Fluctuations of the log-gamma polymer free energy with general parameters and slopes

Abstract

We prove that the free energy of the log-gamma polymer between lattice points (1, 1) and (M, N) converges to the GUE Tracy–Widom distribution in the $M^{1/3}$ scaling, provided that N/M remains bounded away from zero and infinity. We prove this result for the model with inverse gamma weights of any shape parameter $\theta >0$ and furthermore establish a moderate deviation estimate for the upper tail of the free energy in this case. Finally, we consider a non i.i.d. setting where the weights on finitely many rows and columns have different parameters, and we show that when these parameters are critically scaled the limiting free energy fluctuations are governed by a generalization of the Baik–Ben Arous–Péché distribution from spiked random matrices with two sets of parameters.

Appendices

Appendix A

In Sects. 2 and 3 we needed several results, whose proofs require a careful analysis of the properties of the function $G_{M,N}$, recalled in this section in equation (5.3). In this section we study some of these properties, focusing on the way the real part of $G_{M,N}$ varies along different contours.

1.1 Definition and basic properties

Let $\Psi (x)$ denote the digamma function, i.e.

$$\begin{aligned} \Psi (x) = \frac{\Gamma '(x)}{\Gamma (x)} = - \gamma _{E} + \sum _{n = 0}^\infty \left( \frac{1}{n + 1} - \frac{1}{n+z} \right) , \end{aligned}$$

(5.1)

where $\gamma _{E}$ is the Euler constant. Suppose that $M, N \ge 1$ and $\theta > 0$ are given. We let $z_{c}(M,N,\theta )$ denote the maximizer of

$$\begin{aligned} W_{M,N}(x) := N \Psi (x) + M \Psi (\theta - x) \end{aligned}$$

on the interval $(0, \theta )$. As mentioned in Sect. 3.1 the above expression converges to $- \infty$ as $x \rightarrow 0+$ or $x \rightarrow \theta -$ and also the function $W_{M,N}(x)$ is strictly concave, hence the maximum exists and is unique. Since $z_c$ is the maximizer we have

$$\begin{aligned} 0 = W'_{M,N}(z_c) = N \sum _{n = 0}^{\infty } \frac{1}{(n+z_c)^2} - M \sum _{n = 0}^\infty \frac{1}{(n+\theta - z_c)^2}. \end{aligned}$$

(5.2)

The main object of interest in this section is the following function

$$\begin{aligned} G_{M,N}(z) = N \log \Gamma (z) - M \log \Gamma (\theta - z) - W_{M,N}(z_{c}) z - C_{M,N}, \end{aligned}$$

(5.3)

where the constant $C_{M,N}$ is such that $G_{M,N}(z_{c}) = 0$. We also define $\alpha = N/M$ and $G_\alpha (z)$ through the equation

$$\begin{aligned} G_\alpha (z) = M^{-1} \cdot G_{M,N}(z). \end{aligned}$$

In the remainder of this section we derive a few basic properties for the function $G_{M,N}$ or equivalently $G_\alpha$. From (5.2) we know that $z_c(\alpha )$ is the unique number in $(0, \theta )$ that satisfies

$$\begin{aligned} 0 = \alpha \sum _{n = 0}^\infty \frac{1}{(n+z_c)^2} - \sum _{n =0}^\infty \frac{1}{(n+\theta - z_c)^2}. \end{aligned}$$

In particular, for $z \in (0, \theta )$ the function

$$\begin{aligned} g(z) = \frac{\sum _{n =0}^\infty \frac{1}{(n+\theta - z)^2}}{ \sum _{n = 0}^\infty \frac{1}{(n+z)^2}} \end{aligned}$$

satisfies $g(z_c) = \alpha$. The numerator of g(z) is clearly increasing, while the denominator is decreasing, so that function g(z) is a continuous strictly increasing bijection between $(0, \theta )$ and $(0, \infty )$. Consequently, $g^{-1}$ is also a strictly increasing continuous bijection between $(0,\infty )$ and $(0,\theta )$. We conclude that $g^{-1}([\delta , 1]) = [a,b]$ where $a = g^{-1}(\delta )$ and $b = g^{-1}(1)$.

One directly computes $G_\alpha (z_c) = \partial _zG_\alpha (z_c) = \partial _z^2 G_\alpha (z_c) = 0$, while $\partial ^3_z G_\alpha (z_c) = 2 \sigma _\alpha ^3$ with

$$\begin{aligned} \sigma _\alpha := \left( \sum _{n = 0}^\infty \frac{\alpha }{(n+z_c)^3} + \sum _{n = 0}^\infty \frac{1}{(n+\theta - z_c)^3} \right) ^{1/3}. \end{aligned}$$

From the discussion above we see that

$$\begin{aligned} \left( \sum _{n = 0}^\infty \frac{\delta }{(n+b)^3} + \sum _{n = 0}^\infty \frac{1}{(n+\theta - a)^3} \right) ^{1/3} \le \sigma _{\alpha } \le \left( \sum _{n = 0}^\infty \frac{1}{(n+a)^3} + \sum _{n = 0}^\infty \frac{1}{(n+\theta - b)^3} \right) ^{1/3} \end{aligned}$$

(5.4)

We end this section with a formula for the derivative of $G_{M,N}$ along a generic contour. Denote $z(r) = z_c + x(r) + {\mathsf {i}}y(r)$. Then we have

$$\begin{aligned} \begin{aligned}&\frac{d}{dr} G_{M,N}(z(r)) = (x'(r) + {\mathsf {i}}y'(r)) \cdot \left( N \Psi (z(r)) + M \Psi (\theta - z(r)) - W_{M,N}(z_{c}) \right) \\&\quad = (x'(r) + {\mathsf {i}}y'(r))\cdot \left( N \sum _{n = 0 }^\infty \left( \frac{1}{n + z_{c}} - \frac{1}{n+z(r)} \right) + M \sum _{n = 0 }^\infty \left( \frac{1}{n + \theta - z_{c} }- \frac{1}{n+\theta - z(r)} \right) \right) \\&\quad = (x'(r) + {\mathsf {i}}y'(r)) N \sum _{n = 0 }^\infty \frac{|n + z(r)|^2 - (n+z_{c})(n + {\overline{z}}(r))}{(n + z_{c})|n+z(r)|^2} \\&\qquad + (x'(r) + {\mathsf {i}}y'(r)) M \sum _{n = 0 }^\infty \frac{|n+\theta - z(r)|^2 - (n+ \theta - z_{c})(n+\theta - {\overline{z}}(r))}{(n + \theta - z_{c})|n+\theta - z(r)|^2 }. \end{aligned}\nonumber \\ \end{aligned}$$

Taking the real part of the above equation we obtain

$$\begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] = N \sum _{n = 0 }^\infty \frac{ x'(r)(x^2(r) + y^2(r))}{(n + z_{c})|n+z(r)|^2} + M\sum _{n = 0}^{\infty } \frac{ x'(r)(x^2(r) + y^2(r))}{(n + \theta - z_{c})|n+\theta - z(r)|^2} \nonumber \\&\quad + \left( N \sum _{n = 0}^\infty \frac{x(r)x'(r) - y(r)y'(r)}{|n+z(r)|^2} - M \sum _{n = 0}^\infty \frac{x(r)x'(r) - y(r)y'(r)}{|n + \theta - z(r)|^2} \right) . \end{aligned}$$

(5.5)

1.2 Estimates along $C_{z_c, \phi }$

In this section we obtain estimates on $Re \left[ G_{M,N}(z)\right]$ when $\mathsf {arg}(z)$ is bounded away from $0, \pi /2, \pi , 3\pi /2$. The main result we prove is the following.

Lemma 5.1

Suppose that $M, N \ge 1$, $\theta > 0$ and $\delta _1 \in (0, \pi /4)$. Then we can find constants $C,R > 0$ that depend on M, N, $\theta$ and $\delta _1$ such that if $z = re^{{\mathsf {i}}\phi }$ with $r \ge R$ we have

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(re^{{\mathsf {i}}\phi }) \right] \le - C r \log (r) \text{ if } \phi \in [\pi /2 + \delta _1, \pi -\delta _1] \cup [\pi + \delta _1, 3\pi /2 - \delta _1]; \\&\quad {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}( re^{{\mathsf {i}}\phi }) \right] \ge C r \log (r) \text{ if } \phi \in [\delta _1, \pi /2 - \delta _1] \cup [-\pi /2 + \delta _1, -\delta _1]. \end{aligned}\nonumber \\ \end{aligned}$$

(5.6)

Proof

Our approach is to study ${\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}( z_{c} + re^{{\mathsf {i}}\phi }) \right]$. Let us fix $\theta > 0$ and $\delta _1 > 0$ as in the statement of the lemma. Setting $z(r) = z_c + re^{{\mathsf {i}}\phi }$ and applying (5.5) we get

$$\begin{aligned} \begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] =&N \sum _{n = 0 }^\infty \frac{r^2 \cos (\phi ) }{(n + z_c)|n+z(r)|^2} - N \sum _{n = 0 }^\infty \frac{r (\sin ^2(\phi ) - \cos ^2(\phi ))}{|n+z(r)|^2} \\&+M\sum _{n = 0 }^\infty \frac{r^2 \cos (\phi ) }{(n + \theta - z_c)|n+\theta - z(r)|^2 } + M \sum _{n = 0 }^\infty \frac{r( \sin ^2(\phi ) - \cos ^2(\phi ))}{|n + \theta - z(r)|^2}. \end{aligned}\nonumber \\ \end{aligned}$$

(5.7)

We study the large positive r behavior of the right side above for different ranges of $\phi$.

