On the location of the zero-free half-plane of a random Epstein zeta function

Abstract

In this note we study, for a random lattice L of large dimension n, the supremum of the real parts of the zeros of the Epstein zeta function $E_n(L,s)$ and prove that this random variable scaled by $n^{-1}$ has a limit distribution, which we give explicitly. This limit distribution is studied in some detail; in particular we give an explicit formula for its distribution function. Furthermore, we obtain a limit distribution for the frequency of zeros of $E_n(L,s)$ in vertical strips contained in the half-plane $\mathfrak {R}s>\frac{n}{2}$.

Appendix A. Residue calculus and numerical computation of the density

In this appendix we discuss the evaluation of the integrals in (5.20) and (6.1) using the residue theorem, resulting in alternative formulas for $\mathbf {P}\bigl (\sigma _{\{T_j\}}>c\bigr )$ and the corresponding density. These formulas turn out to be useful for numerical computation, something which we discuss briefly towards the end of the appendix (see also supplementary material 3).

We now write z in place of y. By (5.7) we have $\Phi _a(z)=z^{-\frac{1}{a}}(e^{iz}-iz\int _0^1 e^{izt}t^{-\frac{1}{a}}\,dt)$, and here the expression in the parenthesis is clearly an entire function of z. Hence

$$\begin{aligned} \Psi _a(z):=\frac{e^{-iz}z^{-1-\frac{1}{a}}}{\Phi _a(z)}=\frac{e^{-iz}z^{-1}}{e^{iz}-iz\int _0^1 e^{izt}t^{-\frac{1}{a}}\,dt} \end{aligned}$$

(A.1)

is a meromorphic function in all of $\mathbb {C}$. In (5.20) we are integrating $\mathfrak {I}\Psi _a(z)$ along the positive real line; using the symmetry $\Psi _a(-z)=-\overline{\Psi _a(\overline{z})}$, we may rewrite this as

$$\begin{aligned} \mathbf {P}\bigl (\sigma _{\{T_j\}}>c\bigr ) =\frac{1}{2}+\lim _{r\rightarrow 0^+}\frac{1}{2\pi i}\Bigl (\int _{-\infty }^{-r}\Psi _a(y)\,dy+\int _r^{\infty }\Psi _a(y)\,dy\Bigr ). \end{aligned}$$

(A.2)

Let $C'_r$ be the semicircle $\{z\, : \,|z|=r,\,\mathfrak {I}z\le 0\}$, oriented in the direction from $-r$ to r, and let $C_r$ be the contour going from $-\infty$ to $-r$ along ${\mathbb {R}}$, then from $-r$ to r along $C'_r$ and finally from r to $+\infty$ along ${\mathbb {R}}$. Since $\Psi _a(z)$ has a simple pole at $z=0$ with residue 1, we have $\int _{C'_r}\Psi _a(z)\,dz=i\pi +O(r)$ as $r\rightarrow 0$. Thus (A.2) equals $\lim _{r\rightarrow 0^+}\frac{1}{2\pi i}\int _{C_r}\Psi _a(z)\,dz$. However, by Cauchy’s integral theorem, $\int _{C_r}\Psi _a(z)\,dz$ is independent of r for all sufficiently small r. Hence

$$\begin{aligned} \mathbf {P}\bigl (\sigma _{\{T_j\}}>c\bigr )=\frac{1}{2\pi i}\int _{C_r}\Psi _a(z)\,dz, \end{aligned}$$

(A.3)

for any $r>0$ so small that $\Psi _a(z)$ has no pole in the punctured disk $\{z\, : \,0<|z|\le r\}$.

