 0 ≤ 2 3/2<d_0\le2 3/2 < d 0 ≤ 2 π / 2 < arg q < π / 2 + 1 / 20 \pi/2<\arg q<\pi/2+1/20 π /2 < arg q < π /2 + 1/20 ∣ arg A α ∣ < π / 6 |\arg A_{\alpha}|<\pi/6 ∣ arg A α ∣ < π /6
− 3 π 2 < − 7 π 6 − 1 10 < − d 0 arg q + arg A α ( q ) < − 7 π 12 < 0. -\frac{3\pi}{2}<-\frac{7\pi}{6}-\frac{1}{10}<-d_0\arg q+\arg A_{\alpha}(q)<-\frac{7\pi}{12}<0. − 2 3 π < − 6 7 π − 10 1 < − d 0 arg q + arg A α ( q ) < − 12 7 π < 0. The exterior arc continues this phase to zero on the positive axis. The positive axis and the small origin arc, whose image lies in a right half-plane, complete a contour with zero total argument change. The argument principle excludes zeros in the bounded upper slit region. Conjugation and the exterior estimate cover the remaining parts of the stated domain. ∩ევნ
Control of the two time ends
Lemma 4.3 (A signed cut representation). Throughout the range of Lemma 4.2,
K α ( t ) = d 0 e − 2 t + ∫ 0 1 e − ν t w α ( ν ) d ν + E α ( t ) , ∣ E α ( t ) ∣ ≤ B α e − t . (35) K_{\alpha}(t)=d_{0}e^{-2t}+\int_{0}^{1}e^{-\nu t}w_{\alpha}(\nu)\,d\nu+E_{\alpha}(t),\qquad|E_{\alpha}(t)|\le B_{\alpha}e^{-t}. \tag*{(35)} K α ( t ) = d 0 e − 2 t + ∫ 0 1 e − ν t w α ( ν ) d ν + E α ( t ) , ∣ E α ( t ) ∣ ≤ B α e − t . (35) One can take B α < 94 , 37 , 26 B_{\alpha}<94,37,26 B α < 94 , 37 , 26
Proof. Subtract d 0 / ( q + 2 ) d_{0}/(q+2) d 0 / ( q + 2 ) F α F_{\alpha} F α Re q = − 1 \operatorname{Re}q=-1 Re q = − 1 q F α ( q ) → 0 qF_{\alpha}(q)\to0 q F α ( q ) → 0 − 2 -2 − 2
On every finite vertical segment the preceding margin ℓ j > 0 \ell_j>0 ℓ j > 0 ∣ L ∣ ≥ ℓ j |L|\ge\ell_j ∣ L ∣ ≥ ℓ j
B α ≤ 1 π { ∑ j h j ( B j ℓ j + d 0 1 + y j 2 ) + the complete exterior tail } . B_{\alpha}\le\frac{1}{\pi}\left\{\sum_j h_j\left(\frac{B_j}{\ell_j}+\frac{d_0}{\sqrt{1+y_j^2}}\right)+\text{the complete exterior tail}\right\}. B α ≤ π 1 ⎩ ⎨ ⎧ j ∑ h j ℓ j B j + 1 + y j 2 d 0 + the complete exterior tail ⎭ ⎬ ⎫ . For the first two ranges the last tail is at most 1004 / 400 1004/400 1004/400 93.0745928287506103515625 93.0745928287506103515625 93.0745928287506103515625 B α e − t B_{\alpha}e^{-t} B α e − t β \beta β ∣ K α ( t ) ∣ ≤ 1 + 2 β + B α |K_{\alpha}(t)|\le1+2\beta+B_{\alpha} ∣ K α ( t ) ∣ ≤ 1 + 2 β + B α
Lemma 4.4 (Time-end positivity). The following inequalities hold uniformly on the indicated parameter sets:
Table 1 (PDF p. 10).
In particular K α ( t ) > 0 K_{\alpha}(t)>0 K α ( t ) > 0 0 < t ≤ 9 / 200 0<t\le9/200 0 < t ≤ 9/200
Proof. For small time we sharpen the finite expansion by splitting its inversion line. The proof of Lemma 3.1 permits any lower bound μ \mu μ max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) ≤ 1 / sin δ \max(|c_N|,|c_{N-1}|)\le1/\sin\delta max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) ≤ 1/ sin δ q = c + i y q=c+iy q = c + i y c = N / t c=N/t c = N / t ∣ arg q ∣ ≤ π / 3 |\arg q|\le\pi/3 ∣ arg q ∣ ≤ π /3 μ = 3 / 2 \mu=\sqrt{3}/2 μ = 3 /2 ∣ q ∣ ≥ 2 c |q|\ge2c ∣ q ∣ ≥ 2 c μ = 1 / 2 \mu=1/\sqrt{2} μ = 1/ 2 0 < t ≤ T 0<t\le T 0 < t ≤ T C F i n C_F^{\mathrm{in}} C F in C F o u t C_F^{\mathrm{out}} C F out Y = N / T Y=N/T Y = N / T Y = 2 N / T Y=2N/T Y = 2 N / T R q ≥ N / T \mathbb{R}q\ge N/T R q ≥ N / T
∫ ∣ y ∣ ≥ 3 c ∣ c + i y ∣ − N − 1 d y ≤ ( 2 c ) 1 − N ∫ ∣ y ∣ ≥ 3 c d y c 2 + y 2 = c − N π 3 2 1 − N . \int_{|y|\ge\sqrt{3}c}|c+iy|^{-N-1}\,dy\le(2c)^{1-N}\int_{|y|\ge\sqrt{3}c}\frac{dy}{c^2+y^2}=c^{-N}\frac{\pi}{3}2^{1-N}. ∫ ∣ y ∣ ≥ 3 c ∣ c + i y ∣ − N − 1 d y ≤ ( 2 c ) 1 − N ∫ ∣ y ∣ ≥ 3 c c 2 + y 2 d y = c − N 3 π 2 1 − N . Thus the entire inversion error is at most
E N ( T ) = e N ( T / N ) N { C F i n Γ ( N / 2 ) 2 π Γ ( ( N + 1 ) / 2 ) + C F o u t 2 1 − N 6 } . E_N(T)=e^N(T/N)^N\left\{\frac{C_F^{\mathrm{in}}\Gamma(N/2)}{2\sqrt{\pi}\Gamma((N+1)/2)}+\frac{C_F^{\mathrm{out}}2^{1-N}}{6}\right\}. E N ( T ) = e N ( T / N ) N { 2 π Γ (( N + 1 ) /2 ) C F in Γ ( N /2 ) + 6 C F out 2 1 − N } . To retain the positive polynomial terms, let d n d_n d n d n d_n d n
B k = ∑ n = 0 k d ‾ n T n n ! ( k n ) ( N − 1 n ) , 0 ≤ k < N . B_k=\sum_{n=0}^{k}\frac{\underline{d}_nT^n}{n!}\frac{\binom{k}{n}}{\binom{N-1}{n}},\qquad0\leq k<N. B k = n = 0 ∑ k n ! d n T n ( n N − 1 ) ( n k ) , 0 ≤ k < N . The Bernstein basis is nonnegative and sums to one on [ 0 , 1 ] [0,1] [ 0 , 1 ] ∑ n < N d n t n / n ! ≥ min k < N B k \sum_{n<N}d_nt^n/n!\geq\min_{k<N}B_k ∑ n < N d n t n / n ! ≥ min k < N B k 0 ≤ t ≤ T 0\leq t\leq T 0 ≤ t ≤ T
min k < N B k − E N ( T ) > 1 2 . (36) \min_{k<N}B_k-E_N(T)>\frac{1}{2}. \tag*{(36)} k < N min B k − E N ( T ) > 2 1 . (36) They use N = 40 , 32 , 24 , 16 N=40,32,24,16 N = 40 , 32 , 24 , 16 [ 1 , 6 / 5 ] [1,6/5] [ 1 , 6/5 ] [ 6 / 5 , 13 / 10 ] [6/5,13/10] [ 6/5 , 13/10 ] [ 13 / 10 , 3 / 2 ] [13/10,3/2] [ 13/10 , 3/2 ] [ 3 / 2 , 1.6671 ] [3/2,1.6671] [ 3/2 , 1.6671 ] 1 / 20 1/20 1/20 3 / 100 3/100 3/100 1 / 20 1/20 1/20 1 / 20 1/20 1/20
For large time we use one cut-mass estimate on all three parameter ranges. Put b = 1 / 20 b=1/20 b = 1/20 χ = cos δ \chi=\cos\delta χ = cos δ λ = ( 1 − 2 χ 2 ) / ( 2 ( 1 + χ ) ) \lambda=(1-2\chi^2)/(2(1+\chi)) λ = ( 1 − 2 χ 2 ) / ( 2 ( 1 + χ ))
1 1 + 2 χ v + v 2 ≤ 1 − χ v − λ v 2 , 0 ≤ v ≤ 1 , \frac{1}{1+2\chi v+v^2}\leq1-\chi v-\lambda v^2,\qquad0\leq v\leq1, 1 + 2 χ v + v 2 1 ≤ 1 − χ v − λ v 2 , 0 ≤ v ≤ 1 , follows by multiplying through: the difference is v ( 1 − v ) { χ + ( χ + λ + 2 χ λ ) v + λ v 2 } v(1-v)\{\chi+(\chi+\lambda+2\chi\lambda)v+\lambda v^2\} v ( 1 − v ) { χ + ( χ + λ + 2 χ λ ) v + λ v 2 } λ ≥ 0 \lambda\geq0 λ ≥ 0 λ < 0 \lambda<0 λ < 0 χ \chi χ 1 + 2 χ − 2 χ 2 1+2\chi-2\chi^2 1 + 2 χ − 2 χ 2 v = min ( u α , u − α ) v=\min(u^\alpha,u^{-\alpha}) v = min ( u α , u − α )
H α ( x ) ≤ H ^ α ( x ) : = I α ( x ) − χ I 2 α ( x ) − λ I 3 α ( x ) , H_\alpha(x)\leq\widehat{H}_\alpha(x):=I_\alpha(x)-\chi I_{2\alpha}(x)-\lambda I_{3\alpha}(x), H α ( x ) ≤ H α ( x ) := I α ( x ) − χ I 2 α ( x ) − λ I 3 α ( x ) , I r ( x ) : = ∫ 0 1 e − x u u r d u + ∫ 1 ∞ e − x u u − r d u = x r − 1 γ ( r + 1 , x ) + x r − 1 Γ ( 1 − r , x ) . I_r(x):=\int_0^1e^{-xu}u^r\,du+\int_1^\infty e^{-xu}u^{-r}\,du
=x^{r-1}\gamma(r+1,x)+x^{r-1}\Gamma(1-r,x). I r ( x ) := ∫ 0 1 e − xu u r d u + ∫ 1 ∞ e − xu u − r d u = x r − 1 γ ( r + 1 , x ) + x r − 1 Γ ( 1 − r , x ) . The lower and upper incomplete Gamma functions here represent convergent real integrals, including when 1 − r ≤ 0 1-r\leq0 1 − r ≤ 0
The function H α ( x ) H_\alpha(x) H α ( x ) x x x α \alpha α 2 cosh ( α log u ) + 2 cos δ 2\cosh(\alpha\log u)+2\cos\delta 2 cosh ( α log u ) + 2 cos δ A = [ a 0 , a 1 ] A=[a_0,a_1] A = [ a 0 , a 1 ]
x 0 = b 1 / a 0 , x 1 = b 1 / a 1 , η ˉ A = a 1 sin δ ( a 0 ) π , n A = η ˉ A H ^ a 0 ( x 0 ) − sin x 0 . x_0=b^{1/a_0},\qquad x_1=b^{1/a_1},\qquad\bar{\eta}_A=\frac{a_1\sin\delta(a_0)}{\pi},\qquad n_A=\bar{\eta}_A\widehat{H}_{a_0}(x_0)-\sin x_0. x 0 = b 1/ a 0 , x 1 = b 1/ a 1 , η ˉ A = π a 1 sin δ ( a 0 ) , n A = η ˉ A H a 0 ( x 0 ) − sin x 0 . On the boxes below n A > 0 n_A>0 n A > 0 x 0 ≤ x ≤ x 1 x_0\leq x\leq x_1 x 0 ≤ x ≤ x 1 η α ≤ η ˉ A \eta_\alpha\leq\bar{\eta}_A η α ≤ η ˉ A
W α ( b ) : = ∫ 0 b w α ( v ) d v ≥ 1 a 1 − 1 2 − 1 π arctan ( n A cos x 1 ) . W_\alpha(b):=\int_0^b w_\alpha(v)\,dv\geq\frac{1}{a_1}-\frac{1}{2}-\frac{1}{\pi}\arctan\left(\frac{n_A}{\cos x_1}\right). W α ( b ) := ∫ 0 b w α ( v ) d v ≥ a 1 1 − 2 1 − π 1 arctan ( cos x 1 n A ) . Only the following scalar inequalities are required:
Table 1 (PDF p. 11).