Suppose first that $\phi \in [\pi /2 + \delta _1/2, \pi -\delta _1/2] \cup [\pi + \delta _1/2, 3\pi /2 - \delta _1/2]$. Below C will stand for a positive constant, which depends on $\theta$ and $\delta _1$, and whose value may change from line to line.

$$\begin{aligned} \begin{aligned}&\sum _{n = 0 }^\infty \frac{r}{|n+ z(r)|^2} = \sum _{n = 0 }^\infty \frac{ r }{(n+ z_c + r\cos (\phi ))^2 + r^2 \sin ^2(\phi )} \\&\quad = \sum _{n = 0 }^\infty \frac{r }{(n+ z_c)^2 +2r(n+ z_c) \cos (\phi ) + r^2 }\\&\quad \le \sum _{n = 0 }^\infty \frac{r }{(n+ z_c)^2 + 2r(n+ z_c) \cos (\pi + \delta _1/2) + r^2 } \\&\quad \le \sum _{n = 0 }^\infty \frac{ C r }{(n+z_c)^2 + r^2 } \le \frac{ C r}{ z_c^2 + r^2} + \int _0^{\infty } \frac{ C rdx}{x^2 + r^2} = \frac{ C r}{ z_c^2 + r^2} + \frac{ C\pi }{2}. \end{aligned} \end{aligned}$$

(5.8)

Analogously, one shows that

$$\begin{aligned} \begin{aligned}&\sum _{n = 0 }^\infty \frac{r}{|n+\theta - z(r)|^2} \le \frac{C r}{(\theta - z_c)^2 + r^2} + \frac{C\pi }{2}. \end{aligned} \end{aligned}$$

(5.9)

Furthermore we have

$$\begin{aligned} \begin{aligned}&\sum _{n = 0 }^\infty \frac{r^2 \cos (\phi ) }{(n + z_c)|n+z(r)|^2} \le \sum _{n = 0 }^\infty \frac{ \cos (\pi /2 + \delta _1/2) r^2 }{(n +z_c) ((n+z_c)^2 + 2(n+z_c)r\cos (\phi ) + r^2)}\\&\quad \le \sum _{n = 0 }^\infty \frac{ \cos (\pi /2 + \delta _1/2) r^2 }{(n + z_c) ((n+z_c)^2 + r^2)} \le -C r^2\int _{z_c}^{\infty } \frac{dx}{(x + z_{c})(x^2 + r^2)} \\&\quad = \frac{ -Cr^2}{r^2 + z_c^2} \left( \frac{z_c }{r} \left( \frac{\pi }{2} - \tan ^{-1}(z_c/r) \right) + \log \left( \frac{\sqrt{z_c^2 + r^2}}{2z_c}\right) \right) \\&\quad \le \frac{ -Cr^2}{r^2 + z_c^2}\log \left( \frac{\sqrt{z_c^2 + r^2}}{2z_c}\right) \le -C\log (r), \end{aligned} \end{aligned}$$

(5.10)

where the last inequality holds for all $r \ge r_0$, where $r_0$ depends on $\theta$ alone. Analogous considerations show that

$$\begin{aligned} \begin{aligned}&\sum _{n = 0 }^\infty \frac{r^2 \cos (\phi ) }{(n + \theta - z_{c})|n+\theta - z(r)|^2 }\le -C \log (r), \end{aligned} \end{aligned}$$

(5.11)

where the last inequality holds for all $r \ge r_0$, where $r_0$ depends on $\theta$ alone. Combining (5.8), (5.9), (5.10) and (5.11) with (5.7) we conclude that there exist $r_0$ and $c_0$ depending on $\theta$ and $\delta _1$ such if $\phi \in [\pi /2 + \delta _1/2, \pi -\delta _1/2] \cup [\pi + \delta _1/2, 3\pi /2 - \delta _1/2]$ and $r \ge r_0$ we have

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z_c + re^{{\mathsf {i}}\phi }) \right] \le - c_0 (M+N)\log (r). \end{aligned}$$

(5.12)

The first equation in (5.6) is now an immediate consequence of (5.12) since $G_{M,N}(z_c) = 0$.

A similar argument for $\phi \in [ \delta _1/2, \pi /2 - \delta _1/2] \cup [-\pi /2 + \delta _1/2, - \delta _1/2]$ reveals that there exists $r_0$ and $c_0$ depending on $\theta$ and $\delta _1$ such that if $r \ge r_0$ then

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z_c + re^{{\mathsf {i}}\phi }) \right] \ge c_0 (M+N)\log (r). \end{aligned}$$

(5.13)

The second equation in (5.6) is now an immediate consequence of (5.13) since $G_{M,N}(z_c) = 0$. $\square$

1.3 Estimates along vertical contours

In this section we study the properties of ${\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r))\right]$ along contours of the form $C_{z_c, \pi /2}$, i.e. when we move vertically up or down from the point $z_c$. The main result we prove is as follows.

Lemma 5.2

Suppose that $M \ge N \ge 1$. If we set $z(r) = z_c + {\mathsf {i}}r$ we claim

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z_c + {\mathsf {i}}r) \right] \ge 0 \text{ when } r \ge 0 \text{ and } {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z_c + {\mathsf {i}}r) \right] \le 0 \text{ when } r \le 0. \end{aligned}\nonumber \\ \end{aligned}$$

(5.14)

Proof

Notice that since $M\ge N$ we have $\theta \ge 2z_c$. In view of (5.5) we can reduce (5.14) to establishing

$$\begin{aligned} N \sum _{n = 0}^\infty \frac{1}{|n+z(r)|^2} - M \sum _{n = 0}^\infty \frac{1}{|n + \theta - z(r)|^2} \le 0 \end{aligned}$$

(5.15)

for $r \ge 0$. Recall from (5.2) that

$$\begin{aligned} N \sum _{n = 0}^\infty \frac{1}{(n+z_c)^2} = M \sum _{n = 0}^\infty \frac{1}{(n+\theta -z_c)^2}, \end{aligned}$$

and so it suffices to show that

$$\begin{aligned} \sum _{m \ge 0} \sum _{n \ge 0} \frac{1}{(m+\theta - z_c)^2} \frac{1}{|n+z(r)|^2} - \sum _{m \ge 0} \sum _{n \ge 0}\frac{1}{(m+z_c)^2} \frac{1}{|n + \theta - z(r)|^2} \le 0.\nonumber \\ \end{aligned}$$

(5.16)

If $m = n$ the above reads

$$\begin{aligned}&\frac{1}{(n + \theta - z_c)^2} \frac{1}{(n+z_c )^2 +r^2} - \frac{1}{(n+z_c)^2} \frac{1}{(n+\theta - z_c)^2 + r^2} = \\&\quad \frac{-r^2(\theta -2z_c)(2n + \theta )}{(n+\theta -z_c)^2((n+z_c)^2 + r^2)(n+z_c)^2((n+\theta - z_c)^2 + r^2)}, \end{aligned}$$

which is non-positive as $\theta \ge 2z_c$.

We split the remaining sum as follows

$$\begin{aligned}&\sum _{m,n \ge 0, m\ne n} \frac{1}{(m+\theta - z_c)^2} \frac{1}{|n+z(r)|^2} - \sum _{m,n \ge 0, m\ne n}\frac{1}{(m+z_c)^2} \frac{1}{|n + \theta - z(r)|^2}\\&\quad = \sum _{n = 0}^\infty \sum _{m = n+ 1}^\infty \frac{1}{(m+\theta -z_c)^2} \frac{1}{(n+z_c)^2 + r^2} + \frac{1}{(n+\theta -z_c)^2} \frac{1}{(m+z_c)^2 + r^2} \\&\qquad - \frac{1}{(m+z_c)^2}\frac{1}{(n + \theta - z_c)^2 + r^2} - \frac{1}{(n+z_c)^2}\frac{1}{(m + \theta - z_c)^2 + r^2}\\&\quad = \sum _{n = 0}^\infty \sum _{m = n+ 1}^\infty \frac{-r^2(n+m+\theta )(m-n +\theta -2z_c)}{(m+\theta -z_c)^2(n+z_c)^2((n+z_c)^2+r^2)((m+\theta -z_c)^2 + r^2)} + \\&\quad \frac{r^2(n+m+\theta )(m -n + 2z_c - \theta )}{(m+z_c)^2(n+\theta -z_c)^2((m+z_c)^2 + r^2)((n+\theta -z_c)^2 + r^2)}. \end{aligned}$$

We want to show that each summand above is negative. We see that the first term is always negative and so if we increase its denominator we will make the summand larger. Notice that

$$\begin{aligned} \begin{aligned}&(m+\theta -z_c)^2(n+z_c)^2((n+z_c)^2+r^2)((m+\theta -z_c)^2 + r^2) \\&\quad - (m+z_c)^2(n+\theta -z_c)^2((m+z_c)^2 + r^2)((n+\theta -z_c)^2 + r^2) \\&\quad = -(\theta - 2z_c)(m-n) (a r^4 + b r^2 +c), \end{aligned} \end{aligned}$$

(5.17)

where

$$\begin{aligned} a= & {} 2mn + \theta (m+n) + 2\theta z_c - 2z_c^2> 0, b= (2n +\theta )(m-n)^3 \\&+ 4(n+z_c)(n+\theta -z_c) (m-n)^2 + \\&(2n + \theta )(2n^2 + 2n \theta + \theta ^2 - 2\theta z_c + 2z_c^2)(m-n) \\&+ 2(n+z_c)(n + \theta - z_c)(n^2 + n\theta +\theta ^2 - 3\theta z_c + 2z_c^2)> 0, \\ c= & {} (2mn + m\theta + n \theta + 2\theta z_c - 2z_c^2)( (2n^2 + 2n\theta + \theta ^2 - 2\theta z_c + 2z_c^2) (m-n)^2 + \\&+ 2(n + z_c)(2n + \theta )(n+\theta - z_c) + 2 (n+z_c)^2 (n + \theta -z_c)^2 > 0. \end{aligned}$$

We conclude that the second denominator is larger than the first and so by replacing it we make the summands larger. But then the summand becomes

$$\begin{aligned} \frac{r^2(n+m+\theta )( 4z_c - 2\theta )}{(m+z_c)^2(n+\theta -z_c)^2((m+z_c)^2 + r^2)((n+\theta -z_c)^2 + r^2)}, \end{aligned}$$

which is clearly non-positive. We conclude that the sum is negative (unless $M = N$ in which case it is 0) and this proves (5.16). $\square$

As an immediate corollary we have the following result.