We wish to replace $C_r$ in (A.3) by a contour over z’s with large negative imaginary part. In order to do so, we first need to understand the poles of $\Psi _a(z)$ in the lower half plane. Numerics indicate that there is exactly one simple pole in the infinite vertical strip $\{z\, : \,(2n-1)\pi<\mathfrak {R}z<(2n+1)\pi ,\,\mathfrak {I}z<0\}$ for each integer n; cf. Fig. 2 below. However, for technical reasons it seems easier to prove a corresponding statement instead for certain “curved vertical strips”, as follows. For each $n\in {\mathbb {Z}}^+$, we let $\Gamma _n$ be the curve in the complex plane given by

$$\begin{aligned} x\mapsto c_n(x)=x-ix\tan \bigl ((n-\tfrac{1}{4})\pi -\tfrac{1}{2}x\bigr ), \qquad \bigl (2n-\tfrac{3}{2}\bigr )\pi <x\le \bigl (2n-\tfrac{1}{2}\bigr )\pi . \end{aligned}$$

(A.4)

One notes that $\mathfrak {I}c_n(x)\rightarrow -\infty$ as $x\rightarrow (2n-\frac{3}{2})\pi ^+$, that $\mathfrak {I}c_n((2n-\frac{1}{2})\pi )=0$ and that $0<\arg c_n'(x)<\frac{\pi }{2}$ for all $(2n-\tfrac{3}{2})\pi<x<(2n-\tfrac{1}{2})\pi$. Hence $\Gamma _n$ and $\Gamma _{n+1}$, together with the real interval $[(2n-\tfrac{1}{2})\pi ,(2n+\tfrac{3}{2})\pi ]$, bound a curved vertical strip, which we call $S_n$ (we take $S_n$ to be closed). We also let $S_{-n}=\{-\overline{z}\, : \,z\in S_n\}$ be the reflection of $S_n$ in the imaginary axis, and we let $S_0$ be the curved vertical strip bounded by the curves $\Gamma _1$, $\{-\overline{z}\, : \,z\in \Gamma _1\}$ and $[-\frac{3}{2}\pi ,\frac{3}{2}\pi ]$. Now the union of all $S_n$ ($n\in {\mathbb {Z}}$) equals the negative half plane, $\{z\, : \,\mathfrak {I}z\le 0\}$, and the $S_n$’s have pairwise disjoint interiors.

Proposition 3

Let $a>1$ be given. For each $n\in {\mathbb {Z}}$, the function $z\Psi _a(z)$ has a unique pole in the strip $S_n$. This pole is simple, and lies in the interior of $S_n$.

For the proof we need the following lemma. We will use the definition (5.4) of $\Gamma (s,z)$ for general $z\in \mathbb {C}\setminus {\mathbb {R}}_{\le 0}$, the integral being over the infinite ray $u\in z+{\mathbb {R}}_{>0}$.

Lemma 9

For any $s\in [1,2]$ and any $z=-x+iy\in \mathbb {C}$, satisfying either $\frac{1}{2}\pi \le |y|\le \frac{1}{2}x$, $\frac{3}{4}\pi \le |y|\le x$ or $[x\ge 0$ and $|y|\ge \pi ]$, we have

$$\begin{aligned} \bigl |\Gamma (-s,z)\bigr |<s^{-1}|z|^{-s}e^x. \end{aligned}$$

(A.5)

Proof

Take s and $z=-x+iy$ satisfying the assumptions. By symmetry, we may assume $y>0$. We may deform the contour of integration in (5.4) to be the ray $\{z+t(1+ki)\, : \,t\ge 0\}$, where k is any fixed non-negative number. This ray intersects the imaginary axis at $(y+kx)i$, and thus $|u|\ge (y+kx)(1+k^2)^{-1/2}$ holds for every point u on the ray, and

$$\begin{aligned} \bigl |\Gamma (-s,z)\bigr | \le \frac{(1+k^2)^{\frac{s+1}{2}}}{(y+kx)^{s+1}}\int _0^\infty e^{x-t}(1+k^2)^{\frac{1}{2}}\,dt =\frac{(1+k^2)^{1+\frac{s}{2}}}{(y+kx)^{s+1}}e^x. \end{aligned}$$

(A.6)