The exact endpoints of these nonuniform partitions are listed in the scalar certificates. These whole-box bounds use real incomplete Gamma and elementary functions, with no transform quadrature. The final time comparisons follow from e 627 / 100 > 523 e^{627/100}>523 e 627/100 > 523 e 57 / 10 > 298 e^{57/10}>298 e 57/10 > 298
899 5000 523 > 94 , 1351 10000 298 > 37 , 441 5000 298 > 26. \frac{899}{5000}523>94,\qquad\frac{1351}{10000}298>37,\qquad\frac{441}{5000}298>26. 5000 899 523 > 94 , 10000 1351 298 > 37 , 5000 441 298 > 26. The two exponential bounds already follow from their degree-16 Taylor sums. Lemma 4.3 now gives K α ( t ) ≥ W α ( b ) e − b t − B α e − t > 0 K_\alpha(t)\geq W_\alpha(b)e^{-bt}-B_\alpha e^{-t}>0 K α ( t ) ≥ W α ( b ) e − b t − B α e − t > 0 t ≥ T t\geq T t ≥ T W α ( b ) e ( 1 − b ) t W_\alpha(b)e^{(1-b)t} W α ( b ) e ( 1 − b ) t t t t
Validated inversion on compact rectangles
We use the rational model R p R_p R p r p r_p r p R = R p R=R_p R = R p r = r p r=r_p r = r p G = F − R G=F-R G = F − R g = K − r g=K-r g = K − r G = O ( q − p − 1 ) G=O(q^{-p-1}) G = O ( q − p − 1 ) P > t P>t P > t ω = 2 π / P \omega=2\pi/P ω = 2 π / P J ω ≥ Y > h J\omega\ge Y>h J ω ≥ Y > h
Lemma 4.5 (Complete Fourier error). If ∣ g ( t ) ∣ ≤ M |g(t)|\le M ∣ g ( t ) ∣ ≤ M t > 0 t>0 t > 0
H J ( t ) = r ( t ) + e σ t P { G ( σ ) + 2 Re ∑ j = 1 J e i j ω t G ( σ + i j ω ) } \mathcal{H}_J(t)=r(t)+\frac{e^{\sigma t}}{P}\left\{G(\sigma)+2\operatorname{Re}\sum_{j=1}^{J}e^{ij\omega t}G(\sigma+ij\omega)\right\} H J ( t ) = r ( t ) + P e σ t { G ( σ ) + 2 Re j = 1 ∑ J e ij ω t G ( σ + ij ω ) } satisfies, for 0 < t < P 0<t<P 0 < t < P
∣ K ( t ) − H J ( t ) ∣ ≤ M e − σ P 1 − e − σ P + e σ t π { ∑ n = p N − 1 ∣ d n − r n ∣ n Y n + C F + C R N Y N } , |K(t)-\mathcal{H}_J(t)|\le\frac{M e^{-\sigma P}}{1-e^{-\sigma P}}+\frac{e^{\sigma t}}{\pi}\left\{\sum_{n=p}^{N-1}\frac{|d_n-r_n|}{nY^n}+\frac{C_F+C_R}{NY^N}\right\}, ∣ K ( t ) − H J ( t ) ∣ ≤ 1 − e − σ P M e − σ P + π e σ t { n = p ∑ N − 1 n Y n ∣ d n − r n ∣ + N Y N C F + C R } , where r n r_n r n R R R
C R = ∑ k = 1 p ∣ b k ∣ ( N k − 1 ) h N + 1 − k ( 1 − h / Y ) k . C_R=\sum_{k=1}^{p}|b_k|\binom{N}{k-1}\frac{h^{N+1-k}}{(1-h/Y)^k}. C R = k = 1 ∑ p ∣ b k ∣ ( k − 1 N ) ( 1 − h / Y ) k h N + 1 − k . The same assertion holds for ∣ g ( t ) ∣ ≤ M e λ t |g(t)|\le M e^{\lambda t} ∣ g ( t ) ∣ ≤ M e λ t σ > λ \sigma>\lambda σ > λ M e λ t e − ( σ − λ ) P / ( 1 − e − ( σ − λ ) P ) M e^{\lambda t}e^{-(\sigma-\lambda)P}/(1-e^{-(\sigma-\lambda)P}) M e λ t e − ( σ − λ ) P / ( 1 − e − ( σ − λ ) P )
Proof. Extend g g g e − σ t g ( t ) e^{-\sigma t}g(t) e − σ t g ( t ) G ( σ + i j ω ) / P G(\sigma+ij\omega)/P G ( σ + ij ω ) / P 0 < t < P 0<t<P 0 < t < P
g ( t ) = e σ t P ∑ j ∈ Z e i j ω t G ( σ + i j ω ) − ∑ ℓ ≥ 1 e − σ ℓ P g ( t + ℓ P ) . g(t)=\frac{e^{\sigma t}}{P}\sum_{j\in\mathbb{Z}}e^{ij\omega t}G(\sigma+ij\omega)-\sum_{\ell\ge1}e^{-\sigma\ell P}g(t+\ell P). g ( t ) = P e σ t j ∈ Z ∑ e ij ω t G ( σ + ij ω ) − ℓ ≥ 1 ∑ e − σ ℓ P g ( t + ℓ P ) . The last sum has exactly the stated alias bound. Expand the rational model to the same finite order as F F F C R ∣ q ∣ − N − 1 C_R|q|^{-N-1} C R ∣ q ∣ − N − 1 ∑ j > J j − n − 1 ≤ J − n / n \sum_{j>J}j^{-n-1}\le J^{-n}/n ∑ j > J j − n − 1 ≤ J − n / n J ω ≥ Y J\omega\ge Y J ω ≥ Y 1 / P 1/P 1/ P 1 / π 1/\pi 1/ π
The bound in Lemma 4.3 supplies M M M
sup t ≥ 0 ∣ r ( t ) ∣ ≤ ∣ b 1 ∣ + ∑ k = 2 p ∣ b k ∣ ( k − 1 ) ! ( k − 1 e h ) k − 1 . \sup_{t\ge0}|r(t)|\le|b_1|+\sum_{k=2}^{p}\frac{|b_k|}{(k-1)!}\left(\frac{k-1}{eh}\right)^{k-1}. t ≥ 0 sup ∣ r ( t ) ∣ ≤ ∣ b 1 ∣ + k = 2 ∑ p ( k − 1 )! ∣ b k ∣ ( e h k − 1 ) k − 1 . A real parameter mean-value enclosure is applied to the finite function H J \mathcal{H}_J H J t m t_m t m ∣ e i j ω t ∣ = 1 |e^{ij\omega t}|=1 ∣ e ij ω t ∣ = 1
For the earlier ranges one uses directly the absolutely convergent Bromwich integral after subtracting d 0 / q d_0/q d 0 / q q − 2 q^{-2} q − 2 2 e c t / ( π t Y ) 2e^{ct}/(\pi tY) 2 e c t / ( π t Y ) e c t C F / ( π N Y N ) e^{ct}C_F/(\pi N Y^N) e c t C F / ( π N Y N ) w = e i ϕ u w=e^{i\phi}u w = e i ϕ u z = w α / q z=w^\alpha/q z = w α / q
A = e i ( α + 1 ) ϕ Γ ( α + 1 ) ∫ 0 ∞ e − e i ϕ u u α D δ ( z ) d u , F = β q N A . A=\frac{e^{i(\alpha+1)\phi}}{\Gamma(\alpha+1)}\int_0^\infty\frac{e^{-e^{i\phi}u}u^\alpha}{D_\delta(z)}\,du,\qquad F=\frac{\beta}{q}\frac{N}{A}. A = Γ ( α + 1 ) e i ( α + 1 ) ϕ ∫ 0 ∞ D δ ( z ) e − e i ϕ u u α d u , F = q β A N . where N N N w w w
log u + i ϕ − ψ ( α + 1 ) − ∂ α D δ ( z ) D δ ( z ) , ∂ α D δ ( z ) = − π sin δ z − 2 cos δ z log w + 2 z 2 log w . \log u+i\phi-\psi(\alpha+1)-\frac{\partial_\alpha D_\delta(z)}{D_\delta(z)},\qquad\partial_\alpha D_\delta(z)=-\pi\sin\delta\,z-2\cos\delta\,z\log w+2z^2\log w. log u + i ϕ − ψ ( α + 1 ) − D δ ( z ) ∂ α D δ ( z ) , ∂ α D δ ( z ) = − π sin δ z − 2 cos δ z log w + 2 z 2 log w . Thus ∂ α F = − β F + ( β / q ) ( N α / A − N A α / A 2 ) \partial_\alpha F=-\beta F+(\beta/q)(N_\alpha/A-NA_\alpha/A^2) ∂ α F = − β F + ( β / q ) ( N α / A − N A α / A 2 )
Lemma 4.6 (The lower parameter ranges). For 1 < α ≤ 33 / 20 1<\alpha\le33/20 1 < α ≤ 33/20 K α ( t ) > 0 K_\alpha(t)>0 K α ( t ) > 0 t > 0 t>0 t > 0
Proof. The compact parts not covered by Lemma 4.4 have the following finite, closed-rectangle covers. The bounds in the last column are conservative rational lower bounds, restricted to the displayed compact time domains.
Table 2.
The first two rows use c = 1 / 8 c=1/8 c = 1/8 Y = 40 Y=40 Y = 40 N = 12 N=12 N = 12 c = 1 / 32 c=1/32 c = 1/32 Y = 120 Y=120 Y = 120 N = 8 N=8 N = 8 h = 1 h=1 h = 1
K α ( t ) ∈ K α m ( t m ) + ( t − t m ) ∂ t K α m ( T ) + ( α − α m ) ∂ α K A ( T ) . K_\alpha(t)\in K_{\alpha_m}(t_m)+(t-t_m)\,\partial_tK_{\alpha_m}(T)+(\alpha-\alpha_m)\,\partial_\alpha K_A(T). K α ( t ) ∈ K α m ( t m ) + ( t − t m ) ∂ t K α m ( T ) + ( α − α m ) ∂ α K A ( T ) . The last row uses Lemma 4.5 with σ = 1 / 8 \sigma=1/8 σ = 1/8 P = 64 P=64 P = 64 J = 4096 J=4096 J = 4096 Y = 400 Y=400 Y = 400 N = 18 N=18 N = 18 p = 4 p=4 p = 4 h = 20 h=20 h = 20 α = 1 \alpha=1 α = 1
The contact and completion of the positive range
Put
α 0 = 1.66706599 , u = 10 − 8 , t 0 = .1628 , v = 10 − 6 , \alpha_0=1.66706599,\qquad u=10^{-8},\qquad t_0=.1628,\qquad v=10^{-6}, α 0 = 1.66706599 , u = 1 0 − 8 , t 0 = .1628 , v = 1 0 − 6 , A ∗ = [ α 0 − u , α 0 + u ] , I ∗ = [ t 0 − v , t 0 + v ] . A_*=[\alpha_0-u,\alpha_0+u],\qquad I_*=[t_0-v,t_0+v]. A ∗ = [ α 0 − u , α 0 + u ] , I ∗ = [ t 0 − v , t 0 + v ] . Lemma 4.7 (One local contact). There is exactly one pair ( α ∗ , t ∗ ) ∈ A ∗ × I ∗ (\alpha_*,t_*) \in A_* \times I_* ( α ∗ , t ∗ ) ∈ A ∗ × I ∗ K α ∗ ( t ∗ ) = ∂ t K α ∗ ( t ∗ ) = 0 K_{\alpha_*}(t_*) = \partial_t K_{\alpha_*}(t_*) = 0 K α ∗ ( t ∗ ) = ∂ t K α ∗ ( t ∗ ) = 0
1.6670659903 < α ∗ < 1.6670659915 , .162799 < t ∗ < .162801. 1.6670659903 < \alpha_* < 1.6670659915,\qquad.162799 < t_* < .162801. 1.6670659903 < α ∗ < 1.6670659915 , .162799 < t ∗ < .162801. Each K α K_\alpha K α α ∈ A ∗ \alpha\in A_* α ∈ A ∗ I ∗ I_* I ∗ α \alpha α
Proof. The interval inversion gives the following rational inequalities:
3 ⋅ 10 − 9 < K α 0 ( t 0 ) < 13 ⋅ 10 − 9 , ∣ ∂ t K α 0 ( t 0 ) ∣ < 57 ⋅ 10 − 6 , 3 \cdot10^{-9} < K_{\alpha_0}(t_0) < 13 \cdot10^{-9},\qquad\lvert\partial_t K_{\alpha_0}(t_0)\rvert< 57 \cdot10^{-6}, 3 ⋅ 1 0 − 9 < K α 0 ( t 0 ) < 13 ⋅ 1 0 − 9 , ∣ ∂ t K α 0 ( t 0 )∣ < 57 ⋅ 1 0 − 6 , 1402 < ∂ t 2 K α ( t ) < 1421 , − 9.068 < ∂ α K α ( t ) < − 9.059 ( ( α , t ) ∈ A ∗ × I ∗ ) . 1402 < \partial_t^2 K_\alpha(t) < 1421,\qquad-9.068 < \partial_\alpha K_\alpha(t) < -9.059\qquad((\alpha,t) \in A_* \times I_*). 1402 < ∂ t 2 K α ( t ) < 1421 , − 9.068 < ∂ α K α ( t ) < − 9.059 (( α , t ) ∈ A ∗ × I ∗ ) . For these local derivative evaluations, the line is Re q = 20 \operatorname{Re} q = 20 Re q = 20 Y = 1000 Y = 1000 Y = 1000 N = 18 N = 18 N = 18 q j F q^jF q j F d j / q d_j/q d j / q d j d_j d j
At α 0 \alpha_0 α 0 − 57 ⋅ 10 − 6 v + 701 v 2 -57 \cdot10^{-6}v + 701v^2 − 57 ⋅ 1 0 − 6 v + 701 v 2 ∣ α − α 0 ∣ ≤ u \lvert\alpha-\alpha_0\rvert\le u ∣ α − α 0 ∣ ≤ u u ( 9.068 − 9.059 ) u(9.068-9.059) u ( 9.068 − 9.059 )
− 57 ⋅ 10 − 12 + 701 ⋅ 10 − 12 − 90 ⋅ 10 − 12 = 554 ⋅ 10 − 12 > 0. -57 \cdot10^{-12} + 701 \cdot10^{-12} - 90 \cdot10^{-12} = 554 \cdot10^{-12} > 0. − 57 ⋅ 1 0 − 12 + 701 ⋅ 1 0 − 12 − 90 ⋅ 1 0 − 12 = 554 ⋅ 1 0 − 12 > 0. Strict convexity therefore gives a unique interior minimizer for every α ∈ A ∗ \alpha\in A_* α ∈ A ∗ m ( α ) = min t ∈ I ∗ K α ( t ) m(\alpha) = \min_{t \in I_*} K_\alpha(t) m ( α ) = min t ∈ I ∗ K α ( t ) a < b a < b a < b A ∗ A_* A ∗
− 9.068 ( b − a ) ≤ m ( b ) − m ( a ) ≤ − 9.059 ( b − a ) . (37) -9.068(b-a) \le m(b)-m(a) \le-9.059(b-a). \tag*{(37)} − 9.068 ( b − a ) ≤ m ( b ) − m ( a ) ≤ − 9.059 ( b − a ) . (37) In particular m m m
m ( α 0 ) > 3 ⋅ 10 − 9 − ( 57 ⋅ 10 − 6 ) 2 2 ⋅ 1402 > 0 , m ( α 0 + u ) < 13 ⋅ 10 − 9 − 9.059 ⋅ 10 − 8 < 0. m(\alpha_0) > 3 \cdot10^{-9} - \frac{(57 \cdot10^{-6})^2}{2 \cdot1402} > 0,\qquad m(\alpha_0+u) < 13 \cdot10^{-9} - 9.059 \cdot10^{-8} < 0. m ( α 0 ) > 3 ⋅ 1 0 − 9 − 2 ⋅ 1402 ( 57 ⋅ 1 0 − 6 ) 2 > 0 , m ( α 0 + u ) < 13 ⋅ 1 0 − 9 − 9.059 ⋅ 1 0 − 8 < 0. Consequently m m m (37) implies
3 ⋅ 10 − 9 − ( 57 ⋅ 10 − 6 ) 2 / ( 2 ⋅ 1402 ) 9.068 < α ∗ − α 0 < 13 ⋅ 10 − 9 9.059 . \frac{3 \cdot10^{-9} - (57 \cdot10^{-6})^2/(2 \cdot1402)}{9.068} < \alpha_*-\alpha_0 < \frac{13 \cdot10^{-9}}{9.059}. 9.068 3 ⋅ 1 0 − 9 − ( 57 ⋅ 1 0 − 6 ) 2 / ( 2 ⋅ 1402 ) < α ∗ − α 0 < 9.059 13 ⋅ 1 0 − 9 . These bounds give the stated enclosure. The unique interior minimizer at α ∗ \alpha_* α ∗ t ∗ t_* t ∗ A ∗ × I ∗ A_* \times I_* A ∗ × I ∗
Lemma 4.8 (Complete positive side). For every 1 < α < α ∗ 1 < \alpha< \alpha_* 1 < α < α ∗ K α ( t ) > 0 K_\alpha(t) > 0 K α ( t ) > 0 t > 0 t > 0 t > 0 α ∗ \alpha_* α ∗ t ∗ t_* t ∗
Proof. By Lemmas 4.6 and 4.4, it remains to consider 33 / 20 ≤ α ≤ α ∗ 33/20 \le\alpha\le\alpha_* 33/20 ≤ α ≤ α ∗ .045 ≤ t ≤ 6 .045 \le t \le6 .045 ≤ t ≤ 6 a − = α 0 − u = 1.66706598 a_- = \alpha_0-u = 1.66706598 a − = α 0 − u = 1.66706598
First let a − ≤ α ≤ α ∗ a_- \le\alpha\le\alpha_* a − ≤ α ≤ α ∗ A ∗ × [ .145 , .175 ] A_* \times[.145,.175] A ∗ × [ .145 , .175 ] ∂ t 2 K α > 242 \partial_t^2 K_\alpha> 242 ∂ t 2 K α > 242 I ∗ I_* I ∗ ( α , t ) = ( α ∗ , t ∗ ) (\alpha,t) = (\alpha_*,t_*) ( α , t ) = ( α ∗ , t ∗ )
A ∗ × ( [ .045 , .145 ] ∪ [ .175 , 6 ] ) , A_* \times([.045,.145] \cup[.175,6]), A ∗ × ([ .045 , .145 ] ∪ [ .175 , 6 ]) , 891 adjacent time rectangles give K α > 10 − 5 K_\alpha> 10^{-5} K α > 1 0 − 5 σ = 1 / 2 \sigma= 1/2 σ = 1/2 P = 32 P = 32 P = 32 J = 2048 J = 2048 J = 2048 Y = 400 Y = 400 Y = 400 N = 18 N = 18 N = 18 p = 4 p = 4 p = 4 h = 20 h = 20 h = 20
Now let 33 / 20 ≤ α ≤ a − 33/20 \le\alpha\le a_- 33/20 ≤ α ≤ a − [ 33 / 20 , a − ] × [ .15 , .17 ] [33/20,a_-] \times[.15,.17] [ 33/20 , a − ] × [ .15 , .17 ] ∂ α K α < − 1 \partial_\alpha K_\alpha< -1 ∂ α K α < − 1 Z ( q ) = L α ( q ) ( q + 2 ) d 0 / ( κ α Γ ( α + 1 ) ) Z(q) = L_\alpha(q)(q+2)^{d_0}/(\kappa_\alpha\Gamma(\alpha+1)) Z ( q ) = L α ( q ) ( q + 2 ) d 0 / ( κ α Γ ( α + 1 )) ∣ L α ( − v + i 0 ) ∣ ≥ sin ( π β ) cos 1 |L_\alpha(-v+i0)| \ge\sin(\pi\beta)\cos1 ∣ L α ( − v + i 0 ) ∣ ≥ sin ( π β ) cos 1 ∣ Z ∣ > 0.1476 |Z| > 0.1476 ∣ Z ∣ > 0.1476 1 / Z 1/Z 1/ Z Re q ≥ 1 \operatorname{Re} q \ge1 Re q ≥ 1 ∣ A α ∣ > 0.0252 |A_\alpha| > 0.0252 ∣ A α ∣ > 0.0252 ∣ log u ∣ ≤ u + u − 1 |\log u| \le u+u^{-1} ∣ log u ∣ ≤ u + u − 1 ∣ ∂ α F α ( q ) ∣ < 4.178 ⋅ 10 9 / ∣ q ∣ |\partial_\alpha F_\alpha(q)| < 4.178 \cdot10^9/|q| ∣ ∂ α F α ( q ) ∣ < 4.178 ⋅ 1 0 9 /∣ q ∣ − β 2 / ( q + 2 ) -\beta^2/(q+2) − β 2 / ( q + 2 ) ∣ y ∣ ≥ 400 |y| \ge400 ∣ y ∣ ≥ 400
∣ ∂ α K α ( t ) ∣ < 5.319 ⋅ 10 11 e t . |\partial_\alpha K_\alpha(t)| < 5.319 \cdot10^{11}e^t. ∣ ∂ α K α ( t ) ∣ < 5.319 ⋅ 1 0 11 e t . The exponential-growth version of Lemma 4.5, with σ = 5 \sigma= 5 σ = 5 P = 8 P = 8 P = 8 J = 512 J = 512 J = 512 α \alpha α
K α ( t ) ≥ K a − ( t ) + ( a − − α ) > 0 , .15 ≤ t ≤ .17. K_\alpha(t) \ge K_{a_-}(t) + (a_- - \alpha) > 0, \qquad.15 \le t \le.17. K α ( t ) ≥ K a − ( t ) + ( a − − α ) > 0 , .15 ≤ t ≤ .17. where positivity at a − a_- a − [ .045 , .15 ] ∪ [ .17 , 6 ] [.045,.15] \cup[.17,6] [ .045 , .15 ] ∪ [ .17 , 6 ] K α > 2 ⋅ 10 − 4 K_\alpha> 2 \cdot10^{-4} K α > 2 ⋅ 1 0 − 4
The negative-side argument is needed separately to establish that this locally unique contact is the unique classification threshold. No claim that other contacts cannot occur at larger indices is used or implied.
The non-infinitely-divisible range
The local minimum constructed below the critical index already gives a negative value of K α K_\alpha K α F α F_\alpha F α
Lemma 5.1 (Gaussian necessary condition). Suppose a nonnegative random variable has an infinitely divisible law with Laplace transform L L L F = − L ′ / L F=-L'/L F = − L ′ / L σ , h > 0 \sigma,h>0 σ , h > 0 t 0 ∈ R t_0\in\mathbb{R} t 0 ∈ R Ω > 0 \Omega>0 Ω > 0
J ( σ , t 0 , h , Ω ) = 2 π h R ∫ 0 Ω e − h 2 y 2 / 2 − i y t 0 F ( σ − i y ) d y . (38) J(\sigma,t_0,h,\Omega)=\sqrt{\frac{2}{\pi}}h\mathfrak{R}\int_0^\Omega e^{-h^2y^2/2-iyt_0}F(\sigma-iy)\,dy. \tag*{(38)} J ( σ , t 0 , h , Ω ) = π 2 h R ∫ 0 Ω e − h 2 y 2 /2 − i y t 0 F ( σ − i y ) d y . (38) Then
J ( σ , t 0 , h , Ω ) + 2 π e − h 2 Ω 2 / 2 h Ω F ( σ ) ≥ 0. (39) J(\sigma,t_0,h,\Omega)+\sqrt{\frac{2}{\pi}}\frac{e^{-h^2\Omega^2/2}}{h\Omega}F(\sigma)\ge0. \tag*{(39)} J ( σ , t 0 , h , Ω ) + π 2 h Ω e − h 2 Ω 2 /2 F ( σ ) ≥ 0. (39) Proof. By (3.20), F ( q ) = ∫ [ 0 , ∞ ) e − q t μ ( d t ) F(q)=\int_{[0,\infty)}e^{-qt}\,\mu(dt) F ( q ) = ∫ [ 0 , ∞ ) e − q t μ ( d t ) μ = δ δ 0 + τ ν ( d t ) \mu=\delta\delta_0+\tau\nu(dt) μ = δ δ 0 + τ ν ( d t ) ∫ e − σ t μ ( d t ) = F ( σ ) < ∞ \int e^{-\sigma t}\mu(dt)=F(\sigma)<\infty ∫ e − σ t μ ( d t ) = F ( σ ) < ∞ ∣ F ( σ − i y ) ∣ ≤ F ( σ ) |F(\sigma-iy)|\le F(\sigma) ∣ F ( σ − i y ) ∣ ≤ F ( σ )
0 ≤ ∫ [ 0 , ∞ ) e − σ t − ( t − t 0 ) 2 / ( 2 h 2 ) μ ( d t ) = 2 π h R ∫ 0 ∞ e − h 2 y 2 / 2 − i y t 0 F ( σ − i y ) d y . (40) 0\le\int_{[0,\infty)}e^{-\sigma t-(t-t_0)^2/(2h^2)}\,\mu(dt)
=\sqrt{\frac{2}{\pi}}h\mathfrak{R}\int_0^\infty e^{-h^2y^2/2-iyt_0}F(\sigma-iy)\,dy. \tag*{(40)} 0 ≤ ∫ [ 0 , ∞ ) e − σ t − ( t − t 0 ) 2 / ( 2 h 2 ) μ ( d t ) = π 2 h R ∫ 0 ∞ e − h 2 y 2 /2 − i y t 0 F ( σ − i y ) d y . (40) The absolute value of the omitted part is at most
2 π h F ( σ ) ∫ Ω ∞ e − h 2 y 2 / 2 d y ≤ 2 π F ( σ ) e − h 2 Ω 2 / 2 h Ω . \sqrt{\frac{2}{\pi}}hF(\sigma)\int_{\Omega}^{\infty}e^{-h^{2}y^{2}/2}\,dy\le\sqrt{\frac{2}{\pi}}F(\sigma)\frac{e^{-h^{2}\Omega^{2}/2}}{h\Omega}. π 2 h F ( σ ) ∫ Ω ∞ e − h 2 y 2 /2 d y ≤ π 2 F ( σ ) h Ω e − h 2 Ω 2 /2 . This proves the assertion. Notice that the estimate is used under the infinite-divisibility hypothesis; no signed inverse kernel is needed. □