Lemma 5.3

Suppose that $M \ge N \ge 1$ and $\theta > 0$. Then for any $x > 0$ and $y \in {\mathbb {R}}$ with $|y| \ge x$

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z_c + x +{\mathsf {i}}y) \right] \ge 0. \end{aligned}$$

(5.18)

Proof

Let us briefly explain the idea of the proof. We will construct $y_0 \in {\mathbb {R}}$ and a contour connecting the points $z_c + {\mathsf {i}}y_0$ to $z_c + x + {\mathsf {i}}y$ such that ${\mathsf {R}}{\mathsf {e}}[G_{M,N}(z)]$ increases along this contour. This would imply through (5.14) that

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z_c + x +{\mathsf {i}}y) \right] \ge {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z_c +{\mathsf {i}}y_0) \right] \ge {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z_c) \right] = 0. \end{aligned}$$

(5.19)

Let $C =y^2 - x^2 \ge 0$ and set $y_0 = \sqrt{C}$ if $y \ge 0$ and $y_0 = -\sqrt{C}$ if $y < 0$. Consider the curve $z(r) =z_c+ r \pm {\mathsf {i}}\sqrt{C + r^2}$, where the positive sign is taken when $y \ge 0$ and the negative is taken otherwise; here r varies in [0, x]. The curve z(r) connects $y_0$ to (x, y) as r ranges from 0 to x. In addition, writing $z(r) = z_c + x(r) + {\mathsf {i}}y(r)$ we see that $x'(r) x(r) - y'(r)y(r) = 0$, which in view of (5.5) shows that ${\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r)) \right]$ increases on [0, x]. From this we conclude (5.19) where the second inequality follows from (5.14) and last equality is true by the definition of $z_c$. $\square$

1.4 Proof of Lemmas 3.5 and  3.7

In this section we give the proof of Lemmas 3.5 and 3.7 , whose statements are recalled here for the reader’s convenience. We follow the same notation as in Sect. 3.2.

Lemma 5.4

Fix $\theta > 0$, $\delta \in (0,1)$ and assume that $M \ge N \ge 1$, $N/M \in [\delta , 1]$. There exist constants $C > 0$ and $r > 0$ depending on $\theta$ and $\delta$ such that $G_{\alpha }$ is analytic in the disc $|z - z_c| < r$ and the following hold for $|z - z_c| \le r$:

$$\begin{aligned} \begin{aligned}&\left| G_\alpha (z)+ (z-z_c)^3\sigma ^3_\alpha /3 \right| \le C |z-z_c|^4 \text{ whenever } |z - z_c| < r;\\&\quad {\mathsf {R}}{\mathsf {e}}[G_\alpha (z) ] \ge (\sqrt{2}/2)^3|z-z_c|^3\sigma ^3_\alpha /6 \text{ when } z \in C_{z_c, \phi } \text{ with } \phi = \pi /4; \\&\quad {\mathsf {R}}{\mathsf {e}}[G_\alpha (z) ] \le - (\sqrt{2}/2)^3|z-z_c|^3 \sigma ^3_\alpha /6 \text{ when } z \in C_{z_c, \phi } \text{ with } \phi = 3\pi /4. \end{aligned} \end{aligned}$$

(5.20)

In the above equations $z_c, \sigma _\alpha$ are as in Definition 3.1 and $C_{a, \phi }$ is as in Definition 2.10.

Proof

As explained in the beginning of this section $z_c(\alpha )$ defines an increasing continuous between $(0,\infty )$ and $(0, \theta )$. We let [a, b] denote the preimage of $[\delta , 1]$ under this bijection.

Let $r_0 > 0$ be sufficiently small, depending on $\theta$ and $\delta$, such that $\theta -b \ge 2r_0$ and $a \ge 2r_0$. The singularities of $G_{\alpha }$ come from the gamma functions in its definition, and are at $z = 0, -1, -2, \dots$ and $z = \theta , \theta + 1, \theta + 2, \dots$. In particular, by our choice of $r_0$ we know that the closure of $B_{r_0}(z_c) := \{z \in {\mathbb {C}}: |z - z_c| < r_0\}$ is disjoint from those sets. By Hadamard theorem, see [38, Theorem 4.4, Chapter 2], we can expand $G_{\alpha }$ in absolutely convergent power series inside $B_{r_0}(z_c)$. From the definition of $G_\alpha$ we have that $G_\alpha (z_c) = \partial _zG_\alpha (z_c) = \partial _z^2 G_\alpha (z_c) = 0$, while $\partial ^3_z G_\alpha (z_c) = 2 \sigma _\alpha ^3.$ The latter implies that

$$\begin{aligned} G_\alpha (z) = \frac{1}{3} \cdot \sigma _\alpha ^3 \cdot (z- z_c)^3 + \sum _{ n = 4}^\infty a_n(\alpha ) (z-z_c)^n. \end{aligned}$$

Define $K = \{z \in {\mathbb {C}}: d(z, [a,b]) \le r_0\}$. Since $G_\alpha (z)$ is jointly continuous on $(z,\alpha ) \in K \times [\delta , 1]$, which is compact, we know that we can find a constant $A > 0$, depending on $\delta$ and $\theta$ such that

$$\begin{aligned} \sup _{\alpha \in [\delta , 1], z \in K} |G_\alpha (z)| \le A. \end{aligned}$$

Using the Cauchy inequalities, see [38, Corollary 4.3, Chapter 2], we know that

$$\begin{aligned} |a_n(\alpha )| \le Ar_0^{-n}. \end{aligned}$$

The above inequalities show that if $|z - z_c| \le r_0/2$, then

$$\begin{aligned} \left| G_\alpha (z) - \frac{1}{3} \cdot \sigma _\alpha ^3 \cdot (z- z_c)^3\right|\le & {} \sum _{n = 4}^\infty |a_n(\alpha )| \cdot |z - z_c|^n \\\le & {} A |z-z_c|^4 r_0^{-4} \cdot \sum _{n = 4}^\infty (1/2)^{n-4} = 2A |z-z_c|^4 r_0^{-4} , \end{aligned}$$

which proves the first line in (5.20) for any $r \in (0, r_0/2)$ and $C = 2A r_0^{-4}$.

Next we pick $r \le r_0$ sufficiently small so that $Cr < W = (2\sqrt{2})\sigma ^3_\alpha /48$ for all $\alpha \in [\delta , 1]$. Then

$$\begin{aligned}&\left| {\mathsf {R}}{\mathsf {e}}[G_\alpha (z) + (z-z_c)^3\sigma ^3_\alpha /3 ] \right| \le \left| G_\alpha (z)+ (z-z_c)^3\sigma ^3_\alpha /3 \right| \\&\quad \le C |z-z_c|^4 \le Cr |z-z_c|^3 \le W |z-z_c|^3. \end{aligned}$$

The above implies for $z \in C_{z_c, \phi }$ with $\phi = \pi /4$ that

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}[G_\alpha (z) ]\ge & {} -{\mathsf {R}}{\mathsf {e}}[ (z-z_c)^3]\sigma ^3_\alpha /3 - W |z-z_c|^3 \\= & {} |z-z_c|^3 \cdot \left( (2 \sqrt{2}) \sigma ^3_\alpha /24 - W \right) = (\sqrt{2}/2)^3|z-z_c|^3\sigma ^3_\alpha /6. \end{aligned}$$

Similarly if $z \in C_{z_c, \phi }$ with $\phi = 3\pi /4$ we get

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}[G_\alpha (z) ]\le -{\mathsf {R}}{\mathsf {e}}[ (z-z_c)^3]\sigma ^3_\alpha /3+ W |z-z_c|^3 = - (\sqrt{2}/2)^3|z-z_c|^3\sigma ^3_\alpha /6. \end{aligned}$$

The latter two inequalities imply the remainder of (5.20). $\square$

Lemma 5.5

Fix $\theta > 0$, $\delta \in (0,1)$ and assume that $M \ge N \ge 1$, $N/M \in [\delta , 1]$. Let $z(r) = z_c + r e^{{\mathsf {i}}\phi }$. Then we have

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] \le 0, \text{ provided } \text{ that } \phi = 3\pi /4 \text{ or } \phi = 5\pi /4,\\&\quad {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] \ge 0, \text{ provided } \text{ that } \phi = \pi /4 \text{ or } \phi = -\pi /4. \end{aligned}\nonumber \\ \end{aligned}$$

(5.21)

In addition, if $\delta _1 > 0$ is given then there is a constant $c > 0$ depending on $\delta _1, \delta , \theta$ such that

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r)) \right] \le -c (M+N) r \log (1 + r) \text{ for } \text{ any } r \ge \delta _1 \text{ provided } \\&\phi = 3\pi /4 \text{ or } \phi = 5\pi /4,\\&\quad {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r)) \right] \ge c (M+N) r \log (1 + r) \text{ for } \text{ any } r \ge \delta _1 \text{ provided } \\&\phi = \pi /4 \text{ or } \phi = -\pi /4,\\ \end{aligned} \end{aligned}$$

(5.22)

Proof

From (5.7) we get whenever $\phi = \pi /4$ or $3\pi /4$ or $5\pi /4$ or $7\pi /4$ that

$$\begin{aligned} \begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] =&\sum _{n = 0 }^\infty \frac{N r^2 \cos (\phi )}{(n + z_c)|n+z(r)|^2}+ \sum _{n = 0 }^\infty \frac{M r^2 \cos (\phi )}{(n + \theta - z_c)|n+\theta - z(r)|^2 }, \end{aligned}\nonumber \\ \end{aligned}$$

(5.23)

which automatically proves (5.21).