Applying this with $k=1$, we see that (A.5) holds whenever $s\bigl (\frac{\sqrt{2} |z|}{x+y}\bigr )^s<\frac{x+y}{2}$. But $\frac{\sqrt{2} |z|}{x+y}\ge 1$ for all non-zero z and thus the inequality holds for all $s\in [1,2]$ if and only if it holds for $s=2$, i.e., if and only if $\frac{x^2+y^2}{(x+y)^3}<\frac{1}{8}$. However, it is easily verified that $\frac{x^2+y^2}{(x+y)^3}$ is a decreasing function of $x>0$ for any fixed $y\ge 0$. Hence, if $x\ge y\ge \frac{3}{4}\pi$, then $\frac{x^2+y^2}{(x+y)^3}\le \frac{1}{4y}\le \frac{1}{4\cdot \frac{3}{4}\pi }<\frac{1}{8}$; similarly, if $x\ge 2y\ge \pi$, then $\frac{x^2+y^2}{(x+y)^3}\le \frac{5}{27\cdot \frac{1}{2}\pi }<\frac{1}{8}$, and if $y\ge \pi$ and $x\ge \frac{3}{4}y$, then $\frac{x^2+y^2}{(x+y)^3}\le \frac{100}{343y}\le \frac{100}{343\pi }<\frac{1}{8}$. To treat the remaining case, when $y\ge \pi$ and $0\le x<\frac{3}{4}y$, we apply (A.6) with $k=0$; from this we see that (A.5) holds whenever $s(|z|/y)^s<y$. However, if $y\ge \pi$ and $0\le x<\frac{3}{4}y$, then $s(|z|/y)^s\le 2(|z|/y)^2<\frac{25}{8}<\pi \le y$, and we are done. $\square$

We also record the following bound, which follows from (A.6) with $k=1$:

Lemma 10

The bound $\bigl |\Gamma (-s,z)\bigr |\ll |z|^{-s-1}e^{-\mathfrak {R}z}$ holds uniformly for all $s\in [1,2]$ and all $z\in \mathbb {C}$ with $\mathfrak {R}z\le 0$, $\mathfrak {I}z\ne 0$.

Proof of Proposition 3

Let $\eta _a(z)=z^{\frac{1}{a}}\Phi _a(z)=e^{iz}-iz\int _0^1 e^{izt}t^{-\frac{1}{a}}\,dt$ and note that $\eta _a$ is an entire function. By (A.1), our task is to prove that for each n, $\eta _a(z)$ has a unique zero in $S_n$, which is simple and lies in the interior of $S_n$. Using (5.21) and applying the recursion formula $\Gamma (s,z)=e^{-z}z^{s-1}+(s-1)\Gamma (s-1,z)$ twice, we find that for z with $\mathfrak {R}z>0$, we have

$$\begin{aligned} \eta _a(z)&=w_1+w_2+w_3 \quad \text {with }\, {\left\{ \begin{array}{ll} w_1=(-iz)^{\frac{1}{a}}\Gamma (1-a^{-1}) \\ w_2=a^{-1}(-iz)^{-1}e^{iz} \\ w_3=-\frac{a+1}{a^2}(-iz)^{\frac{1}{a}}\Gamma \bigl (-1-a^{-1},-iz\bigr ), \end{array}\right. } \end{aligned}$$

(A.7)

wherein $(-iz)^{\frac{1}{a}}=\exp (\frac{1}{a}\log (-iz))$ with the principal branch of the logarithm; $-\pi<\mathfrak {I}\log (-iz)<0$.