Lemma 5.2 (Rotated moments and their error bounds). Let k k k s = q β s=q^\beta s = q β ℜ q > 0 \Re q>0 ℜ q > 0
J k ( s ) = ∫ 0 ∞ v α + k e − s v D δ ( v α ) d v . J_k(s)=\int_0^\infty\frac{v^{\alpha+k}e^{-sv}}{D_\delta(v^\alpha)}\,dv. J k ( s ) = ∫ 0 ∞ D δ ( v α ) v α + k e − s v d v . On a ray v = e i γ u v=e^{i\gamma}u v = e iγ u ℜ ( s e i γ ) = λ > 0 \Re(se^{i\gamma})=\lambda>0 ℜ ( s e iγ ) = λ > 0
J k ( s ) = P k ( s ) + e i ( α + k + 1 ) γ ∫ 0 ∞ u α + k e − s e i γ u D δ ( e i α γ u α ) d u . (41) J_k(s)=P_k(s)+e^{i(\alpha+k+1)\gamma}\int_0^\infty\frac{u^{\alpha+k}e^{-se^{i\gamma}u}}{D_\delta(e^{i\alpha\gamma}u^\alpha)}\,du. \tag*{(41)} J k ( s ) = P k ( s ) + e i ( α + k + 1 ) γ ∫ 0 ∞ D δ ( e i α γ u α ) u α + k e − s e iγ u d u . (41) Here P k = 0 P_k=0 P k = 0 0 ≤ α γ < δ 0\le\alpha\gamma<\delta 0 ≤ α γ < δ δ < α γ \delta<\alpha\gamma δ < α γ α γ + δ < π \alpha\gamma+\delta<\pi α γ + δ < π
P k ( s ) = π p k + 1 e − p s α sin δ , p = e i δ / α . (42) P_k(s)=\frac{\pi p^{k+1}e^{-ps}}{\alpha\sin\delta},\qquad p=e^{i\delta/\alpha}. \tag*{(42)} P k ( s ) = α sin δ π p k + 1 e − p s , p = e i δ / α . (42) In either case a denominator lower bound is
∣ D δ ( e i α γ u α ) ∣ ≥ D ∗ : = sin ∣ α γ − δ ∣ sin ( α γ + δ ) > 0. (43) \left|D_\delta(e^{i\alpha\gamma}u^\alpha)\right|\ge D_*:=\sin|\alpha\gamma-\delta|\sin(\alpha\gamma+\delta)>0. \tag*{(43)} D δ ( e i α γ u α ) ≥ D ∗ := sin ∣ α γ − δ ∣ sin ( α γ + δ ) > 0. (43) For 0 < u 0 < U 0<u_0<U 0 < u 0 < U ( 0 , u 0 ) (0,u_0) ( 0 , u 0 ) ( U , ∞ ) (U,\infty) ( U , ∞ )
u 0 α + k + 1 ( α + k + 1 ) D ∗ , U α + k e − λ U D ∗ [ λ − ( α + k ) / U ] . (44) \frac{u_0^{\alpha+k+1}}{(\alpha+k+1)D_*},\qquad\frac{U^{\alpha+k}e^{-\lambda U}}{D_*[\lambda-(\alpha+k)/U]}. \tag*{(44)} ( α + k + 1 ) D ∗ u 0 α + k + 1 , D ∗ [ λ − ( α + k ) / U ] U α + k e − λ U . (44) provided the denominator in the latter bound is positive. Moreover F α ( q ) = β s J 1 ( s ) / ( q J 0 ( s ) ) F_\alpha(q)=\beta sJ_1(s)/(qJ_0(s)) F α ( q ) = β s J 1 ( s ) / ( q J 0 ( s ))
Proof. Rotation of the original contour proves (5.3). In the second case only v = p v=p v = p p k + 1 e − p s / ( 2 i α sin δ ) p^{k+1}e^{-ps}/(2i\alpha\sin\delta) p k + 1 e − p s / ( 2 i α sin δ ) (42) . The factorization of D δ D_\delta D δ ∣ 1 − x e i θ ∣ ≥ ∣ sin θ ∣ |1-xe^{i\theta}|\ge|\sin\theta| ∣1 − x e i θ ∣ ≥ ∣ sin θ ∣ x ≥ 0 x\ge0 x ≥ 0 (43) . The first estimate in (44) follows by dropping the exponential. For the second, write u = U + v u=U+v u = U + v u p ≤ U p e p v / U u^p\le U^pe^{pv/U} u p ≤ U p e p v / U p = α + k p=\alpha+k p = α + k (9) under the integral gives the stated formula for F F F
For completeness, the parameter differentiation used in the Gaussian cover is specified here. The ray is held fixed while α \alpha α θ = π α / 2 \theta=\pi\alpha/2 θ = π α /2 c = cos θ c=\cos\theta c = cos θ z = v α z=v^\alpha z = v α
s α = − s log q / α 2 , c ′ = − ( π / 2 ) sin θ , (45) s_\alpha=-s\log q/\alpha^2,\qquad c'=-(\pi/2)\sin\theta, \tag*{(45)} s α = − s log q / α 2 , c ′ = − ( π /2 ) sin θ , (45) ( D δ ( z ) ) α = 2 z [ ( z + c ) log v + c ′ ] , (46) (D_\delta(z))_\alpha=2z[(z+c)\log v+c'], \tag*{(46)} ( D δ ( z ) ) α = 2 z [( z + c ) log v + c ′ ] , (46) ∂ α ( v α + k e − s v D δ ( z ) ) = v α + k e − s v D δ ( z ) ( log v − s α v − ( D δ ( z ) ) α D δ ( z ) ) . (47) \partial_\alpha\left(\frac{v^{\alpha+k}e^{-sv}}{D_\delta(z)}\right)=\frac{v^{\alpha+k}e^{-sv}}{D_\delta(z)}\left(\log v-s_\alpha v-\frac{(D_\delta(z))_\alpha}{D_\delta(z)}\right). \tag*{(47)} ∂ α ( D δ ( z ) v α + k e − s v ) = D δ ( z ) v α + k e − s v ( log v − s α v − D δ ( z ) ( D δ ( z ) ) α ) . (47) If Q k Q_k Q k
∂ α F α ( q ) = F α ( q ) ( 1 α − log q α 2 ) + β s q ( Q 1 J 0 − J 1 Q 0 J 0 2 ) . (48) \partial_\alpha F_\alpha(q)=F_\alpha(q)\left(\frac{1}{\alpha}-\frac{\log q}{\alpha^2}\right)+\frac{\beta s}{q}\left(\frac{Q_1}{J_0}-\frac{J_1Q_0}{J_0^2}\right). \tag*{(48)} ∂ α F α ( q ) = F α ( q ) ( α 1 − α 2 log q ) + q β s ( J 0 Q 1 − J 0 2 J 1 Q 0 ) . (48) For the pole term, p α / p = − i π / α 2 p_{\alpha}/p=-i\pi/\alpha^{2} p α / p = − iπ / α 2
∂ α P k = P k { − α − 1 + π 2 cot δ + ( k + 1 ) p α p − p α s − p s α } . \partial_{\alpha}P_k=P_k\left\{-\alpha^{-1}+\frac{\pi}{2}\cot\delta+(k+1)\frac{p_{\alpha}}{p}-p_{\alpha}s-ps_{\alpha}\right\}. ∂ α P k = P k { − α − 1 + 2 π cot δ + ( k + 1 ) p p α − p α s − p s α } . These identities include the dependence of q 1 1 / α q_1^{1/\alpha} q 1 1/ α α \alpha α (47) are obtained from
∫ 0 u 0 u p ( ∣ log u ∣ + γ ) d u = u 0 p + 1 ( − log u 0 + γ p + 1 + 1 ( p + 1 ) 2 ) , u 0 ≤ 1 , (49) \int_{0}^{u_0}u^p(|\log u|+\gamma)\,du=u_0^{p+1}\left(\frac{-\log u_0+\gamma}{p+1}+\frac{1}{(p+1)^2}\right),\qquad u_0\le1, \tag*{(49)} ∫ 0 u 0 u p ( ∣ log u ∣ + γ ) d u = u 0 p + 1 ( p + 1 − log u 0 + γ + ( p + 1 ) 2 1 ) , u 0 ≤ 1 , (49) ∣ log ( u e i γ ) ∣ ≤ 2 u ( u ≥ 100 , 0 ≤ γ < 1 ) , (50) |\log(ue^{i\gamma})|\le2u\qquad(u\ge100,0\le\gamma<1), \tag*{(50)} ∣ log ( u e iγ ) ∣ ≤ 2 u ( u ≥ 100 , 0 ≤ γ < 1 ) , (50) together with (44) , applied also to the powers 3 α + k 3\alpha+k 3 α + k 2 α + k 2\alpha+k 2 α + k
Lemma 5.3 (The finite Gaussian cover; computer-assisted). For every α ∈ [ 1.6671 , 1.7386 ] \alpha\in[1.6671,1.7386] α ∈ [ 1.6671 , 1.7386 ] T 1 T_1 T 1
Proof. All decimals in this proof are exact rationals. For the first part take σ = 1 \sigma=1 σ = 1 t 0 = .163 t_0=.163 t 0 = .163 h = 1 / 3000 h=1/3000 h = 1/3000 Ω = 7 / h = 21000 \Omega=7/h=21000 Ω = 7/ h = 21000 α = 1.6671 \alpha=1.6671 α = 1.6671
[ − 1.403446562329232 ⋅ 10 − 7 ± 3.90 ⋅ 10 − 10 ] . [-1.403446562329232\cdot10^{-7}\pm3.90\cdot10^{-10}]. [ − 1.403446562329232 ⋅ 1 0 − 7 ± 3.90 ⋅ 1 0 − 10 ] . For the same finite function (38) , the derivative intervals on the four successive parameter intervals of radius .0001 and centers 1.6672, 1.6674, 1.6676, 1.6678 are contained respectively in
[ − .00963 , − .00304 ] , [ − .00962 , − .00301 ] , [ − .00960 , − .00299 ] , [ − .00956 , − .00299 ] . [-.00963,-.00304],\qquad[-.00962,-.00301],\qquad[-.00960,-.00299],\qquad[-.00956,-.00299]. [ − .00963 , − .00304 ] , [ − .00962 , − .00301 ] , [ − .00960 , − .00299 ] , [ − .00956 , − .00299 ] . Their exact unions cover [ 1.6671 , 1.6679 ] [1.6671,1.6679] [ 1.6671 , 1.6679 ] − 1.39 ⋅ 10 − 7 -1.39\cdot10^{-7} − 1.39 ⋅ 1 0 − 7
For the rest of the cover, 33 overlapping parameter intervals cover [ 1.6679 , 1.7386 ] [1.6679,1.7386] [ 1.6679 , 1.7386 ] α ∈ [ a − r , a + r ] \alpha\in[a-r,a+r] α ∈ [ a − r , a + r ] J ( a ) J(a) J ( a ) B B B ∣ ∂ α J ∣ ≤ M |\partial_{\alpha}J|\le M ∣ ∂ α J ∣ ≤ M
sup B + r M + R < 0 (51) \sup B+rM+R<0 \tag*{(51)} sup B + r M + R < 0 (51) certifies the entire interval, where R R R (39) . The parameters, input balls, integral enclosures, derivative balls and tails of every interval are supplied with the ancillary certificate. The exact coverage test is to sort the rational endpoints, check that the first left endpoint is at most 1.6679, each subsequent left endpoint is at most the current right endpoint, and the final right endpoint is at least 1.7386. All 33 inequalities (51) are strict. Two of the intervals use more of the parameter range already covered by their derivative enclosures:
Table 3.