Below C stands for a positive constant, whose value depends on $\theta$ and $\delta$ alone, and whose value may change from line to line. By repeating the same arguments as in (5.10) we see that if $\phi = 3\pi /4$ or $5\pi /4$ we have

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] \le -C (M+N) \log (r), \end{aligned} \end{aligned}$$

(5.24)

where the last inequality holds for all $r \ge r_0$, where $r_0$ depends on $\theta$ alone. Furthermore, from (5.23) we have for any $r > 0$ that

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dr} G_{M,N}(z(r)) \right] \le \frac{-N(\sqrt{2}/2) r^2}{z_c (z_c^2 + r^2 )}\\&\quad + \frac{-Mr^2(\sqrt{2}/2)}{(\theta - z_c)((\theta - z_c)^2 + r^2)} \le -C(M+N)r^2. \end{aligned} \end{aligned}$$

(5.25)

Let $R_0 \ge r_0$ be sufficiently large so that for any $r \ge R_0$ we have

$$\begin{aligned} \int _{r_0}^r \log x dx = (r\log r - r) - (r_0 \log r_0 - r_0) \ge (1/2) r\log r. \end{aligned}$$

Then, using (5.21) we see that if $r \in [\delta _1, R_0]$ we have

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r)) \right]= & {} \int _0^{r} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dx} G_{M,N}(z(x)) \right] dx \le \int _{0}^{r} -C(M+N)x^2dx \\\le & {} -(C/3)(M+N)\delta _1^3, \end{aligned}$$

while if $r \ge R_0$ we have

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(z(r)) \right]= & {} \int _0^{r} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dx} G_{M,N }(z(x)) \right] dx \le \int _{r_0}^{r} {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dx} G_{M,N}(z(x)) \right] dx \le \\\le & {} \int _{r_0}^{r} -C(M+N) \log (x)dx = - (C/2)(M+N)r \log r . \end{aligned}$$

The last two inequalities imply the first line in (5.22). The second one is derived analogously. $\square$

1.5 Proof of Lemma 3.9

In this section we give the proof of Lemma 3.9, whose statement is recalled here for the reader’s convenience. We follow the same notation as in Sect. 3.2.

Lemma 5.6

Fix $\theta , A > 0$, $\delta \in (0,1)$ and assume that $M \ge N \ge 1$, $N/M \in [\delta , 1]$. There exist constants $M_0, C_0 > 0$ depending on $\delta , \theta , A$ such that if $M \ge M_0$, $x \in [-A, A]$ and $z(r) = z_c + r e^{i\phi }$ with $z_c$ as in Definition 3.1, $r \ge 0$ and $\phi \in \{\pi /4, 3\pi /4, 5\pi /4, 7\pi /4\}$ we have

$$\begin{aligned} \left| {\mathsf {R}}{\mathsf {e}}[G_{M,N}(z(r) + xM^{-1/3}) - G_{M,N}(z(r))] \right| \le C_0 M^{2/3} \cdot ( 1 + r). \end{aligned}$$

(5.26)

Proof

Let us denote $z(r,w) = z_c + r e^{{\mathsf {i}}\phi } + w = x(w) + {\mathsf {i}}y(w)$ for $w M^{1/3} \in [-A, A]$]. From equation (5.5) we have

$$\begin{aligned} \begin{aligned}&{\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dw} G_{M,N}(z(r,w)) \right] = N \sum _{n = 0 }^\infty \frac{ (z_c + w)^2 + 2r \cos (\phi ) (z_c+ w) +r^2 }{(n + z_{c})((n + z_c + w)^2 + 2r (n + z_c + w) \cos (\phi ) +r^2 )} \\&\quad + M\sum _{n = 0}^{\infty } \frac{(z_c + w)^2 + 2r \cos (\phi ) (z_c+ w) +r^2}{(n + \theta - z_{c})((n + \theta - z_c - w)^2 -2 r(n + \theta - z_c - w) \cos (\phi ) +r^2)} \\&\quad + N \sum _{n = 0}^\infty \frac{z_c + w + r \cos (\phi ) }{(n + z_c + w)^2 + 2r (n + z_c + w) \cos (\phi ) +r^2}\\&\quad - M \sum _{n = 0}^\infty \frac{z_c + w + r \cos (\phi )}{(n + \theta - z_c - w)^2 -2 r(n + \theta - z_c - w) \cos (\phi ) +r^2}. \end{aligned} \end{aligned}$$

(5.27)

Below C will stand for a positive constant, which depends on $\theta , A, \delta$. We first note that for M large enough

$$\begin{aligned} \begin{aligned}&\sum _{n = 0 }^\infty \frac{ 1}{(n + z_c + w)^2 + 2r (n + z_c + w) \cos (\phi ) +r^2 } \le \sum _{n = 0 }^\infty \frac{ C }{(n+z_c + w)^2 + r^2 }\\&\quad \le \frac{ C }{ (z_c + w)^2 + r^2} + \frac{ C }{ (z_c + w + 1)^2 + r^2} + \int _1^{\infty } \frac{ C dx}{x^2 + r^2} \le \frac{ C }{ 1 + r} . \end{aligned}\nonumber \\ \end{aligned}$$

(5.28)

Analogously, one shows that

$$\begin{aligned} \begin{aligned}&\sum _{n = 0}^\infty \frac{1}{(n + \theta - z_c - w)^2 -2 r(n + \theta - z_c - w) \cos (\phi ) +r^2} \le \frac{C}{1 + r}. \end{aligned} \end{aligned}$$

(5.29)

Combining (5.27) with (5.28) and (5.29) we conclude that for all large enough M

$$\begin{aligned} \left| {\mathsf {R}}{\mathsf {e}}\left[ \frac{d}{dw} G_{M,N}(z(r,w)) \right] \right| \le C(1+r), \end{aligned}$$

and then (5.26) follows by the mean value theorem. $\square$

Appendix B

In this section we provide the proofs of various results we used in Sects. 2 and 3.

We summarize several estimates about special functions that will be required in the proofs below. Note that if $\text{ dist }(s, {\mathbb {Z}}) \ge c$ for some fixed constant $c> 0$ and $s = x+ {\mathsf {i}}y$ then there exists a constant $c' > 0$ such that

$$\begin{aligned} \left| \Gamma (-s) \Gamma (1 + s) \right| = \left| \frac{\pi }{ \sin (\pi s)} \right| \le c' e^{-\pi |y|}. \end{aligned}$$

(6.1)

It follows from [30, (2), pp. 32] that if $z \in {\mathbb {C}}$ and $|\mathsf {arg}(z)| \le \pi - \epsilon$ for some $\epsilon \in (0, \pi )$ then

$$\begin{aligned} \Gamma (z) = e^{-z} z^{z - 1/2} (2\pi )^{1/2} \cdot \left( 1 + O(z^{-1}) \right) , \end{aligned}$$

(6.2)

where the constant in the big O notation depends on $\epsilon$ and we take the principal branch of the logarithm. Also by [38, Theorem 1.6, Chapter 6] there are positive constants $c_1, c_2$ such that

$$\begin{aligned} \left| \frac{1}{\Gamma (z)} \right| \le c_1 e^{c_2 |z| \log |z|}. \end{aligned}$$

(6.3)

1.1 Proof of Propositions 2.15 and 2.19

In this section we give the proofs of Propositions 2.15 and 2.19 , whose statements are recalled here for the reader’s convenience.

Proposition 6.1

Fix $M, N \ge 1$, $\vec {a} = (a_1, \dots , a_N) \in {\mathbb {R}}^N$, $\vec {\alpha }= (\alpha _1, \dots , \alpha _M) \in {\mathbb {R}}^M$, $T \ge 0$, a compact set $K \subset {\mathbb {C}}$ and $v,u\in {\mathbb {C}}$ with ${\mathsf {R}}{\mathsf {e}}(u) > 0$. Then there exist positive constants $L_0, C_0, c_0$, depending on $M, N, \vec {a}, \vec {\alpha }, T, v,u$ and K, such that if $\tau \in [0,T]$, $x,a \in K$, $w = a + z$, $\mathsf {arg}(z) \in [\pi /4, \pi /3] \cup [-\pi /3, -\pi /4]$, $|w| \ge L_0$

$$\begin{aligned} \left| \frac{\pi }{\sin [(\pi (v-w))} \frac{\prod _{m = 1}^M \Gamma (x+ \alpha _m - w )}{\prod _{n = 1}^N\Gamma (w - a_n) } u^{w-v} e^{\tau (w^2 - v^2)/2} \right| \le C_0 e^{- c_0 |w| \log |w|}.\nonumber \\ \end{aligned}$$

(6.4)

Proposition 6.2

Fix $M, N \ge 1$, $\vec {a} = (a_1, \dots , a_N) \in {\mathbb {R}}^N$, $\vec {\alpha }= (\alpha _1, \dots , \alpha _M) \in {\mathbb {R}}^M$, $\tau > 0$, a compact set $K \subset {\mathbb {C}}$ and $v,u\in {\mathbb {C}}$ with ${\mathsf {R}}{\mathsf {e}}(u) > 0$. Then there exist positive constants $L_0, C_0, c_0$, depending on $M, N, \vec {a}, \vec {\alpha }, \tau , v,u$ and K, such that if $w = a + z$ with $a \in K$, $\mathsf {arg}(z) \in [\pi /4, \pi /2] \cup [-\pi /2, -\pi /4]$, $|w| \ge L_0$ we have

$$\begin{aligned} \left| \frac{\pi }{\sin (\pi (v-w))} \frac{\prod _{m = 1}^M \Gamma ( \alpha _m - w )}{\prod _{n = 1}^N\Gamma (w - a_n) } u^{w-v}e^{\tau (w^2 - v^2)/2} \right| \le C_0 e^{- c_0 |w| \log |w|}. \end{aligned}$$

(6.5)

As the proofs of the above propositions are quite similar, we will combine them.