Let $n\in {\mathbb {Z}}^+$ and $z=x-iy\in \Gamma _n$. We wish to apply Lemma 9 with $s=1+a^{-1}$ and with $-iz$ in place of z. In order to justify this application, we have to check that either $x\ge \pi$, $y\ge x\ge \frac{3}{4}\pi$ or $y\ge 2x\ge \pi$; this is clear if $n\ge 2$, since then $x>\pi$, and if $n=1$, then the claim follows using (A.4), $\tan (\frac{1}{4}\pi )=1$ and $\tan (\frac{3}{8}\pi )>2$. The conclusion from Lemma 9 is that $|w_3|<|w_2|$ holds in (A.7). We also note that $\arg \bigl (w_1/w_2\bigr )\in (1+a^{-1})(-\tfrac{1}{2}\pi +\arg (z))-x+2\pi {\mathbb {Z}}$, and by (A.4), we have $x\in ((2n-\frac{3}{2})\pi ,(2n-\frac{1}{2})\pi ]$ and $\arg (z)=-(n-\frac{1}{4})\pi +\frac{1}{2}x\in (-\frac{1}{2}\pi ,0]$; together these imply that $\arg \bigl (-ize^{-iz}w_1\bigr )$ lies in $\bigl [-\frac{1}{2}a^{-1}\pi ,(\frac{1}{2}-a^{-1})\pi \bigr )\subset (-\frac{1}{2}\pi ,\frac{1}{2}\pi )$, i.e., that $\mathfrak {R}(w_1/w_2)>0$. Moreover, $|w_3|<|w_2|$ forces $\mathfrak {R}((w_2+w_3)/w_2)>0$; hence we conclude that $\mathfrak {R}((w_1+w_2+w_3)/w_2)>0$, i.e., that

$$\begin{aligned} \mathfrak {R}(-ize^{-iz}\eta _a(z))>0\quad \text {for all }\, z\in \Gamma _n. \end{aligned}$$

(A.8)

This shows that $\eta _a(z)$ has no zeros along $\Gamma _n$, and also gives a precise control on the variation of $\arg \eta _a(z)$ along $\Gamma _n$.

Next, from (A.7) and Lemma 10, we see that for $z=x-iy$ with y large and $x>0$ bounded, we have $\eta _a(z)=w_1+w_2+w_3=w_2(1+O(y^{-1}))$, and thus $\arg \eta _a(z)\in \pi +x+O(y^{-1})+2\pi {\mathbb {Z}}$. Also note that $\mathfrak {R}\eta _a(z)>0$ for all $z\ge 0$, since $\mathfrak {R}\Phi _a(z)>0$ for all $z>0$ (as noted previously) and $\eta _a(0)=1$. Using these facts together with (A.8) (applied both for n and $n+1$), we conclude that for any $n\in {\mathbb {Z}}^+$ and any sufficiently large $Y>0$ (depending on both a and n), $\arg \eta _a(z)$ increases by $2\pi$ as z travels around the boundary of $S_n\cap \{\mathfrak {I}z\ge -Y\}$ in the positive direction. Hence, by the argument principle, $\eta _a(z)$ has a unique simple zero in the interior of $S_n$. Using the symmetry $\eta _a(-\overline{z})=\overline{\eta _a(z)}$, one proves the same fact also for $S_0$ and any $S_n$, $n<0$. This completes the proof of the proposition. $\square$

From now on, we write $\zeta _n=\zeta _n(a)$ for the unique pole of $z\Psi _a(z)$ in $S_n$ ($n\in {\mathbb {Z}}$). By symmetry we have $\zeta _{-n}=-\overline{\zeta _n}$ for all n, and in particular $\zeta _0$ lies on the negative imaginary axis. Figure 2 below shows the curves traced by $\zeta _0,\ldots ,\zeta _4$ as a varies.

The next lemma gives an asymptotic formula for $\zeta _n$ ($n>0$) with an error which is small whenever at least one of n, a and $(a-1)^{-1}$ is large.

Lemma 11

We have, uniformly over all $a>1$ and all $n\in {\mathbb {Z}}^+$,

$$\begin{aligned} \zeta _n=(2n-a^{-1})\pi +(1+a^{-1})\arctan \Bigl (\frac{2\pi n}{Y_n}\Bigr ) -iY_n+O\biggl (\frac{1}{n+\log \bigl |\Gamma (-a^{-1})\bigr |}\biggr ), \end{aligned}$$

(A.9)

where $Y_n$ equals the unique root $y>0$ of the equation

$$\begin{aligned} y-{\textstyle \frac{1}{2}}(1+a^{-1})\log \bigl ((2\pi n)^2+y^2\bigr )=\log \bigl |\Gamma (-a^{-1})\bigr |. \end{aligned}$$

(A.10)

(Regarding the error term in (A.9), we remark that $|\Gamma (-a^{-1})|>3$, and thus that $\log |\Gamma (-a^{-1})|>1$, for all $a>1$.)