Recombining the same point, derivative and tail balls on these radii still gives strict negativity. Their overlap makes one former bridging interval unnecessary; two further intervals are redundant. The active list and its exact endpoint checks are supplied in the ancillary material. For example, the last interval [ 1.7356 , 1.7386 ] [1.7356,1.7386] [ 1.7356 , 1.7386 ] [ − .00515256635094 ± .001260000000001 ] [-.00515256635094\pm.001260000000001] [ − .00515256635094 ± .001260000000001 ]
Here are the evaluation details needed to reproduce the inequalities. The radial integral is regularized by u = x 4 u=x^4 u = x 4 u 0 = 2 − 64 u_0=2^{-64} u 0 = 2 − 64 U = 100 U=100 U = 100 (44) and its differentiated versions. The noncrossing ray is γ = π ( 1 / α 0 − 1 / 2 ) / 2 \gamma= \pi(1/\alpha_0 - 1/2)/2 γ = π ( 1/ α 0 − 1/2 ) /2 γ = π / 4 \gamma= \pi/4 γ = π /4 (42) . Every ray inequality is checked on the whole parameter interval. High frequencies in the point tests use the finite identity (17) , with N = 18 N = 18 N = 18 h Ω = 7 h\Omega= 7 h Ω = 7
Lemma 5.4 (A uniform zero certificate). Let I ⊂ ( 1 , 2 ) I \subset(1, 2) I ⊂ ( 1 , 2 ) G α ( s ) = sin δ J 0 ( s ) G_\alpha(s) = \sin\delta\, J_0(s) G α ( s ) = sin δ J 0 ( s ) ∣ s − z 0 ∣ ≤ r |s-z_0| \le r ∣ s − z 0 ∣ ≤ r ℜ s > 0 \Re s > 0 ℜ s > 0 α ∈ I \alpha\in I α ∈ I D 0 D_0 D 0
sup α ∈ I ∣ G α ( z 0 ) ∣ + r sup α ∈ I ∣ G α ′ ( z 0 ) − D 0 ∣ + r 2 2 sup α ∈ I ∣ s − z 0 ∣ ≤ r ∣ G α ′ ′ ( s ) ∣ < r ∣ D 0 ∣ , (52) \sup_{\alpha\in I}|G_\alpha(z_0)| + r\sup_{\alpha\in I}|G'_\alpha(z_0)-D_0| + \frac{r^2}{2}\sup_{\substack{\alpha\in I\\ |s-z_0|\le r}}|G''_\alpha(s)| < r|D_0|, \tag*{(52)} α ∈ I sup ∣ G α ( z 0 ) ∣ + r α ∈ I sup ∣ G α ′ ( z 0 ) − D 0 ∣ + 2 r 2 α ∈ I ∣ s − z 0 ∣ ≤ r sup ∣ G α ′′ ( s ) ∣ < r ∣ D 0 ∣ , (52) and ℜ ( s α ) > 0 \Re(s^\alpha) > 0 ℜ ( s α ) > 0 α ∈ I \alpha\in I α ∈ I T 1 T_1 T 1 α ∈ I \alpha\in I α ∈ I
Proof. Taylor’s formula with integral remainder gives
G α ( z 0 + z ) = G α ( z 0 ) + z G α ′ ( z 0 ) + z 2 ∫ 0 1 ( 1 − u ) G α ′ ′ ( z 0 + u z ) d u . G_\alpha(z_0+z)=G_\alpha(z_0)+zG'_\alpha(z_0)+z^2\int_0^1(1-u)G''_\alpha(z_0+uz)\,\mathrm{d}u. G α ( z 0 + z ) = G α ( z 0 ) + z G α ′ ( z 0 ) + z 2 ∫ 0 1 ( 1 − u ) G α ′′ ( z 0 + u z ) d u . On ∣ z ∣ = r |z|=r ∣ z ∣ = r (52) compares this expression with D 0 z D_0z D 0 z ∣ arg s ∣ < π / 2 |\arg s|<\pi/2 ∣ arg s ∣ < π /2 α < 2 \alpha<2 α < 2 ℜ ( s α ) > 0 \Re(s^\alpha)>0 ℜ ( s α ) > 0 ∣ α arg s ∣ < π / 2 |\alpha\arg s|<\pi/2 ∣ α arg s ∣ < π /2 q = s α q=s^\alpha q = s α ℜ q > 0 \Re q>0 ℜ q > 0 q 1 / α = s q^{1/\alpha}=s q 1/ α = s L α L_\alpha L α (27) shows that a nonnegative infinitely divisible law has a zero-free Laplace transform throughout that half-plane, a contradiction. □
Lemma 5.5 (The finite zero cover; computer-assisted). The distribution of T 1 T_1 T 1 1.7386 ≤ α ≤ 15 / 8 1.7386 \le\alpha\le15/8 1.7386 ≤ α ≤ 15/8
Proof. The active certificate contains 102 parameter intervals and disks to which Lemma 5.4 applies. Their rational endpoints cover the stated closed interval with overlaps. The moment enclosures in Lemma 5.2, on the ray γ = π / 4 \gamma=\pi/4 γ = π /4
G α = sin δ J 0 , G α ′ = − sin δ J 1 , G α ′ ′ = sin δ J 2 . G_\alpha=\sin\delta J_0,\qquad G'_\alpha=-\sin\delta J_1,\qquad G''_\alpha=\sin\delta J_2. G α = sin δ J 0 , G α ′ = − sin δ J 1 , G α ′′ = sin δ J 2 . For the last expression the complex rectangle [ ℜ z 0 − r , ℜ z 0 + r ] + i [ ℑ z 0 − r , ℑ z 0 + r ] [\Re z_0-r,\Re z_0+r]+i[\Im z_0-r,\Im z_0+r] [ ℜ z 0 − r , ℜ z 0 + r ] + i [ ℑ z 0 − r , ℑ z 0 + r ] ℜ z 0 − r > 5.63 \Re z_0-r>5.63 ℜ z 0 − r > 5.63 exp ( α log s ) \exp(\alpha\log s) exp ( α log s ) (52) hold strictly. Recombining the saved balls gives a maximum left/right ratio below .999668 .999668 .999668 ℜ ( s α ) \Re(s^\alpha) ℜ ( s α ) .0266059 .0266059 .0266059 z 0 z_0 z 0
Lemma 5.6 (An analytic zero for every 15 / 8 < α < 2 15/8 < \alpha< 2 15/8 < α < 2 For 15 / 8 ≤ α < 2 15/8 \le\alpha< 2 15/8 ≤ α < 2 L α L_\alpha L α ℜ q > 0 \Re q>0 ℜ q > 0 T 1 T_1 T 1
Proof. Put ε = 2 − α \varepsilon=2-\alpha ε = 2 − α 0 < ε ≤ 1 / 8 0<\varepsilon\le1/8 0 < ε ≤ 1/8 δ = π ε / 2 ≤ π / 16 \delta=\pi\varepsilon/2\le\pi/16 δ = π ε /2 ≤ π /16
B = κ α Γ ( m ) , p = e − i δ / α , d = π κ α α sin δ , A = d p , ℓ = log ( d / B ) , ν = π + δ . B=\kappa_\alpha\Gamma(m),\qquad p=e^{-i\delta/\alpha},\qquad d=\frac{\pi\kappa_\alpha}{\alpha\sin\delta},\qquad A=dp,\qquad\ell=\log(d/B),\qquad\nu=\pi+\delta. B = κ α Γ ( m ) , p = e − i δ / α , d = α sin δ π κ α , A = d p , ℓ = log ( d / B ) , ν = π + δ . We shall use the elementary digamma bound
ψ ( t ) > log ( t − 1 / 2 ) , t > 1 / 2. (53) \psi(t)>\log(t-1/2),\qquad t>1/2. \tag*{(53)} ψ ( t ) > log ( t − 1/2 ) , t > 1/2. (53) Indeed, subtracting the integral representations of its two sides gives
ψ ( t ) − log ( t − 1 / 2 ) = ∫ 0 ∞ e − ( t − 1 / 2 ) u ( 1 u − 1 2 sinh ( u / 2 ) ) d u > 0. \psi(t)-\log(t-1/2)=\int_{0}^{\infty}e^{-(t-1/2)u}\left(\frac{1}{u}-\frac{1}{2\sinh(u/2)}\right)\,du>0. ψ ( t ) − log ( t − 1/2 ) = ∫ 0 ∞ e − ( t − 1/2 ) u ( u 1 − 2 sinh ( u /2 ) 1 ) d u > 0. Since sin δ ≤ π ε / 2 \sin\delta\leq\pi\varepsilon/2 sin δ ≤ π ε /2 (53) gives
ℓ ≥ − log ( ε ( 2 − ε ) ) + ∫ 3 − ε 3 log ( t − 1 / 2 ) d t . \ell\geq-\log\bigl(\varepsilon(2-\varepsilon)\bigr)+\int_{3-\varepsilon}^{3}\log(t-1/2)\,dt. ℓ ≥ − log ( ε ( 2 − ε ) ) + ∫ 3 − ε 3 log ( t − 1/2 ) d t . This lower bound decreases with ε ∈ ( 0 , 1 / 8 ] \varepsilon\in(0,1/8] ε ∈ ( 0 , 1/8 ] − 1 / ε + 1 / ( 2 − ε ) + log ( 5 / 2 − ε ) < 0 -1/\varepsilon+1/(2-\varepsilon)+\log(5/2-\varepsilon)<0 − 1/ ε + 1/ ( 2 − ε ) + log ( 5/2 − ε ) < 0
ℓ ≥ log 64 15 + 5 2 log 5 2 − 19 8 log 19 8 − 1 8 > 31 20 . (54) \ell\geq\log\frac{64}{15}+\frac{5}{2}\log\frac{5}{2}-\frac{19}{8}\log\frac{19}{8}-\frac{1}{8}>\frac{31}{20}. \tag*{(54)} ℓ ≥ log 15 64 + 2 5 log 2 5 − 8 19 log 8 19 − 8 1 > 20 31 . (54) The duplication formula and (53) also give
d d α log Γ ( 2 α + 1 ) Γ ( α + 1 ) = log 4 + ψ ( α + 1 / 2 ) > log ( 4 α ) ≥ log 15 2 > 2. \frac{d}{d\alpha}\log\frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}
=\log4+\psi(\alpha+1/2)>\log(4\alpha)\geq\log\frac{15}{2}>2. d α d log Γ ( α + 1 ) Γ ( 2 α + 1 ) = log 4 + ψ ( α + 1/2 ) > log ( 4 α ) ≥ log 2 15 > 2. Here e < 87 / 32 e<87/32 e < 87/32 ( 87 / 32 ) 2 < 15 / 2 (87/32)^2<15/2 ( 87/32 ) 2 < 15/2
Γ ( 2 α + 1 ) Γ ( α + 1 ) ≤ 12 e − 2 ε . (55) \frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}\leq12e^{-2\varepsilon}. \tag*{(55)} Γ ( α + 1 ) Γ ( 2 α + 1 ) ≤ 12 e − 2 ε . (55) We first locate an exact zero of the exponential model. For x > m x>m x > m y ( x ) > 0 y(x)>0 y ( x ) > 0
y − m arctan ( y / x ) = ν . y-m\arctan(y/x)=\nu. y − m arctan ( y / x ) = ν . Existence and uniqueness follow because the left side increases from zero to infinity, with derivative at least 1 − m / x > 0 1-m/x>0 1 − m / x > 0
y ′ ( x ) = − m y x 2 + y 2 − m x , d d x ( x − m 2 log ( x 2 + y 2 ) ) = ( x − m ) 2 + y 2 x 2 + y 2 − m x > 0. y'(x)=-\frac{my}{x^2+y^2-mx},\qquad
\frac{d}{dx}\left(x-\frac{m}{2}\log(x^2+y^2)\right)=\frac{(x-m)^2+y^2}{x^2+y^2-mx}>0. y ′ ( x ) = − x 2 + y 2 − m x m y , d x d ( x − 2 m log ( x 2 + y 2 ) ) = x 2 + y 2 − m x ( x − m ) 2 + y 2 > 0. The expression differentiated on the left is negative as x ↓ m x\downarrow m x ↓ m x → ∞ x\to\infty x → ∞ (54) , there is therefore exactly one x > m x>m x > m
w 0 = x + i y ( x ) , w 0 − m log w 0 = ℓ + i ν . (56) w_0=x+iy(x),\qquad w_0-m\log w_0=\ell+i\nu. \tag*{(56)} w 0 = x + i y ( x ) , w 0 − m log w 0 = ℓ + i ν . (56) For useful uniform bounds, put m 0 = 23 / 8 m_0=23/8 m 0 = 23/8 h ( x ) h(x) h ( x ) h − m 0 arctan ( h / x ) = π h-m_0\arctan(h/x)=\pi h − m 0 arctan ( h / x ) = π x > m 0 x>m_0 x > m 0 y ( x ) ≥ h ( x ) y(x)\geq h(x) y ( x ) ≥ h ( x ) x 0 = 79 / 10 x_0=79/10 x 0 = 79/10
14 3 − 23 8 arctan 140 237 < π , 79 10 − 23 16 log 75769 900 < 31 20 (57) \frac{14}{3}-\frac{23}{8}\arctan\frac{140}{237}<\pi,\qquad
\frac{79}{10}-\frac{23}{16}\log\frac{75769}{900}<\frac{31}{20} \tag*{(57)} 3 14 − 8 23 arctan 237 140 < π , 10 79 − 16 23 log 900 75769 < 20 31 (57) show that h ( x 0 ) > 14 / 3 h(x_0)>14/3 h ( x 0 ) > 14/3 (56) cannot hold for x ≤ x 0 x\leq x_0 x ≤ x 0 x − m 0 log x 2 + h ( x ) 2 x-m_0\log\sqrt{x^2+h(x)^2} x − m 0 log x 2 + h ( x ) 2 31 / 20 31/20 31/20 x 0 x_0 x 0 x 2 + h ( x ) 2 \sqrt{x^2+h(x)^2} x 2 + h ( x ) 2 x x x arctan ( 51 / 79 ) < 7 / 12 \arctan(51/79)<7/12 arctan ( 51/79 ) < 7/12 ν ≤ 17 π / 16 < 67 / 20 = 51 / 10 − 3 ( 7 / 12 ) \nu\leq17\pi/16<67/20=51/10-3(7/12) ν ≤ 17 π /16 < 67/20 = 51/10 − 3 ( 7/12 ) y ( x ) < 51 / 10 y(x)<51/10 y ( x ) < 51/10
x > 79 10 , ∣ w 0 ∣ > 55 6 , π < y < 51 10 . (58) x>\frac{79}{10},\qquad|w_0|>\frac{55}{6},\qquad\pi<y<\frac{51}{10}. \tag*{(58)} x > 10 79 , ∣ w 0 ∣ > 6 55 , π < y < 10 51 . (58) Put r 0 = 13 / 10 r_0=13/10 r 0 = 13/10 ∣ w − w 0 ∣ ≤ r 0 |w-w_0|\leq r_0 ∣ w − w 0 ∣ ≤ r 0 U = ℜ w U=\Re w U = ℜ w V = ℑ w V=\Im w V = ℑ w λ = 7 / 8 \lambda=7/8 λ = 7/8 λ x − y > 29 / 16 \lambda x-y>29/16 λ x − y > 29/16
r 0 1 + λ 2 < 29 16 . r_0\sqrt{1+\lambda^2}<\frac{29}{16}. r 0 1 + λ 2 < 16 29 . Cauchy–Schwarz therefore gives U − V > 0 \mathcal{U}-V>0 U − V > 0 V > π − r 0 > 0 V>\pi-r_0>0 V > π − r 0 > 0
θ = π 2 α , γ = θ − δ α = π ( α − 1 ) 2 α . \theta=\frac{\pi}{2\alpha},\qquad\gamma=\theta-\frac{\delta}{\alpha}=\frac{\pi(\alpha-1)}{2\alpha}. θ = 2 α π , γ = θ − α δ = 2 α π ( α − 1 ) . Since
0 < arg w < arctan 7 8 < 73 100 < 7 π 30 ≤ γ , 0<\arg w<\arctan\frac{7}{8}<\frac{73}{100}<\frac{7\pi}{30}\leq\gamma, 0 < arg w < arctan 8 7 < 100 73 < 30 7 π ≤ γ , the point s = w / p s=w/p s = w / p 0 < arg s < θ 0<\arg s<\theta 0 < arg s < θ q = s α q=s^\alpha q = s α ℜ q > 0 \Re q>0 ℜ q > 0 q 1 / α = s q^{1/\alpha}=s q 1/ α = s a = π / 4 − γ = δ / ( 2 α ) ≤ π / 60 a=\pi/4-\gamma=\delta/(2\alpha)\leq\pi/60 a = π /4 − γ = δ / ( 2 α ) ≤ π /60
U − V U + V > 1 15 > 1 30 > tan ( a / 2 ) . \frac{U-V}{U+V}>\frac{1}{15}>\frac{1}{30}>\tan(a/2). U + V U − V > 15 1 > 30 1 > tan ( a /2 ) . Expanding cos ( π / 4 − a ) \cos(\pi/4-a) cos ( π /4 − a ) sin ( π / 4 − a ) \sin(\pi/4-a) sin ( π /4 − a )
ℜ ( s e − i θ ) = ℜ ( w e − i γ ) ≥ U + V 2 . (59) \Re(se^{-i\theta})=\Re(we^{-i\gamma})\geq\frac{U+V}{\sqrt{2}}. \tag*{(59)} ℜ ( s e − i θ ) = ℜ ( w e − iγ ) ≥ 2 U + V . (59) Rotate the r r r (9) downward through angle θ \theta θ p p p A e − p s Ae^{-ps} A e − p s 0 < arg s < θ < π / 2 0<\arg s<\theta<\pi/2 0 < arg s < θ < π /2
G ( s ) : = L α ( s α ) = A e − p s + J ( s ) , (60) G(s):=L_\alpha(s^\alpha)=Ae^{-ps}+J(s), \tag*{(60)} G ( s ) := L α ( s α ) = A e − p s + J ( s ) , (60) J ( s ) = κ α e − i m θ ∫ 0 ∞ e − s e − i θ u u α D δ ( − i u α ) d u . (61) J(s)=\kappa_\alpha e^{-im\theta}\int_0^\infty\frac{e^{-se^{-i\theta}u}u^\alpha}{D_\delta(-iu^\alpha)}\,du. \tag*{(61)} J ( s ) = κ α e − im θ ∫ 0 ∞ D δ ( − i u α ) e − s e − i θ u u α d u . (61) For z = − i b z=-ib z = − ib b ≥ 0 b\geq0 b ≥ 0
4 ∣ D δ ( − i b ) ∣ 2 − ∣ 2 cos δ + i b ∣ 2 = 4 sin 2 δ + ( 16 cos 2 δ − 9 ) b 2 + 4 b 4 ≥ 0. 4\lvert D_\delta(-ib)\rvert^2-\lvert2\cos\delta+ib\rvert^2=4\sin^2\delta+(16\cos^2\delta-9)b^2+4b^4\geq0. 4 ∣ D δ ( − ib ) ∣ 2 − ∣ 2 cos δ + ib ∣ 2 = 4 sin 2 δ + ( 16 cos 2 δ − 9 ) b 2 + 4 b 4 ≥ 0. Here cos δ > 3 / 4 \cos\delta>3/4 cos δ > 3/4 D δ ( z ) − 1 = 1 + z ( 2 cos δ − z ) / D δ ( z ) D_\delta(z)^{-1}=1+z(2\cos\delta-z)/D_\delta(z) D δ ( z ) − 1 = 1 + z ( 2 cos δ − z ) / D δ ( z )
J ( s ) = B s − m ( 1 + R 0 ( s ) ) , ∣ R 0 ( w / p ) ∣ ≤ 2 Γ ( 2 α + 1 ) Γ ( α + 1 ) ∣ w ∣ m ℜ ( w e − i γ ) 2 α + 1 . (62) J(s)=Bs^{-m}(1+R_0(s)),\qquad\left\lvert R_0(w/p)\right\rvert\leq2\frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}\frac{\lvert w\rvert^m}{\Re(we^{-i\gamma})^{2\alpha+1}}. \tag*{(62)} J ( s ) = B s − m ( 1 + R 0 ( s )) , ∣ R 0 ( w / p ) ∣ ≤ 2 Γ ( α + 1 ) Γ ( 2 α + 1 ) ℜ ( w e − iγ ) 2 α + 1 ∣ w ∣ m . (62) The remainder is analytic on a neighborhood of the disk.