Proof

We consider the two cases $|\mathsf {arg}(z)| \in [\pi /4, \pi /3]$ and $|\mathsf {arg}(z)| \in [\pi /3, \pi /2]$ separately, with the second one only being relevant for Proposition 6.2. Below we fix an arbitrary $\theta >0$, say $\theta = 1$.

Assume first that $|\mathsf {arg}(z)| \in [\pi /4, \pi /3]$. Below $c_i$ denote positive constants. Observe that

$$\begin{aligned} \frac{\Gamma (\theta - w)^M}{\Gamma (w)^N} = \exp \left( - G_{M,N}(w) + W_{M,N}(z_{c})w + C_{M,N} \right) . \end{aligned}$$

In particular, by Lemma 5.1 we have for all |w| sufficiently large that

$$\begin{aligned} \left| \frac{\Gamma (\theta - w)^M}{\Gamma (w)^N} \right| \le \exp \left( - c_1|w| \log |w| \right) . \end{aligned}$$

(6.6)

Furthermore by (6.2) we have for all $x \in K$ and |w| sufficiently large that

$$\begin{aligned} \left| \frac{\prod _{m = 1}^M\Gamma (x + \alpha _m - w)}{\Gamma (\theta - w)^M} \cdot \frac{\Gamma (w)^N}{\prod _{n = 1}^N\Gamma ( w - a_n)} \right| \le |w|^{c_2} . \end{aligned}$$

(6.7)

In addition, by (6.1) we have for all |w| sufficiently large that

$$\begin{aligned} \left| \frac{\pi }{\sin (\pi (v-w))} \right| \le c_3. \end{aligned}$$

(6.8)

Finally, we have for all |w| sufficiently large and $\tau \in [0,T]$ that

$$\begin{aligned} \left| u^{w-v} e^{\tau (w^2-v^2)/2} \right| \le c_4 e^{c_5 |w|}, \end{aligned}$$

(6.9)

where we used the fact that

$$\begin{aligned} |e^{\tau z^2/2}| = \exp \left( \tau {\mathsf {R}}{\mathsf {e}}[z^2]/2 \right) \le 1, \end{aligned}$$

since ${\mathsf {R}}{\mathsf {e}}[z^2] \le 0$ by our assumption that $|\mathsf {arg}(z)| \in [\pi /4, \pi /3]$. Combining (6.6), (6.7), (6.8) and (6.9) we conclude (6.4) and so Proposition 6.1 is proved. Also (6.6), (6.8) and (6.9) prove Proposition 6.2 in the case $|\mathsf {arg}(z)| \in [\pi /4, \pi /3]$.

We next suppose that $|\mathsf {arg}(z)| \in [\pi /3, \pi /2]$ and finish the proof of Proposition 6.2 in this case. Combining (6.1) and (6.3) we see that for all |w| sufficiently large

$$\begin{aligned} \left| \frac{\prod _{m = 1}^M \Gamma ( \alpha _m - w )}{\prod _{n = 1}^N\Gamma (w - a_n) } \right| \le \exp \left( c_6|w| \log |w| \right) . \end{aligned}$$

(6.10)

Also, since $\tau > 0$, we have

$$\begin{aligned} \left| u^{w-v} e^{v\tau (w-v)+ \tau (w-v)^2/2}\right| \le c_7 e^{c_8 |w| - \tau |w|^2/4}, \end{aligned}$$

(6.11)

where we used

$$\begin{aligned} |e^{\tau z^2/2}| = \exp \left( \tau {\mathsf {R}}{\mathsf {e}}[z^2]/2 \right) \le \exp \left( - \tau |z|^2/4 \right) , \end{aligned}$$

which in turn relies on our assumption that $|\mathsf {arg}(z)| \in [\pi /3, \pi /2]$. Combining (6.8), (6.10) and (6.11) we conclude that for all |w| sufficiently large we have

$$\begin{aligned} \left| \frac{\pi }{\sin (\pi (v-w))} \frac{\prod _{m = 1}^M \Gamma ( \alpha _m - w )}{\prod _{n = 1}^N\Gamma (w - a_n) } u^{w-v}e^{\tau (w^2 - v^2)/2} \right| \le c_9 \exp ( -c_{10} |w|^2), \end{aligned}$$

which certainly implies (6.5). This concludes the proof of Proposition 6.2. $\square$

1.2 Proof of Proposition 2.17

In this section we give the proof of Proposition 2.17, whose statement is recalled here for the reader’s convenience.

Proposition 6.3

Fix $M, N \ge 1$, $T \ge 0$, $\tau \in [0,T]$, $\vec {a} = (a_1, \dots , a_N) \in {\mathbb {R}}^N$, $\vec {\alpha }= (\alpha _1, \dots , \alpha _M) \in {\mathbb {R}}_{>0}^M$ and $u \in {\mathbb {C}}$ with ${\mathsf {R}}{\mathsf {e}}(u) > 0$. Put $\theta _0 = \min (\vec {\alpha }) - \max (\vec {a})$ and assume that $\theta _0 > 0$ and $\delta _0 \in (0, \min (1/4, \theta _0 /16))$. Suppose that $v, v' \in C_{a, \phi }$ as in Definition 2.10 with $a \in [ \max (\vec {a}) + \delta _0, \min (\vec {\alpha }) - 5\delta _0]$ and $\phi \in [3\pi /4, 5\pi /6]$. Finally, fix $b \in [a+ 2\delta _0, \min (\vec {\alpha }) - 3\delta _0]$ and denote by $D_v$ the contour $D_v(b, \pi /4, \delta _0)$ as in Definition 2.11. Then there exists a positive constant $C_0$ depending on $\vec {a}, \vec {\alpha }, \delta _0, u, N, M, T$ (and not $\tau$) such that if $x \in {\mathbb {C}}$ with $d(x, [ 0, 1]) \le \delta _0$ then we have

$$\begin{aligned} \left| \int _{D_v} \frac{\pi d\mu (w)}{\sin (\pi (v-w))} \prod _{n = 1}^N \frac{\Gamma (v - a_n)}{\Gamma (w - a_n)} \prod _{m = 1}^M \frac{\Gamma (x + \alpha _m - w)}{\Gamma (x + \alpha _m - v)} \frac{u^{w-v} e^{\tau (w^2 - v^2)/2}}{w - v'} \right| \le \frac{C_0}{1 + |v|^2}.\nonumber \\ \end{aligned}$$

(6.12)

Proof

In the arguments below, $c_i$ will denote positive constants that depend on the parameters in the statement of the proposition. Let us put $\theta = \min (\vec {\alpha })$. We notice that we can rewrite the integrand as

$$\begin{aligned} F(w)= \frac{H(w)}{H(v)}\cdot \frac{\pi \exp \left[ G_{M,N}(v) - G_{M,N}(w) \right] }{\sin ( \pi (v-w))} \frac{u^{w-v} e^{\tau (w^2-v^2)/2} }{w - v'}, \end{aligned}$$

(6.13)

where $G_{M,N}$ is as in (5.3) and

$$\begin{aligned} H(w) = \exp (- W_{M,N}(z_{c})w ) \cdot \prod _{n = 1}^N \frac{\Gamma (w)}{\Gamma (w - a_n)} \prod _{m = 1}^M \frac{\Gamma (x + \alpha _m - w)}{\Gamma (\theta - w)}. \end{aligned}$$

We split the integral over the two pieces $D_v^1$ and $D_v^2$ (see Definition 2.11). If $w \in D_v^1$ and |w| is sufficiently large we may apply Lemma 5.1 to get

$$\begin{aligned} \left| \exp ( - G_{M,N}(w))\right| \le \exp \left( - c_1 |w|\log |w| \right) . \end{aligned}$$

(6.14)

On the other hand, using (6.1), we have for all $w \in D_v^1$ and $v,v' \in C_{a, \phi }$

$$\begin{aligned} \left| \frac{\pi }{\sin (\pi (v-w))} \right| \le c_2 \text{ and } \left| \frac{1}{ w - v'}\right| \le c_3. \end{aligned}$$

(6.15)

Setting $v = a + r_v \cos (\phi ) + {\mathsf {i}}r_v \sin (\phi )$ and $w = b + r_w \cos (\pi /4) \pm {\mathsf {i}}r_w \sin (\pi /4)$, we get

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}[w^2 - v^2] =(-\cos ^2(\phi ) + \sin ^2(\phi ))r_v^2 - 2a \cos (\phi ) r_v + b^2 - a^2 + 2b \cos (\pi /4) r_w, \end{aligned}$$

where we used that $\cos (\pi /4) = \sin (\pi /4)$. In particular, by our assumption on $\phi$ we have $(-\cos ^2(\phi ) + \sin ^2(\phi )) \le 0$ and we conclude that

$$\begin{aligned} \left| e^{\tau (w^2-v^2)/2} \right| \le c_4 \cdot e^{c_5(|w|+|v|)}. \end{aligned}$$

(6.16)

Also, from (6.2) we have

$$\begin{aligned} \left| \frac{H(w)}{H(v)} \right| \le c_6 \cdot e^{c_7 (|w| + |v|)}. \end{aligned}$$

(6.17)

Combining (6.14), (6.15), (6.16) and (6.17) we conclude that

$$\begin{aligned} |F(w)| \le c_8 e^{ c_9(|w| + |v|) - c_{10} (M+N) |w|\log |w|} \cdot \exp \left( {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(v) \right] \right) . \end{aligned}$$

(6.18)

From the above we conclude that

$$\begin{aligned} \int _{D^1_v}| F(w) | |dw| \le c_{11} e^{c_9 |v|} \cdot \exp \left( {\mathsf {R}}{\mathsf {e}}\left[ G_{M,N}(v) \right] \right) \le c_{12}\exp ( - c_{13} |v| \log |v|),\nonumber \\ \end{aligned}$$

(6.19)

where the last inequality follows from Lemma 5.1.