Proof

Using (A.7) and Lemma 10, together with the fact that $\eta _a(\zeta _n)=0$, we get

$$\begin{aligned} \Gamma (-a^{-1})=(-i\zeta _n)^{-1-\frac{1}{a}}e^{i\zeta _n}(1+O(|\zeta _n|^{-1})), \qquad \forall a>1,\, n\in {\mathbb {Z}}^+. \end{aligned}$$

(A.11)

Writing $\zeta _n=x_n-iy_n$ ($x_n,y_n>0$) and taking absolute values in (A.11), we get

$$\begin{aligned} \bigl |\Gamma (-a^{-1})\bigr |=|\zeta _n|^{-1-\frac{1}{a}}e^{y_n}(1+O(|\zeta _n|^{-1})). \end{aligned}$$

(A.12)

Now, using the facts that $|\zeta _n|\ge x_n>(2n-\frac{3}{2})\pi \gg n$ and $|\Gamma (-a^{-1})|\rightarrow \infty$ as $a\rightarrow 1^+$ or $a\rightarrow \infty$, we conclude that $y_n$ must be large whenever at least one of n, a and $(a-1)^{-1}$ is large; and due to the form of the error term in (A.9), we may without loss of generality restrict to the case when this holds. Note that also $|\zeta _n|$ must be large, since $|\zeta _n|\ge y_n$.

In more precise terms, we have, considering the logarithm of equation (A.12),

$$\begin{aligned} y_n=\tfrac{1}{2}(1+a^{-1})\log (x_n^2+y_n^2)+\log \bigl |\Gamma (-a^{-1})\bigr |+O(|\zeta _n|^{-1}). \end{aligned}$$

(A.13)

In particular, using $(2n-\frac{3}{2})\pi<x_n<(2n+\frac{3}{2})\pi$ and also $\log (x_n^2+y_n^2)\le \frac{1}{2}y_n+2\log n$ (which holds since $y_n$ is large), we conclude that

$$\begin{aligned} y_n\asymp \log n+\log |\Gamma (-a^{-1})|; \quad \text {and thus }\, |\zeta _n|\asymp x_n+y_n\asymp n+\log \bigl |\Gamma (-a^{-1})\bigr |. \end{aligned}$$

(A.14)

(Note: “$\asymp$” means “both $\ll$ and $\gg$”.) Now $x_n^2+y_n^2=((2\pi n)^2+y_n^2)(1+O(|\zeta _n|^{-1}))$, and thus, in (A.13), we may replace “$\log (x_n^2+y_n^2)$” by “$\log ((2\pi n)^2+y_n^2)$”; the error from this operation is subsumed in the error term $O(|\zeta _n|^{-1})$. We also note that the expression in the left-hand side of (A.10) is an increasing function of $y>0$, which is negative for small y and the derivative of which lies in the interval $(1-(2\pi )^{-1},1]$, for all $y>0$. It follows that $Y_n$ (in the statement of the lemma) is well-defined, and also that

$$\begin{aligned} y_n=Y_n+O\bigl (|\zeta _n|^{-1}\bigr ). \end{aligned}$$

(A.15)

Next, taking the argument of both sides of (A.11), we get

$$\begin{aligned} x_n=(1-a^{-1})\tfrac{\pi }{2}+(1+a^{-1})\arg (\zeta _n)+2k\pi +O(|\zeta _n|^{-1}) \quad \text {for some }\,k\in {\mathbb {Z}}, \end{aligned}$$