To bound (62) , define
Q α ( U , V ) = ( U 2 + V 2 ) ( α + 1 ) / 2 ( U + V ) 2 α + 1 . Q_\alpha(U,V)=\frac{(U^2+V^2)^{(\alpha+1)/2}}{(U+V)^{2\alpha+1}}. Q α ( U , V ) = ( U + V ) 2 α + 1 ( U 2 + V 2 ) ( α + 1 ) /2 . This function decreases in each coordinate on ( 0 , ∞ ) 2 (0,\infty)^2 ( 0 , ∞ ) 2 ( U , V ) = ( x − r 0 , h ( x ) − r 0 ) (U,V)=(x-r_0,h(x)-r_0) ( U , V ) = ( x − r 0 , h ( x ) − r 0 ) x ≥ 79 / 10 x\geq79/10 x ≥ 79/10
∣ h ′ ( x ) ∣ ≤ m 0 2 x ( x − m 0 ) < 1 4 , ∣ ∂ U log Q α ∣ > 1 4 ∣ ∂ V log Q α ∣ . \lvert h'(x)\rvert\leq\frac{m_0}{2\sqrt{x(x-m_0)}}<\frac{1}{4},\qquad\lvert\partial_U\log Q_\alpha\rvert>\frac{1}{4}\lvert\partial_V\log Q_\alpha\rvert. ∣ h ′ ( x )∣ ≤ 2 x ( x − m 0 ) m 0 < 4 1 , ∣ ∂ U log Q α ∣ > 4 1 ∣ ∂ V log Q α ∣ . The second inequality follows from positivity of ( 2 α − 1 ) U 2 − 3 ( α + 1 ) U V + ( 7 α + 4 ) V 2 (2\alpha-1)U^2-3(\alpha+1)UV+(7\alpha+4)V^2 ( 2 α − 1 ) U 2 − 3 ( α + 1 ) U V + ( 7 α + 4 ) V 2 − 47 α 2 + 14 α + 25 < 0 -47\alpha^2+14\alpha+25<0 − 47 α 2 + 14 α + 25 < 0 1 − 2 α − 7 α 2 < 0 1-2\alpha-7\alpha^2<0 1 − 2 α − 7 α 2 < 0 u 0 = 33 / 5 u_0=33/5 u 0 = 33/5 v 0 = 101 / 30 v_0=101/30 v 0 = 101/30
Q α ( U , V ) ≤ Q α ( x − r 0 , y − r 0 ) ≤ Q α ( x − r 0 , h ( x ) − r 0 ) ≤ Q α ( u 0 , v 0 ) . Q_\alpha(U,V)\leq Q_\alpha(x-r_0,y-r_0)\leq Q_\alpha(x-r_0,h(x)-r_0)\leq Q_\alpha(u_0,v_0). Q α ( U , V ) ≤ Q α ( x − r 0 , y − r 0 ) ≤ Q α ( x − r 0 , h ( x ) − r 0 ) ≤ Q α ( u 0 , v 0 ) . Put H = ( u 0 + v 0 ) 2 u 0 2 + v 0 2 H=(u_0+v_0)^2\sqrt{u_0^2+v_0^2} H = ( u 0 + v 0 ) 2 u 0 2 + v 0 2 e > 163 / 60 e>163/60 e > 163/60
H < 59 4 < 2 e 2 , 24 32 Q 2 ( u 0 , v 0 ) < 9 16 . H<\frac{59}{4}<2e^2,\qquad24\sqrt{32}\,Q_2(u_0,v_0)<\frac{9}{16}. H < 4 59 < 2 e 2 , 24 32 Q 2 ( u 0 , v 0 ) < 16 9 . Combining (55) , (59) and (62) , and writing R ( w ) = R 0 ( w / p ) R(w)=R_0(w/p) R ( w ) = R 0 ( w / p )
∣ R ( w ) ∣ ≤ 24 32 Q 2 ( u 0 , v 0 ) ( H 2 e 2 ) ε < 9 16 . (63) |R(w)| \le24\sqrt{32}\,Q_2(u_0,v_0)\left(\frac{H}{2e^2}\right)^{\varepsilon}<\frac{9}{16}. \tag*{(63)} ∣ R ( w ) ∣ ≤ 24 32 Q 2 ( u 0 , v 0 ) ( 2 e 2 H ) ε < 16 9 . (63) Now let Ψ ( w ) = w − m log w − ℓ − i ν \Psi(w)=w-m\log w-\ell-i\nu Ψ ( w ) = w − m log w − ℓ − i ν (56) implies, on ∣ w − w 0 ∣ = r 0 |w-w_0|=r_0 ∣ w − w 0 ∣ = r 0
∣ Ψ ( w ) − ( w − w 0 ) ∣ ≤ − m log ( 1 − r 0 ∣ w 0 ∣ − 1 ) < 3 log 275 236 < 47 100 . (64) |\Psi(w)-(w-w_0)|\le-m\log(1-r_0|w_0|^{-1})<3\log\frac{275}{236}<\frac{47}{100}. \tag*{(64)} ∣Ψ ( w ) − ( w − w 0 ) ∣ ≤ − m log ( 1 − r 0 ∣ w 0 ∣ − 1 ) < 3 log 236 275 < 100 47 . (64) Since ∣ R ∣ < 9 / 16 |R|<9/16 ∣ R ∣ < 9/16 Ξ ( w ) = Ψ ( w ) + log ( 1 + R ( w ) ) \Xi(w)=\Psi(w)+\log(1+R(w)) Ξ ( w ) = Ψ ( w ) + log ( 1 + R ( w ))
∣ log ( 1 + R ( w ) ) ∣ < log 16 7 < 83 100 . |\log(1+R(w))|<\log\frac{16}{7}<\frac{83}{100}. ∣ log ( 1 + R ( w )) ∣ < log 7 16 < 100 83 . Thus ∣ Ξ ( w ) − ( w − w 0 ) ∣ < 13 / 10 = r 0 |\Xi(w)-(w-w_0)|<13/10=r_0 ∣Ξ ( w ) − ( w − w 0 ) ∣ < 13/10 = r 0 w ∗ w_* w ∗ Ξ \Xi Ξ B p m / A = e − ℓ − i δ Bp^m/A=e^{-\ell-i\delta} B p m / A = e − ℓ − i δ
G ( w / p ) B p m w − m = 1 + R ( w ) − e − Ψ ( w ) . \frac{\mathcal{G}(w/p)}{Bp^m w^{-m}}=1+R(w)-e^{-\Psi(w)}. B p m w − m G ( w / p ) = 1 + R ( w ) − e − Ψ ( w ) . At w ∗ w_* w ∗ L α L_\alpha L α q = ( w ∗ / p ) α ∈ { ℜ q > 0 } q=(w_*/p)^\alpha\in\{\Re q>0\} q = ( w ∗ / p ) α ∈ { ℜ q > 0 } (27) excludes infinite divisibility.