We next focus our attention on $D_v^2$. Recall that $D_v^2$ consists of straight oriented segments that connect $z_-$ to $v + 2\delta _0 - i\delta _0$ to $v + 2\delta _0 + i\delta _0$ to $z_+$, where $z_-, z_+$ are the points on $C_{b,\pi /4}$ that have imaginary parts ${\mathsf {I}}{\mathsf {m}}(v) - \delta _0$ and ${\mathsf {I}}{\mathsf {m}}(v) + \delta _0$ respectively. Explicitly, we have $z_- = b + {\mathsf {I}}{\mathsf {m}}(v) -\delta _0 + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(v) - i\delta _0$ and $z_+ = b + {\mathsf {I}}{\mathsf {m}}(v) + \delta _0 + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(v) + {\mathsf {i}}\delta _0$.

Fig. 8

The figure depicts the contour $C_{a, \phi }$ and $D_v = D_v(b,\pi /4, \delta _0)$ with $\min (\vec {\alpha })> b> a > \max (\vec {a})$. The black dots denote the points $v, v+1, v+2, \dots$ and the grey ones denote $\alpha _m + x, \alpha _m + x + 1, \alpha _m + x + 2, \dots$. The contour $C_v$ is the filled contour that connects $z_-$ and $z_+$

If |v| is bounded, we observe that F(w) remains bounded on $D_v^2$ as it is bounded away from all of its poles, so we only need to focus on large |v|. We assume that |v| is sufficiently large so that $D_v^2$ is well separated from the real line and so $z_-, z_+$ lie in the same complex half-plane. For simplicity we will assume that we are in the upper complex half-plane (the other case is completely analogous).

We define a contour connecting $z_-$ and $z_+$, denoted by $C_{v}$ as follows, see also Fig. 8. It starts at $z_-$ and goes horizontally to $v + k_v + 1/2 - {\mathsf {i}}\delta _0$, then up to $v+k_v + 1/2+ {\mathsf {i}}\delta _0$ and then horizontally to $z_+$. Here $k_v$ denotes the largest integer such that ${\mathsf {R}}{\mathsf {e}}[v + k_v ] < - 2a + {\mathsf {I}}{\mathsf {m}}(v)$. We assume that |v| is sufficiently large so that $k_v \ge 1$. Notice that the length of $C_v$ does not exceed $3 + 2\delta _0 + 4|a| + 2|b|$.

An application of the residue theorem gives

$$\begin{aligned} \frac{1}{2\pi {\mathsf {i}}}\int _{D_v^2} F(w) = \sum _{j = 1}^{k_v}R_j(v) + \frac{1}{2\pi {\mathsf {i}}} \int _{C_v} F(w) dw, \end{aligned}$$

(6.20)

where the residue at $w = v + j$ is denoted by $R_j(v)$ and given by the formula

$$\begin{aligned} R_j(v) \!=\! -(-u)^{j} \prod _{n = 1}^N \frac{\Gamma (v - a_n)}{\Gamma (v+ j - a_n)} \prod _{m = 1}^M \frac{\Gamma (x + \alpha _m - v- j)}{\Gamma (x + \alpha _m - v)} \frac{e^{\tau v j+ \tau j^2/2}}{v - v' + j}.\nonumber \\ \end{aligned}$$

(6.21)

From the functional equation for the gamma function $\Gamma (z+1) = z \Gamma (z)$ we know that

$$\begin{aligned}&\left| \frac{\Gamma (v - a_n)}{\Gamma (v+j - a_n)} \right| = \prod _{k = 1}^j \left| \frac{1}{v - a_n + k - 1} \right| \le \frac{1}{|{\mathsf {I}}{\mathsf {m}}(v)|^j} \text{ and } \\&\quad \left| \frac{ \Gamma (x + \alpha _m - v - j)}{\Gamma (x + \alpha _m - v)} \right| = \prod _{k = 1}^j \left| \frac{1}{x + \alpha _m - v - k} \right| \le \frac{1}{(|{\mathsf {I}}{\mathsf {m}}(v)| - \delta _0)^j}, \end{aligned}$$

where in the last inequality we used that ${\mathsf {I}}{\mathsf {m}}(x) \in [-\delta _0, \delta _0]$.

Next we observe that if $v = a + r_v \cos (\phi ) + {\mathsf {i}}r_v \sin (\phi )$ then

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}[v j + j^2/2]= & {} j (a + r_v \cos (\phi )) + j^2/2 \le j (a + r_v \cos (\phi )\\&+ (r_v \sin (\phi ) - 2a - r_v \cos (\phi ) ) /2) \le 0, \end{aligned}$$

where in the last inequality we used that $r_v \cos (\phi ) + r_v \sin (\phi ) \le 0$ by our choice of $\phi$ and the first inequality used $1 \le j \le k_v$ and ${\mathsf {R}}{\mathsf {e}}[v + k_v ] < - 2a + {\mathsf {I}}{\mathsf {m}}(v)$. The conclusion is that for all large enough |v| we have

$$\begin{aligned} \left| e^{\tau v j+ \tau j^2/2} \right| \le 1 \end{aligned}$$

for all $j = 1, \dots , k_v$ and $\tau \ge 0$. Finally, if we assume that |v| is sufficiently large we can ensure that $\frac{\max (1, |u|)}{(|{\mathsf {I}}{\mathsf {m}}(v)|- \delta _0)} \le \frac{1}{2}$. Combining all of the above estimates we see that

$$\begin{aligned} \sum _{j = 1}^{k_v } |R_j(v)|\le & {} \sum _{j = 1}^{k_v } \frac{|u|^j}{(|{\mathsf {I}}{\mathsf {m}}(v)| - \delta _0)^{j(M+N)}} \frac{1}{|v - v' + j|}\nonumber \\\le & {} \frac{c_{14}}{(|{\mathsf {I}}{\mathsf {m}}(v)| - \delta _0)^{M+N}} \le \frac{c_{15}}{1 + |v|^2}, \end{aligned}$$

(6.22)

where in the second inequality we used that $v - v'$ is uniformly bounded away from ${\mathbb {Z}}$ and that the sum of the geometric series is controlled by its first term. The last inequality used that $M + N \ge 2$ and holds for all |v| large enough.

Since the contour $C_v$ is (at least distance $\delta _0$) bounded away from $v + {\mathbb {Z}}$ the estimates in (6.15) hold. Provided that |v| is sufficiently large and $w \in C_v$ the estimates in (6.14) and (6.17) also holds. We next wish to show that (6.16) also holds for $w \in C_v$.

Suppose that $w \in C_v$, and write it as $w = y + r_w \cos (\pi /4) + {\mathsf {i}}r_w \sin (\pi /4)$, where y is a real number between $-2|a| - 2|b| - \delta _0 - 3/2$ and $2|a| + 2|b| + \delta _0 + 3/2$ and $r_w \in [{\mathsf {I}}{\mathsf {m}}(v) - \delta _0, {\mathsf {I}}{\mathsf {m}}(v) + \delta _0]$. As done before in (6.16) we have

$$\begin{aligned} {\mathsf {R}}{\mathsf {e}}[2 v(w-v) + (w-v)^2]= & {} (-\cos ^2(\phi ) + \sin ^2(\phi ))r_v^2 \\&- 2a \cos (\phi ) r_v + y^2 - a^2 + 2y \cos (\pi /4) r_w. \end{aligned}$$

In particular, by our assumption on $\phi$ we have $(-\cos ^2(\phi ) + \sin ^2(\phi )) \le 0$ and so (6.16) also holds. Combining (6.14), (6.15), (6.16) and (6.17) we conclude that for all large v and $w \in C_v$ one has the estimate as in (6.18) and since $C_v$ has length at most $3 + 2\delta _0 + 4|a| + 2|b|$, we conclude from Lemma 5.1 that

$$\begin{aligned} \left| \frac{1}{2\pi {\mathsf {i}}} \int _{C_v} F(w) dw\right| \le c_{17}\exp ( - c_{18} |v| \log |v|). \end{aligned}$$

(6.23)

The proposition now follows from combining (6.19), (6.20),(6.22) and (6.23).

$\square$

1.3 Proof of Proposition 3.11

In this section we give the proof of Proposition 3.11, whose statement is recalled here for the reader’s convenience. We follow the same notation as in Sects. 3.1 and 3.2.

Proposition 6.4

Let $A \ge 0$ and $x\ge -A$ be given and let M, N, u(x, M, N) be as in Definition 3.1. Then for any $\epsilon > 0$ there exist positive constants $M_0, c_0 ,C_0 > 0$ depending on $A,\epsilon$ and the parameters in Definition 3.1 such that if $M \ge M_0$, $v, v' \in C_{a,3\pi /4}$ we have

$$\begin{aligned}&\left| K_u (v,v')\right| \le C_0M^{1/3} \text{ and } \text{ if } |v - a| \ge \epsilon \text{ we } \text{ further } \text{ have } \left| K_u (v,v')\right| \nonumber \\&\quad \le C_0 e^{-c_0 M \log (1 + |v|)}. \end{aligned}$$

(6.24)

If ${\mathsf {r}} = {\mathsf {c}}= 0$ then $M_0, c_0 ,C_0 > 0$ depend only on $\theta$, $\delta$, A and $\epsilon$.