(A.16)

where $-\frac{\pi }{2}<\arg (\zeta _n)<0$. Clearly $(2k-1)\pi -O(|\zeta _n|^{-1})<x_n<(2k+\frac{1}{2})\pi +O(|\zeta _n|^{-1})$, and in fact, since $-\arg (\zeta _n)\gg y_n|\zeta _n|^{-1}$ and $y_n$ is large, we even have $x_n<(2k+\frac{1}{2})\pi$. But also $x_n>(2n-\frac{3}{2})\pi$; hence $k\ge n$. On the other hand, since $\zeta _n$ lies to the left of the curve $\Gamma _{n+1}$, we have $x_n<2\arg (\zeta _n)+(2n+\frac{3}{2})\pi$, and using this fact in (A.16), we get $(1-a^{-1})\arg (\zeta _n)>(2(k-n)-1-\frac{1}{2}a^{-1})\pi -O(|\zeta _n|^{-1})$. This forces $k\le n$, since $\arg (\zeta _n)<0$ and $|\zeta _n|$ is large. Hence we have proved that $k=n$. Finally, using (A.15) and $(2n-\frac{3}{2})\pi<x_n<(2n+\frac{3}{2})\pi$, we get $\bigl |\arg (\zeta _n)+\arctan (\frac{Y_n}{2\pi n})\bigr |\ll Y_n|\zeta _n|^{-2}\ll |\zeta _n|^{-1}$. Now (A.9) follows from (A.15), (A.16) and (A.14). $\square$

We may also remark that $Y_n$, as defined in Lemma 11, satisfies

$$\begin{aligned} Y_n=G+\frac{1+a^{-1}}{2}\biggl (1+\frac{(1+a^{-1})G}{(2\pi n)^2+G^2}\biggr )\log \bigl ((2\pi n)^2+G^2\bigr ) +O\Bigl (\frac{1}{n+G}\Bigr ), \end{aligned}$$

(A.17)

with $G=\log |\Gamma (-a^{-1})|$. This is proved by direct substitution in (A.10), using the properties of the left-hand side in (A.10) noted in the proof of Lemma 11.

We will now change the contour in (A.3). Let $a>1$ be given, and fix $r>0$ sufficiently small so that (A.3) holds. For $n\in {\mathbb {Z}}^+$ and $Y>0$, we let $z_{n,Y}$ be the unique point where $\Gamma _n$ intersects $\{\mathfrak {I}z=-Y\}$, and let $C_{n,Y}$ be the contour going from $-\infty$ to $-(2n-\frac{1}{2})\pi$ along ${\mathbb {R}}$, then along $\Gamma _{-n}:=\{-\overline{z}\, : \,z\in \Gamma _n\}$ to $-\overline{z_{n,Y}}$, then along $\{\mathfrak {I}z=-Y\}$ to $z_{n,Y}$, further along $\Gamma _n$ to $(2n-\frac{1}{2})\pi$, and finally along ${\mathbb {R}}$ to $+\infty$. By the residue theorem and Proposition 3, for every $n\in {\mathbb {Z}}^+$ there is some $Y_0=Y_0(a,n)>0$ such that for $Y>Y_0$, we have

$$\begin{aligned} \frac{1}{2\pi i}\int _{C_r}\Psi _a(z)\,dz =\frac{1}{2\pi i}\int _{C_{n,Y}}\Psi _a(z)\,dz-\sum _{m=1-n}^{n-1}{\text {Res}}_{z=\zeta _m}\Psi _a(z). \end{aligned}$$

(A.18)

Now let $w_1,w_2,w_3$ be as in (A.7). By Lemma 10 there is an $N=N(a)\in {\mathbb {Z}}^+$ such that $|w_3|\le \frac{1}{2}|w_2|$ for all $z\in \Gamma _n$, $n\ge N$. Using also $\mathfrak {R}(w_1/w_2)>0$ for all $z\in \Gamma _n$, we get $|\eta _a(z)|=|w_1+w_2+w_3|\ge \frac{\sqrt{3}}{2}|w_1|$ and thus $|\Psi _a(z)|\ll n^{-1-\frac{1}{a}}e^{\mathfrak {I}z}$ for all $n\ge N$ and $z\in \Gamma _n$. Also, for any fixed a and n, we have $|\eta _a(z)|\gg Y^{-1}e^Y$ for all $z\in C_{n,Y}\cap \{\mathfrak {I}z=-Y\}$ (cf. Lemma 10 and (A.7)); thus $|\Psi _a(z)|\ll e^{-2Y}$ for these z. The above bounds imply $\lim _{n\rightarrow \infty }\bigl (\lim _{Y\rightarrow \infty }\int _{C_{n,Y}}|\Psi _a(z)|\,|dz|\bigr )=0$, and so