All fixed scalar comparisons in this proof reduce to rational arithmetic. One may use 157 / 50 < π < 22 / 7 157/50<\pi<22/7 157/50 < π < 22/7 163 / 60 < e < 87 / 32 163/60<e<87/32 163/60 < e < 87/32 arctan z \arctan z arctan z (54) and (57) follow respectively from
( 64 15 ) 40 ( 5 2 ) 100 ( 8 19 ) 95 > ( 87 32 ) 67 , ( 75769 900 ) 115 > ( 87 32 ) 508 . \left(\frac{64}{15}\right)^{40}\left(\frac{5}{2}\right)^{100}\left(\frac{8}{19}\right)^{95}>\left(\frac{87}{32}\right)^{67},\qquad\left(\frac{75769}{900}\right)^{115}>\left(\frac{87}{32}\right)^{508}. ( 15 64 ) 40 ( 2 5 ) 100 ( 19 8 ) 95 > ( 32 87 ) 67 , ( 900 75769 ) 115 > ( 32 87 ) 508 . The remainder bound is rational after squaring. The two last logarithmic bounds follow respectively from − log ( 1 − t ) ≤ t + t 2 / ( 2 ( 1 − t ) ) -\log(1-t)\le t+t^2/(2(1-t)) − log ( 1 − t ) ≤ t + t 2 / ( 2 ( 1 − t )) t = 39 / 275 t=39/275 t = 39/275 e 83 / 100 e^{83/100} e 83/100
Lemma 5.7 (Completion of the negative side). For every α ∗ < α < 2 \alpha_*<\alpha<2 α ∗ < α < 2 T 1 T_1 T 1
Proof. The contact argument gives a negative minimum for α ∗ < α ≤ 1.667066 \alpha_*<\alpha\le1.667066 α ∗ < α ≤ 1.667066 t 0 = .1628 t_0=.1628 t 0 = .1628
K 1.667066 ( t 0 ) < 13 ⋅ 10 − 9 − 9.059 ⋅ 10 − 8 < 0. K_{1.667066}(t_0)<13\cdot10^{-9}-9.059\cdot10^{-8}<0. K 1.667066 ( t 0 ) < 13 ⋅ 1 0 − 9 − 9.059 ⋅ 1 0 − 8 < 0. On α ∈ [ 1.667066 , 1.6671 ] \alpha\in[1.667066,1.6671] α ∈ [ 1.667066 , 1.6671 ]
∂ α K α ( t 0 ) ∈ [ − 9.06079735696894 ± 2.78000000000001 ] ⊂ ( − ∞ , − 6 ) . \partial_\alpha K_\alpha(t_0)\in[-9.06079735696894\pm2.78000000000001]\subset(-\infty,-6). ∂ α K α ( t 0 ) ∈ [ − 9.06079735696894 ± 2.78000000000001 ] ⊂ ( − ∞ , − 6 ) . It is evaluated by the same regularized inversion and differentiated finite-tail estimates as the contact enclosure. Canonical identification is supplied by the zero-free strip for 1.3 ≤ α ≤ 1.6671 1.3\le\alpha\le1.6671 1.3 ≤ α ≤ 1.6671 [ 1.6671 , 15 / 8 ] [1.6671,15/8] [ 1.6671 , 15/8 ] [ 15 / 8 , 2 ) [15/8,2) [ 15/8 , 2 )
Proof of the main theorem
Proof of Theorem 1.1. Lemma 2.1 reduces the assertion to a = 1 a=1 a = 1 (2) . Lemma 4.8 gives K α > 0 K_\alpha>0 K α > 0 ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) 1 < α < α ∗ 1<\alpha<\alpha_* 1 < α < α ∗ K α ∗ ≥ 0 K_{\alpha_*}\geq0 K α ∗ ≥ 0 α ∈ ( α ∗ , 2 ) \alpha\in(\alpha_*,2) α ∈ ( α ∗ , 2 ) (1) . □
Corollary 6.1. For every 1 < α < 2 1<\alpha<2 1 < α < 2 a ≠ 0 a\ne0 a = 0 T a T_a T a
Proof. This follows from Remark 3.4 and scaling. □
The definition (3) specifies the critical parameter independently of its decimal approximation. The argument uses uniqueness of the contact only in its stated rectangle. It does not assert that the kernel has no other critical points for larger values of α \alpha α
Differentiated remainders and finite inversion
This appendix specifies the remainder bounds used in the critical enclosure. A dot denotes differentiation with respect to the real parameter α \alpha α q q q
Lemma A.1 (Differentiated finite expansion). Fix c , Y > 0 c,Y>0 c , Y > 0 N ≥ 2 N\geq2 N ≥ 2 arctan ( Y / c ) > α π / 4 \arctan(Y/c)>\alpha\pi/4 arctan ( Y / c ) > α π /4 λ Y > 0 \lambda_Y>0 λ Y > 0 q = c ± i y q=c\pm iy q = c ± i y y ≥ Y y\geq Y y ≥ Y (21) satisfy
∣ E ˙ A ( q ) ∣ ≤ C A ′ ∣ q ∣ − N , ∣ E ˙ H ( q ) ∣ ≤ C H ′ ∣ q ∣ − N . |\dot{E}_A(q)| \leq C'_A |q|^{-N}, \qquad|\dot{E}_H(q)| \leq C'_H |q|^{-N}. ∣ E ˙ A ( q ) ∣ ≤ C A ′ ∣ q ∣ − N , ∣ E ˙ H ( q ) ∣ ≤ C H ′ ∣ q ∣ − N . Here primes on error constants denote bounds for differentiated remainders, not derivatives of previously chosen upper bounds. Explicit choices are
G 0 = max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) η , G 1 = max ( ∣ c ˙ N ∣ , ∣ c ˙ N − 1 ∣ ) η , G_0=\frac{\max(|c_N|,|c_{N-1}|)}{\eta}, \qquad G_1=\frac{\max(|\dot{c}_N|,|\dot{c}_{N-1}|)}{\eta}, G 0 = η max ( ∣ c N ∣ , ∣ c N − 1 ∣ ) , G 1 = η max ( ∣ c ˙ N ∣ , ∣ c ˙ N − 1 ∣ ) , D 0 = π sin δ 4 η , D 1 = 2 ( 1 + ∣ cos δ ∣ ) η , D_0=\frac{\pi\sin\delta}{4\eta}, \qquad D_1=\frac{2(1+|\cos\delta|)}{\eta}, D 0 = 4 η π sin δ , D 1 = η 2 ( 1 + ∣ cos δ ∣ ) , B 0 = G 0 ( ∣ ψ ( m ) ∣ + D 0 ) + G 1 , B 1 = G 0 ( N + 1 ) + ∣ c N − 1 ∣ 4 η + G 0 D 1 , B_0=G_0(|\psi(m)|+D_0)+G_1, \qquad B_1=G_0(N+1)+\frac{|c_{N-1}|}{4\eta}+G_0D_1, B 0 = G 0 ( ∣ ψ ( m ) ∣ + D 0 ) + G 1 , B 1 = G 0 ( N + 1 ) + 4 η ∣ c N − 1 ∣ + G 0 D 1 , where ψ = Γ ′ / Γ \psi=\Gamma'/\Gamma ψ = Γ ′ /Γ v = 0 , 1 v=0,1 v = 0 , 1
p v = m + N α + v − 1 , M v = Γ ( p v + 1 ) μ p v + 1 , W v = Γ ( p v + 2 ) μ p v + 2 + Γ ( p v ) μ p v + π 4 M v . p_v=m+N\alpha+v-1,\qquad M_v=\frac{\Gamma(p_v+1)}{\mu^{p_v+1}},\qquad W_v=\frac{\Gamma(p_v+2)}{\mu^{p_v+2}}+\frac{\Gamma(p_v)}{\mu^{p_v}}+\frac{\pi}{4}M_v. p v = m + N α + v − 1 , M v = μ p v + 1 Γ ( p v + 1 ) , W v = μ p v + 2 Γ ( p v + 2 ) + μ p v Γ ( p v ) + 4 π M v . Then C A ′ = ( B 0 M 0 + B 1 W 0 ) / Γ ( m ) C'_A=(B_0M_0+B_1W_0)/\Gamma(m) C A ′ = ( B 0 M 0 + B 1 W 0 ) /Γ ( m ) C H ′ = ( B 0 M 1 + B 1 W 1 ) / Γ ( m ) C'_H=(B_0M_1+B_1W_1)/\Gamma(m) C H ′ = ( B 0 M 1 + B 1 W 1 ) /Γ ( m ) c n c_n c n
c ˙ n = − π 2 sin δ { ( n + 1 ) cos ( ( n + 1 ) δ ) − c n cos δ } (65) \dot{c}_n=-\frac{\pi}{2\sin\delta}\{(n+1)\cos((n+1)\delta)-c_n\cos\delta\} \tag*{(65)} c ˙ n = − 2 sin δ π {( n + 1 ) cos (( n + 1 ) δ ) − c n cos δ } (65) A ˙ n = Γ ( m + n α ) Γ ( m ) { c ˙ n + c n [ ( n + 1 ) ψ ( m + n α ) − ψ ( m ) ] } (66) \dot{A}_n=\frac{\Gamma(m+n\alpha)}{\Gamma(m)}\{\dot{c}_n+c_n[(n+1)\psi(m+n\alpha)-\psi(m)]\} \tag*{(66)} A ˙ n = Γ ( m ) Γ ( m + n α ) { c ˙ n + c n [( n + 1 ) ψ ( m + n α ) − ψ ( m )]} (66) d ˙ 0 = − β 2 , d ˙ n = n A ˙ n − ∑ j = 1 n − 1 ( A ˙ j d n − j + A j d ˙ n − j ) . (67) \dot{d}_0=-\beta^2,\qquad\dot{d}_n=n\dot{A}_n-\sum_{j=1}^{n-1}(\dot{A}_j d_{n-j}+A_j\dot{d}_{n-j}). \tag*{(67)} d ˙ 0 = − β 2 , d ˙ n = n A ˙ n − j = 1 ∑ n − 1 ( A ˙ j d n − j + A j d ˙ n − j ) . (67) For N ≤ n ≤ 2 N − 2 N \le n \le2N-2 N ≤ n ≤ 2 N − 2 V n = ∑ 0 ≤ j < N , 0 ≤ n − j < N A j d n − j V_n=\sum_{0\le j<N,\ 0\le n-j<N}A_jd_{n-j} V n = ∑ 0 ≤ j < N , 0 ≤ n − j < N A j d n − j
U 0 = ∑ n = N 2 N − 2 ∣ V n ∣ Y N − n , U 1 = ∑ n = N 2 N − 2 ∣ V ˙ n ∣ Y N − n , U_0=\sum_{n=N}^{2N-2}|V_n|Y^{N-n},\qquad U_1=\sum_{n=N}^{2N-2}|\dot V_n|Y^{N-n}, U 0 = n = N ∑ 2 N − 2 ∣ V n ∣ Y N − n , U 1 = n = N ∑ 2 N − 2 ∣ V ˙ n ∣ Y N − n , P 0 = ∑ n = 0 N − 1 ∣ d n ∣ Y − n , P 1 = ∑ n = 0 N − 1 ∣ d ˙ n ∣ Y − n , P_0=\sum_{n=0}^{N-1}|d_n|Y^{-n},\qquad P_1=\sum_{n=0}^{N-1}|\dot d_n|Y^{-n}, P 0 = n = 0 ∑ N − 1 ∣ d n ∣ Y − n , P 1 = n = 0 ∑ N − 1 ∣ d ˙ n ∣ Y − n , Z 1 = ∑ n = 0 N − 1 ∣ A ˙ n ∣ Y − n + C A ′ Y − N , Z_1=\sum_{n=0}^{N-1}|\dot A_n|Y^{-n}+C'_A Y^{-N}, Z 1 = n = 0 ∑ N − 1 ∣ A ˙ n ∣ Y − n + C A ′ Y − N , C R = U 0 + β C H + C A P 0 , C R ′ = U 1 + β 2 C H + β C H ′ + C A ′ P 0 + C A P 1 . C_R=U_0+\beta C_H+C_A P_0,\qquad C'_R=U_1+\beta^2 C_H+\beta C'_H+C'_A P_0+C_A P_1. C R = U 0 + β C H + C A P 0 , C R ′ = U 1 + β 2 C H + β C H ′ + C A ′ P 0 + C A P 1 . An explicit differentiated quotient bound is
∣ F ˙ α ( q ) − ∑ n = 0 N − 1 d n q − n − 1 ∣ ≤ C F ′ ∣ q ∣ − N − 1 , C F ′ = C R ′ λ Y + C R Z 1 λ Y 2 . (68) \left|\dot F_\alpha(q)-\sum_{n=0}^{N-1}d_nq^{-n-1}\right|\le C'_F|q|^{-N-1},\qquad C'_F=\frac{C'_R}{\lambda_Y}+\frac{C_RZ_1}{\lambda_Y^2}. \tag*{(68)} F ˙ α ( q ) − n = 0 ∑ N − 1 d n q − n − 1 ≤ C F ′ ∣ q ∣ − N − 1 , C F ′ = λ Y C R ′ + λ Y 2 C R Z 1 . (68) Proof. For q = c − i y q=c-iy q = c − i y w = u e − i π / 4 w=ue^{-i\pi/4} w = u e − iπ /4 A A A H H H z = w α / q z=w^\alpha/q z = w α / q 0 < arg z < δ / 2 0<\arg z<\delta/2 0 < arg z < δ /2 z ˙ = z log w \dot z=z\log w z ˙ = z log w
∣ ∂ α D δ ( z ) D δ ( z ) ∣ ≤ D 0 + D 1 ∣ log w ∣ . \left|\frac{\partial_\alpha D_\delta(z)}{D_\delta(z)}\right|\le D_0+D_1|\log w|. D δ ( z ) ∂ α D δ ( z ) ≤ D 0 + D 1 ∣ log w ∣. Differentiate the exact remainder w α z N ( c N − c N − 1 z ) / ( Γ ( m ) D δ ( z ) ) w^\alpha z^N(c_N-c_{N-1}z)/(\Gamma(m)D_\delta(z)) w α z N ( c N − c N − 1 z ) / ( Γ ( m ) D δ ( z )) u α + N α ∣ q ∣ − N ( B 0 + B 1 ∣ log w ∣ ) / Γ ( m ) u^{\alpha+N\alpha}|q|^{-N}(B_0+B_1|\log w|)/\Gamma(m) u α + N α ∣ q ∣ − N ( B 0 + B 1 ∣ log w ∣ ) /Γ ( m ) ∣ log w ∣ ≤ ∣ log u ∣ + π / 4 ≤ u + u − 1 + π / 4 |\log w|\le|\log u|+\pi/4\le u+u^{-1}+\pi/4 ∣ log w ∣ ≤ ∣ log u ∣ + π /4 ≤ u + u − 1 + π /4 ∣ e − w ∣ ≤ e − μ u |e^{-w}|\le e^{-\mu u} ∣ e − w ∣ ≤ e − μu log q \log q log q κ α \kappa_\alpha κ α F α = β H / ( q A ) F_\alpha=\beta H/(qA) F α = β H / ( q A )
If S = ∑ n < N d n q − n S=\sum_{n<N}d_nq^{-n} S = ∑ n < N d n q − n β H − A S = − ∑ n = N 2 N − 2 V n q − n + β E H − E A S \beta H-AS=-\sum_{n=N}^{2N-2}V_nq^{-n}+\beta E_H-E_AS β H − A S = − ∑ n = N 2 N − 2 V n q − n + β E H − E A S C R ∣ q ∣ − N C_R|q|^{-N} C R ∣ q ∣ − N C R ′ ∣ q ∣ − N C'_R|q|^{-N} C R ′ ∣ q ∣ − N ∣ A ∣ ≥ λ Y |A|\ge\lambda_Y ∣ A ∣ ≥ λ Y ∣ A ˙ ∣ ≤ Z 1 |\dot A|\le Z_1 ∣ A ˙ ∣ ≤ Z 1 q A qA q A
Lemma A.2 (Filtered finite Bromwich formula). Suppose F α F_\alpha F α R q ≥ c \mathbb{R}q\ge c R q ≥ c λ Y > 0 \lambda_Y>0 λ Y > 0 0 ≤ j < p < N 0\le j<p<N 0 ≤ j < p < N h > 0 h>0 h > 0 R p R_p R p r p r_p r p b k b_k b k q y = c + i y q_y=c+iy q y = c + i y
I k ( t ; x , Y ) = ( − t ) k − 1 π ℑ Γ ( 1 − k , − t ( x + i Y ) ) , k ≥ 1 , t > 0 , I_k(t;x,Y)=\frac{(-t)^{k-1}}{\pi}\Im\Gamma(1-k,-t(x+iY)),\qquad k\ge1,\quad t>0, I k ( t ; x , Y ) = π ( − t ) k − 1 ℑΓ ( 1 − k , − t ( x + iY )) , k ≥ 1 , t > 0 , where the upper incomplete gamma function has its principal continuation. Then