Proof

Before we go to the main argument, we give a conceptual roadmap of the proof. For clarity we split the proof into several steps. In the first four steps we prove (6.24) when v is bounded away from a and in the last step we establish it when v is close to a. In both cases we deform the contour $D_v$ in the definition of $K_u(v,v')$ to a suitable contour where we can estimate the integrand in the definition of $K_u(v,v')$ using the results from Sects. 5 and 6. In the case when v is close to a we can deform $D_v$ to a contour that is a horizontal translation of $C_{z_c, \pi /4}$ by $O(M^{-1/3})$ (recall that this contour was defined in Definition 2.10). Ideally, we would prefer to work with the contour $C_{z_c, \pi /4}$, because Lemma 5.4 and Lemma 5.5 give us very good control of the absolute value of the integrand on this contour. A mild translation of $C_{z_c, \pi /4}$ is, however, necessary to ensure that the poles at $\alpha _m$ and $a_n$ are on the correct sides of the contour. The idea then is to combine our good estimates on $C_{z_c, \pi /4}$ with Lemma 5.6, which ensures that slight horizontal movements away from $C_{z_c, \pi /4}$ does not significantly damage our estimates. This idea is also present in the case when v is bounded away from a. The new challenge that arises in this case is that we need to deform $D_v$ to a new contour, which avoids the poles at $v + {\mathbb {Z}}$ coming from the sine function in the definition of $K_u(v,v')$. The idea then is to work with a contour $D_v(b, \pi /4, d)$ as in in Definition 2.11. This contour consists of two parts—$D^1$ and $D^2$. Estimating the absolute value of the integrand on $D^1$ is similar to the case we discussed above and relies only on Lemma 5.4, Lemma 5.5 and Lemma 5.6. The analysis of the integral over $D^2$ is more involved and requires further deforming this contour to the half-plane $\{ {\mathsf {R}}{\mathsf {e}}(z) \ge z_c\}$. Once we are in the half-plane $\{ {\mathsf {R}}{\mathsf {e}}(z) \ge z_c\}$ we can utilize Lemma 5.3 and get the estimates we desire—this is done in Step 3. Unfortunately, in the process of deforming $D^2$ we pick up a large number of poles, whose residues need to be controlled as well—this is done in Step 4 and requires a fairly involved case-by-case analysis. We now turn to the proof.

Step 1 Let r be as in Lemma 5.4 for the choice of $\delta$ and $\theta$ as in the proposition. We also set $r_\epsilon = \min (r/2,\epsilon ,1/2)$, and for $y > 0$ we write $B_y(z_c) := \{z \in {\mathbb {C}}: |z - z_c| < y\}$. We also define $d_\epsilon = \min (r_\epsilon /10,1/20, d_0)$, where $d_0 > 0$ is sufficiently small so that

$$\begin{aligned} d_0\le & {} \frac{\theta - z_c}{4}, d_0 \le \frac{z_c}{4} \text{, } \text{ and } (1- s)\sqrt{(1-s)^2 + (\theta - z_c-s)^2}\nonumber \\\ge & {} 1 + s \text{ for } s \in [0, d_0]. \end{aligned}$$

(6.25)

Notice that all of the above constants can be chosen uniformly in $\theta , \epsilon$ and $\delta$ when ${\mathsf {r}} = {\mathsf {c}}= 0$.

We first suppose that $|v -a| \ge \epsilon$ and establish (6.24) in this and the next three steps. In the sequel we will write G for $G_{\alpha }$, $\sigma$ for $\sigma _\alpha$, d for $d_\epsilon$ and r for $r_\epsilon$ to ease the notation. Then we have by Cauchy’s theorem that if $M_0$ is sufficiently large and $M \ge M_0$ that

$$\begin{aligned} \begin{aligned}&K_u(v,v') = \frac{1}{2\pi {\mathsf {i}}} \int _{D} F(w,v',v)dw, \text{ with } \\&\quad F(w,w',v) = \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n)}{\Gamma (w - a_n)} \prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - w)}{\Gamma (\alpha _m - v)} \\&\quad \times \frac{\pi e^{M[G(v) - G(w)]}e^{M^{1/3} x \sigma (v-z_c)}e^{-M^{1/3} x \sigma (w-z_c)} }{\sin (\pi (v-w)) (w- v')}, \end{aligned} \end{aligned}$$

(6.26)

where D stands for the contour $D_v(b, \pi /4, d)$ in Definition 2.11 with d as above, $b = a + \rho \sigma ^{-1} M^{-1/3}$ and $\rho$ as in Definition 3.1. The decay estminates necessary to deform the contour $D_v$ in the definition of $K_u(v,v')$ to D near infinity come from Proposition 6.1 applied to $K = [0,\theta ]$, $T =0$ and $\vec {a}, \vec {\alpha }, u$ as in the statement of the proposition. In order to ensure that the poles at $\theta , \theta + 1, \theta + 2, \dots$ stay on the right side of D (and hence are not crossed in the process of deformation) we need to ensure that $a + \rho \sigma ^{-1}M^{-1/3} = z_c + ( \mu + \rho ) \sigma ^{-1}M^{-1/3} < \theta$, which is possible by making $M_0$ sufficiently large depending on $\theta$ and $\delta$ alone. Also, by the definition of $\rho$, we have that if ${\mathsf {c}}> 0$ then the poles at $\alpha _i$ for $i = 1, \dots , {\mathsf {c}}$ are also to the right of D for large enough M in view of $|v-a| \ge r$ and $d \le r/10$. We denote by $D^1$ and $D^2$ the portions of D that are part of $C_{b, \pi /4}$ or not respectively, see Definition 2.10.

In the following steps C, c stand for generic positive constants, depending on the parameters in the statement of the proposition, whose value may change from line to line.

Step 2 In this and the next step we assume that $|v-a| \ge r$. Notice that if ${\mathsf {R}}{\mathsf {e}}(z) \le 0$ we have

$$\begin{aligned} |e^{M^{1/3} x \sigma z}| \le 1 \text{ if } x \ge 0 \text{ and } \text{ so } \text{ for } x \ge -A \text{ we } \text{ have } |e^{M^{1/3} x \sigma z}| \le e^{AM^{1/3} \sigma |z|}.\nonumber \\ \end{aligned}$$

(6.27)

The latter equation together with Lemmas 5.5 and 5.6 imply for all large enough M

$$\begin{aligned} \begin{aligned}&| e^{MG(v)} e^{M^{1/3} \sigma (v-z_c) x } | \le Ce^{-c M |v| \log (1 + |v|)} \text{ if } |v - a| \ge r, \text{ and } v \in C_{a, 3\pi /4} \\&\quad | e^{-MG(w)} e^{-M^{1/3} \sigma (w-z_c) x } | \le Ce^{-c M |w| \log (1 + |w|)} \text{ if } |w - b| \ge r/2 \text{ and } w \in C_{b, \pi /4}. \end{aligned}\nonumber \\ \end{aligned}$$

(6.28)

To see why the latter is true, note that Lemma 5.5 implies the above inequalities when a, b are replaced with $z_c$ and then Lemma 5.6 allows us to replace $z_c$ with a, b since such a change affects the right side of the inequalities by $Ce^{c M^{2/3} (1 + |v|)}$ for the first line and $Ce^{c M^{2/3} (1 + |w|)}$ for the second line. Of course, this effect is negligible compared to the exponential in M decay.

In addition, using (6.1) we have for all $w \in D^1$ that

$$\begin{aligned} \left| \frac{1}{\sin (\pi (v-w)) (w- v')} \right| \le CM^{1/3}. \end{aligned}$$

(6.29)

The extra $M^{1/3}$ is coming from the $|w-v'|$ in the denominator, which is lower bounded by $M^{-1/3} \sigma ^{-1} \rho$ as $v'$ is allowed to be close to a unlike v. We also observe that

$$\begin{aligned} \begin{aligned}&\left| \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n)}{\Gamma (w - a_n)} \prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - w)}{\Gamma (\alpha _m - v)}\right| = \left| \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n + 1)}{\Gamma (w - a_n + 1)} \prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - w + 1)}{\Gamma (\alpha _m - v + 1)}\right| \\&\quad \times \ \left| \prod _{n = 1}^{{\mathsf {r}} } \frac{w - a_n}{v - a_n} \prod _{m = 1}^{{\mathsf {c}}} \frac{\alpha _m - v}{\alpha _m - w}\right| \le C \exp \left( c |v| \log (1 + |v|) + c |w| \log (1 + |w|) + c \log M \right) , \end{aligned}\nonumber \\ \end{aligned}$$

(6.30)

where we used (6.2) and (6.3) to upper bound the gamma functions, and the extra $\log M$ term is coming from $\alpha _m - w$ in the denominator, whose absolute value is bounded away from 0 by $(1/2) (\min (\vec {y}) - \mu - \rho ) M^{-1/3} \sigma ^{-1}$ for all large enough M, see Definition 3.1.