$$\begin{aligned} \mathbf {P}\bigl (\sigma _{\{T_j\}}>c\bigr ) =\frac{1}{2\pi i}\int _{C_r}\Psi _a(z)\,dz=-\lim _{n\rightarrow \infty }\sum _{m=-n}^{n}{\text {Res}}_{z=\zeta _m}\Psi _a(z) =\sum _{n=-\infty }^{\infty } ae^{-2i\zeta _n}. \end{aligned}$$

(A.19)

Here the last equality follows from an easy calculation using (5.9) and (A.1), noticing that the sum is absolutely convergent, since, by Lemma 11 and (A.17), we have

$$\begin{aligned} \bigl |ae^{-2i\zeta _n}\bigr |\ll ae^{-2G}(|n|+G)^{-2(1+\frac{1}{a})},\qquad \forall a>1,\, n\in {\mathbb {Z}}\setminus \{0\}. \end{aligned}$$

(A.20)

One also checks that the formula (A.19) may be differentiated termwise with respect to a, yielding

$$\begin{aligned} f(c)=2\sum _{n=-\infty }^\infty e^{-2i\zeta _n}\Bigl (2ai\Bigl (\frac{d}{da}\zeta _n\Bigr )-1\Bigr ) \end{aligned}$$

(A.21)

for the density function (cf. (6.1)).

For c not too large, the formula (A.21) can be used to compute f(c) numerically to a decent precision. We have implemented this in (supplementary material 3). Our experiments indicate that for any given $a>1$ ($a=2c$) and $n\in {\mathbb {Z}}^+$, the asymptotic formula in Lemma 11 is sufficiently accurate so that it can be used as the initial value in the Newton iteration algorithm solving for $\Phi _a(z)=0$, with rapid convergence. Also, $\frac{d}{da}\zeta _n$ is computed using

$$\begin{aligned} \frac{d}{da}\zeta _n&=a\zeta _n^{1+\frac{1}{a}}e^{-i\zeta _n}\Bigl (\frac{\partial }{\partial a}\Phi _a(z)\Bigr )_{|z=\zeta _n}\\&=-a^{-2}\zeta _n^{1+\frac{1}{a}}e^{-i(\zeta _n+\frac{\pi }{2a})}\biggl (\Gamma '(-a^{-1})-\int _{-i\zeta _n}^\infty e^{-u}u^{-1-\frac{1}{a}}(\log u)\,du\biggr ), \end{aligned}$$

which most often can be evaluated very quickly via repeated integration by parts; in the remaining cases we use numerical integration.

The data for the graph in Fig. 1 can be found in (supplementary material 4); it was assembled by computing f(c) ($c=\frac{1}{2}a$) for $a=1+\frac{1}{100}k$, $k=1,2,\ldots ,400$. For each a-value we truncated the sum in (A.21) at $|n|\le 400$ (using also the obvious $n\leftrightarrow -n$ symmetry). It turns out that the terms in (A.21) decay roughly as $n^{-2(1+\frac{1}{a})}$ as $n\rightarrow \infty$ (cf. (A.20)). In particular we have slower convergence for larger a and this is seen in the computations: Our numerics indicate that we obtain the first few f(c)-values to within an absolute error $\lesssim 10^{-11}$, whereas for a near 5 (where $f(c)\approx 0.05$) the error is $\lesssim 10^{-6}$. Of course the precision can be improved by including more terms in (A.21), again cf. (supplementary material 3).

Fig. 2

The curves traced by the poles $\zeta _1,\zeta _2,\zeta _3,\zeta _4$ (and $\zeta _0$) as a varies