K α ( j ) ( t ) = r p ( j ) ( t ) + e c t π ℜ ∫ 0 Y e i y t q y j ( F α ( q y ) − R p ( q y ) ) d y + T j ( t ) + ε j ( t ) . (69) K_\alpha^{(j)}(t)=r_p^{(j)}(t)+\frac{e^{ct}}{\pi}\Re\int_0^Y e^{iyt}q_y^j\bigl(F_\alpha(q_y)-R_p(q_y)\bigr)\,dy+T_j(t)+\varepsilon_j(t). \tag*{(69)} K α ( j ) ( t ) = r p ( j ) ( t ) + π e c t ℜ ∫ 0 Y e i y t q y j ( F α ( q y ) − R p ( q y ) ) d y + T j ( t ) + ε j ( t ) . (69) T j ( t ) = ∑ n = j N − 1 d n I n + 1 − j ( t ; c , Y ) − e − h t ∑ k = 1 p ∑ l = 0 min ( j , k − 1 ) b k ( j l ) ( − h ) j − l I k − l ( t ; c + h , Y ) . (70) T_j(t)=\sum_{n=j}^{N-1}d_nI_{n+1-j}(t;c,Y)-e^{-ht}\sum_{k=1}^{p}\sum_{l=0}^{\min(j,k-1)}b_k\binom{j}{l}(-h)^{j-l}I_{k-l}(t;c+h,Y). \tag*{(70)} T j ( t ) = n = j ∑ N − 1 d n I n + 1 − j ( t ; c , Y ) − e − h t k = 1 ∑ p l = 0 ∑ m i n ( j , k − 1 ) b k ( l j ) ( − h ) j − l I k − l ( t ; c + h , Y ) . (70) ∣ ε j ( t ) ∣ ≤ e c t C F π ( N − j ) Y N − j , r p ( j ) ( t ) = e − h t ∑ k = 1 p ∑ l = 0 min ( j , k − 1 ) ( j l ) ( − h ) j − l b k t k − 1 − l ( k − 1 − l ) ! . (71) |\varepsilon_j(t)|\le\frac{e^{ct}C_F}{\pi(N-j)Y^{N-j}},\qquad r_p^{(j)}(t)=e^{-ht}\sum_{k=1}^{p}\sum_{l=0}^{\min(j,k-1)}\binom{j}{l}(-h)^{j-l}\frac{b_kt^{k-1-l}}{(k-1-l)!}. \tag*{(71)} ∣ ε j ( t ) ∣ ≤ π ( N − j ) Y N − j e c t C F , r p ( j ) ( t ) = e − h t k = 1 ∑ p l = 0 ∑ m i n ( j , k − 1 ) ( l j ) ( − h ) j − l ( k − 1 − l )! b k t k − 1 − l . (71) The alpha derivative is given by the same formulas with F α F_\alpha F α d n d_n d n b k b_k b k r p r_p r p C F C_F C F C F ′ C'_F C F ′
Proof. The coefficients of R p R_p R p F α F_\alpha F α p p p q j ( F α − R p ) q^j(F_\alpha-R_p) q j ( F α − R p ) Y Y Y
I k = π − 1 ℜ ∫ Y ∞ e ( x + i y ) t ( x + i y ) − k d y I_k=\pi^{-1}\mathop{\Re}\int_Y^\infty e^{(x+iy)t}(x+iy)^{-k}\,dy I k = π − 1 ℜ ∫ Y ∞ e ( x + i y ) t ( x + i y ) − k d y follows by substitution into the incomplete gamma integral; for k = 1 k=1 k = 1 q j ( ∑ n < N d n q − n − 1 − R p ) q^j\left(\sum_{n<N}d_nq^{-n-1}-R_p\right) q j ( ∑ n < N d n q − n − 1 − R p ) p > j p>j p > j T j T_j T j ∣ q y ∣ ≥ y |q_y|\ge y ∣ q y ∣ ≥ y Lemma A.1 . □
For whole time intervals one may evaluate T j T_j T j t ∗ t_* t ∗ Y > h Y>h Y > h p ≥ j + 2 p\ge j+2 p ≥ j + 2
a n R = ∑ k = 1 min ( p , n + 1 ) b k ( − h ) n + 1 − k ( n k − 1 ) , Q N = ∑ k = 1 p ∣ b k ∣ ( N k − 1 ) h N + 1 − k ( 1 − h / Y ) k . a_n^R=\sum_{k=1}^{\min(p,n+1)}b_k(-h)^{n+1-k}\binom{n}{k-1},\qquad Q_N=\sum_{k=1}^{p}\frac{|b_k|\binom{N}{k-1}h^{N+1-k}}{(1-h/Y)^k}. a n R = k = 1 ∑ m i n ( p , n + 1 ) b k ( − h ) n + 1 − k ( k − 1 n ) , Q N = k = 1 ∑ p ( 1 − h / Y ) k ∣ b k ∣ ( k − 1 N ) h N + 1 − k . On [ t − , t + ] [t_-,t_+] [ t − , t + ] t − > 0 t_->0 t − > 0 r r r
r e c t + π { ∑ n = p N − 1 ∣ d n − a n R ∣ ( n − j − 1 ) Y n − j − 1 + Q N ( N − j − 1 ) Y N − j − 1 } . (72) \frac{re^{ct_+}}{\pi}\left\{\sum_{n=p}^{N-1}\frac{|d_n-a_n^R|}{(n-j-1)Y^{n-j-1}}+\frac{Q_N}{(N-j-1)Y^{N-j-1}}\right\}. \tag*{(72)} π r e c t + { n = p ∑ N − 1 ( n − j − 1 ) Y n − j − 1 ∣ d n − a n R ∣ + ( N − j − 1 ) Y N − j − 1 Q N } . (72) Indeed the binomial tail of R p R_p R p Q N ∣ q ∣ − N − 1 Q_N|q|^{-N-1} Q N ∣ q ∣ − N − 1 t t t c = h = 20 c=h=20 c = h = 20 Y = 1000 Y=1000 Y = 1000 N = 18 N=18 N = 18 p = 4 p=4 p = 4 j = 2 j=2 j = 2 j = 0 , 1 , 2 j=0,1,2 j = 0 , 1 , 2
K α ( j ) ( t ) = d j + e c t π ℜ ∫ 0 Y e i y t ( d y j F α ( q y ) − ∑ n = 0 j d n q y j − n − 1 ) d y + ∑ n = j + 1 N − 1 d n I n + 1 − j ( t ; c , Y ) + ε j ( t ) . K_\alpha^{(j)}(t)=d_j+\frac{e^{ct}}{\pi}\mathop{\Re}\int_0^Y e^{iyt}\left(d_y^jF_\alpha(q_y)-\sum_{n=0}^{j}d_nq_y^{j-n-1}\right)\,dy+\sum_{n=j+1}^{N-1}d_nI_{n+1-j}(t;c,Y)+\varepsilon_j(t). K α ( j ) ( t ) = d j + π e c t ℜ ∫ 0 Y e i y t ( d y j F α ( q y ) − n = 0 ∑ j d n q y j − n − 1 ) d y + n = j + 1 ∑ N − 1 d n I n + 1 − j ( t ; c , Y ) + ε j ( t ) . The error ε j \varepsilon_j ε j t > 0 t>0 t > 0 d j / q d_j/q d j / q d j d_j d j t ∗ t_* t ∗ [ t − , t + ] [t_-,t_+] [ t − , t + ]
r e c t + π { 2 ∣ d j + 1 ∣ t − Y + ∑ n = j + 2 N − 1 ∣ d n ∣ ( n − j − 1 ) Y n − j − 1 } . \frac{re^{ct_+}}{\pi}\left\{\frac{2|d_{j+1}|}{t_-Y}+\sum_{n=j+2}^{N-1}\frac{|d_n|}{(n-j-1)Y^{n-j-1}}\right\}. π r e c t + { t − Y 2∣ d j + 1 ∣ + n = j + 2 ∑ N − 1 ( n − j − 1 ) Y n − j − 1 ∣ d n ∣ } . Indeed the derivative of the first tail term contains the oscillatory integral of e i y t / ( c + i y ) e^{iyt}/(c+iy) e i y t / ( c + i y ) 2 / ( t Y ) 2/(tY) 2/ ( t Y ) j = 0 j=0 j = 0 F α F_\alpha F α C ˙ F \dot C_F C ˙ F
Computational certificates and reproduction
The numerical part of the proof consists of enclosures of entire closed parameter–time rectangles, together with analytic estimates on the two unbounded time ranges. No inference from a grid of point evaluations is used. The ancillary material contains the actual interval outputs, the programs that produced them, the independent readers, and the analytic error notes. The current entries are PROOF_INDEX.txt, select_positive_records.json and reduced_active_cover.json. The original complete snapshot is preserved; these lists specify the smaller subset required here.
Finite certificates
Put a − = 1.66706598 a_- = 1.66706598 a − = 1.66706598 a + = 1.667066 a_+ = 1.667066 a + = 1.667066
Table 1 (PDF p. 25).
The endpoint α = 1 \alpha= 1 α = 1 α \alpha α
The time-end estimates require 16 finite-expansion parameter boxes and 17 real special-function bounds, without numerical transform integration. They replace 11 164 of the 49 102 compact rectangles in the complete archive. All 115 rectangles crossing a new cutoff are retained and restricted to the required time range; none is discarded by its midpoint.
For the remaining negative side, the saved-data readers verify four Gaussian monotonicity segments covering [1.6671, 1.6679], thirty-three Gaussian certificates covering [1.6679, 1.7386], and 102 Rouché discs covering [1.7386, 15/8]. The local negative minimum and its fixed-time derivative bridge connect these intervals to α ∗ \alpha_* α ∗ Lemma 5.6 . Thus 244 of the 346 archived zero discs and three of the 36 archived Gaussian intervals are unnecessary. Two Gaussian radii are enlarged within their already validated derivative domains as described in Lemma 5.3 ; their strict inequalities are recomputed.
What an enclosure certifies
Most real balls are recorded as a binary midpoint and a nonnegative binary radius. Their endpoints are therefore exact rational numbers. Early [6/5, 13/10] records instead preserve outward decimal enclosures; the active reader checks their displayed lower endpoints directly. The readers check strict signs at these endpoints and verify that the input balls contain their claimed rational rectangles. They also check that the rectangles meet without gaps and that a saved finite approximation is expanded by every stated error term before the sign is tested. In the early slabs where input balls were not saved, input containment instead follows from the exact rational construction in the archived producer. The six archived display-rounding repair records concern only discarded large-time rectangles; every active saved kernel enclosure has a strictly positive lower endpoint without those repairs.
The integration programs use Arb/FLINT ball arithmetic [2 ] and explicit analytic bounds for the omitted integration ranges. For Fourier inversion, the error includes both the entire omitted frequency sum and the periodization error. Time and parameter variations are enclosed over the whole rectangle. Independent finite-sum readers recompute from the persisted transform balls, preserving their original radii and parameter midpoints. Increasing the precision of that last sum does not improve the accuracy of the earlier transform integrations. The Rouché readers recompute the strict boundary inequality from the saved function and first- and second-derivative balls, and check that the entire certified zero disc maps into the right half-plane under the principal power.
The analytic identification of these inversions with the canonical kernel, and the analytic bounds on the unbounded time intervals, are proved in the preceding sections. They are separate obligations from the finite rectangle checks. Likewise, a digest establishes which source and data were used; digest agreement alone does not establish an inequality.
Portable reproduction
The unchanged ancillary snapshot comprises 1262 files and 3450 explicit keyed source/data digest bindings. The current scalar producers and active subset readers are supplied beside it. The launcher reproduce_reduced.py recomputes the time-end scalar bounds, checks the selected saved signs and exact interval geometry, and replays the reduced negative cover. It also checks the fixed rational comparisons in Lemma 5.6 . It does not repeat original transform integrations.
The original relative-path launcher, reproduce.py, retains three verification levels for the complete archived computations:
verify checks archive integrity and the declared reference closure, without numerical computation.
saved reads the actual saved interval balls, recombines the stated budgets, verifies the finite signs and exact interval joins, and rechecks the local contact scalar inequalities. It does not repeat the original transform integrations.
finite-sums independently resums the critical Fourier certificates at the verification precision, including their analytic alias and frequency-tail bounds, using the saved integral balls.
The separate script mode runs an original producer and can perform fresh validated integration. The README gives example commands and the batch specifications needed for a full reproduction. All replay writes are isolated in a temporary copy; the snapshot is unchanged.
The recorded production environment is Python 3.12.14, python-flint 0.9.0, and FLINT 3.6.0. The software version used by an actual replay is recorded in its output. Numerical records and program sources are preserved exactly, apart from relative-path normalization of two metadata strings in each of 33 JSON files and one historical executable path. A change ledger records original and public digests and verifies that no numerical field was modified. The public package requires no access to the original machine or its directory structure.
These certificates constitute a computer-assisted proof with explicit analytic error bounds. They are not a formal proof-kernel verification, and the internal independent reviews should not be confused with external peer review. The distinction between integrity checks, saved-data checks, finite-sum recomputation, and fresh integration is retained in every replay log.