Combining (6.28), (6.29) and (6.30) we conclude that for all large enough M

$$\begin{aligned} \left| \int _{D^1 \cap B^c_{r/2}(b)} F(w,v',v)dw \right| \le C e^{-c M |v| \log (1 + |v|)}. \end{aligned}$$

(6.31)

Suppose next that $w \in D^1 \cap B_{r/2}(b)$ (observe that this piece is separated from $D^2$ by our assumption on v). Using (6.27), Lemma 5.4 and Lemma 5.6 we have

$$\begin{aligned} | e^{-MG(w)} e^{-M^{1/3} \sigma (w-z_c) x } | \le C e^{cM^{2/3}}e^{-M \sigma ^3 (\sqrt{2}/2)^3|w-z_c|^3/6} e^{AM^{1/3} \sigma |w - z_c|}. \end{aligned}$$

In deriving the above inequality we used that our r is smaller than the one in Lemma 5.4 so that the lemma is applicable. Combining the last inequality with the first line in (6.28), (6.29), (6.30) we conclude that for all large enough M

$$\begin{aligned} \begin{aligned}&\left| \int _{D^1 \cap B_{r/2}(b)} F(w,v',v)dw \right| \le C e^{-c M |v| \log (1 + |v|)}. \end{aligned} \end{aligned}$$

(6.32)

Equations (6.31) and (6.32) provide the desired estimates for the integral of $F(w,v',v)$ on $D^1$. We consider the integral over $D^2$ in the next step.

Step 3 In this step we establish the following inequality

$$\begin{aligned} \left| \int _{D^2 } F(w,v',v)dw \right| \le C e^{-c M \log (1 + |v|)}. \end{aligned}$$

(6.33)

Combining (6.31), (6.32) and (6.33) we conclude (6.24) in the case when $|v -a| \ge r$, which of course implies the statement when $|v-a| \ge \epsilon$ as $r \le \epsilon$ by construction.

To prove (6.33) we define a new contour $C_v$ as follows, see also Fig. 9. Let $k_v \in {\mathbb {Z}}_{\ge 0}$ be the unique integer such that ${\mathsf {R}}{\mathsf {e}}(v + k_v) \in [ z_c - d, z_c + d]$ if such an integer exists (the uniqueness follows from our assumption that $d \le 1/20$); if no such integer exits we let $k_v$ be the largest integer such that ${\mathsf {R}}{\mathsf {e}}(v + k_v) \le z_c$. If ${\mathsf {R}}{\mathsf {e}}(v + k_v) \in [ z_c - d, z_c + d]$ we let $C_v$ be the contour that horizontally connects the points $z_-$ to $z_c + 1/2 + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(z_-)$ to $z_c +1/2 + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(z_+)$ to $z_+$, where $z_{\pm }$ are the two points shared by $D^1$ and $D^2$, see the left part of Fig. 9. If ${\mathsf {R}}{\mathsf {e}}(v + k_v) \not \in [ z_c - d, z_c + d]$ we let $C_v$ be the contour that connects $z_-$ to $z_c + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(z_-)$ to $z_c + {\mathsf {i}}{\mathsf {I}}{\mathsf {m}}(z_+)$ to $z_+$, see the right part of Fig. 9. Observe that by construcion we have that if $z = z_c + (\mu + \rho )\sigma ^{-1}M^{-1/3} + f + {\mathsf {i}}g \in C_v$ then $|g| \ge f \ge 0$, $d(z, v + {\mathbb {Z}}) \ge d$ and $d(z, C_{a, 3\pi /4}) \ge d/2$ for large enough M. Also the length of $C_v$ is at most $2 |{\mathsf {I}}{\mathsf {m}}(v)| + 3d$ for large enough M.

Fig. 9

The left part depicts $C_v$ when $Re(v + k_v) \in [ z_c - d, z_c + d]$ and the right part depicts it when $Re(v + k_v) \not \in [ z_c - d, z_c + d].$

An application of the residue theorem, see [38, Corollary 2.3, Chapter 3], gives

$$\begin{aligned} \frac{1}{2\pi {\mathsf {i}}}\int _{D^2} F(w,v',v) = \sum _{j = 1}^{k_v }R_j(v) + \frac{1}{2\pi {\mathsf {i}}} \int _{C_v} F(w,v',v) dw, \end{aligned}$$

(6.34)

where the residue at $w = v + j$ is denoted by $R_j(v)$ and given by the formula

$$\begin{aligned} R_j(v)= & {} -(-e)^{-x \sigma M^{1/3} \cdot j} \frac{\Gamma (v)^N}{\Gamma (v+j)^N} \frac{ \Gamma (\theta - v - j)^M}{\Gamma (\theta - v)^M} \frac{1}{v - v' + j} \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n)}{\Gamma (v + j - a_n)} \\&\prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - v - j)}{\Gamma (\alpha _m - v)}. \end{aligned}$$

Observe that from (6.27), Lemma 5.3 and Lemma 5.6 we have for $w \in C_v$ that

$$\begin{aligned} \left| e^{-MG(w)} e^{-M^{1/3} \sigma (w-z_c) x} \right| \le Ce^{c M^{1/3}A|v| + cM^{2/3} |v| }. \end{aligned}$$

(6.35)

Let us elaborate on equation (6.35) briefly. Equation (6.27) allows us to bound $|e^{-M^{1/3} \sigma (w-z_c) x}|$ by $e^{c M^{1/3}A |v|}$, which in turn is controlled by $e^{c M^{1/3}A |w|}$ as |w| is upper and lower bounded by constant multiples of |v| for $w \in C_v$. Furthermore, Lemma 5.3 bounds $|e^{-MG(w)}|$ by 1 when $|{\mathsf {I}}{\mathsf {m}}[ w- z_c] \ge {\mathsf {R}}{\mathsf {e}}[w - z_c ] \ge 0$. By construction of $C_v$ we know that ${\mathsf {R}}{\mathsf {e}}[w - z_c ] \ge 0$ is satisfied for all points on $C_v$. The inequality $|{\mathsf {I}}{\mathsf {m}}[ w- z_c]| \ge {\mathsf {R}}{\mathsf {e}}[w - z_c ]$ can fail for some points on $C_v$ but only those that are at most distance $(\mu + \rho )\sigma ^{-1}M^{-1/3}$ (if the latter is positive) from the points $z^{\pm }$. For these w we know from Lemma 5.6 that ${\mathsf {R}}{\mathsf {e}}[-MG(w)] = {\mathsf {R}}{\mathsf {e}}[-MG(w -(\mu + \rho )\sigma ^{-1}M^{-1/3} )] + O(M^{2/3}( 1 + |w|)),$ and since by Lemma 5.3 the first term is negative, we see that ${\mathsf {R}}{\mathsf {e}}[-G(w)] \le cM^{2/3}|v|$ for all $w \in C_v$.

Since $C_v$ is bounded away from $v + {\mathbb {Z}}$, and $C_{a, 3\pi /4}$ we know that (6.29) holds, and as it is bounded away from ${\mathbb {R}}$ we also know that (6.30) holds. Combining the first line of (6.28), with (6.29), (6.30), (6.35) and the fact that the length of $C_v$ is at most $2 |{\mathsf {I}}{\mathsf {m}}(v)| + 3d$ we conclude that for all large enough M

$$\begin{aligned} \left| \int _{C_v} F(w,v',v)dw \right| \le C e^{-c M |v| \log (1 + |v|)}. \end{aligned}$$

(6.36)

We now claim that

$$\begin{aligned} \sum _{j = 1}^{k_v}|R_j(v)| \le C e^{-c M \log (1 + |v|)}. \end{aligned}$$

(6.37)

If the latter is true, then (6.34), (6.36) and (6.37) would imply (6.33). We establish (6.37) in the next step.

Step 4 We prove (6.37) by considering the cases when $|{\mathsf {I}}{\mathsf {m}}(v)| \le 1 - d$, $|{\mathsf {I}}{\mathsf {m}}(v)| \in [1-d, 1+ d]$, $|{\mathsf {I}}{\mathsf {m}}(v)| \in [ 1-d, 3]$ and $|{\mathsf {I}}{\mathsf {m}}(v)| \ge 3$. The case when $|{\mathsf {I}}{\mathsf {m}}(v)| \le 1 - d$ is trivial since then $k_v = 0$ and the sum in (6.37) is empty. The focus is on the remaining three cases.

From the functional equation for the gamma function $\Gamma (z+1) = z \Gamma (z)$ we know that

$$\begin{aligned}&\left| \frac{\Gamma (v)}{\Gamma (v+j)} \right| = \prod _{k = 1}^j \left| \frac{1}{v + k - 1} \right| \text{, } \left| \frac{ \Gamma (\theta - v - j)}{\Gamma (\theta - v)} \right| = \prod _{k = 1}^j \left| \frac{1}{\theta - v - k} \right| \text{ and } \nonumber \\&\quad \left| \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n)}{\Gamma (v + j - a_n)} \prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - v - j)}{\Gamma (\alpha _m - v)} \right| \nonumber \\&\quad = \prod _{n = 1}^{{\mathsf {r}} } \prod _{k = 1}^j \left| \frac{1}{v + k - 1 - a_n}\right| \prod _{n = 1}^{{\mathsf {c}}} \prod _{k = 1}^j \left| \frac{1}{\alpha _m- v - k} \right| . \end{aligned}$$

(6.38)

If $|{\mathsf {I}}{\mathsf {m}}(v)| \in [1-d, 1+ d]$ we have that $k_v = 1$ and so the left side of (6.37) becomes

$$\begin{aligned} \begin{aligned} |R_1(v)|&= \frac{|e^{-x\sigma M^{1/3}}|}{|v - v' + 1|} \frac{1}{|v|^N \cdot |\theta - v - 1|^M}\prod _{n = 1}^{{\mathsf {r}} } \left| \frac{1}{v - a_n}\right| \prod _{n = 1}^{{\mathsf {c}}} \left| \frac{1}{\alpha _m- v - 1} \right| \\&\le \frac{C\exp (A \sigma M^{1/3})}{|{\mathsf {I}}{\mathsf {m}}(v)|^N (\sqrt{(1-d)^2 + (\theta - a + |{\mathsf {I}}{\mathsf {m}}(v)| - 1)^2} )^M} \\&\le \frac{C\exp (A \sigma M^{1/3})}{ ((1-d)\sqrt{(1-d)^2 + (\theta - a - d)^2})^M} \end{aligned}\nonumber \\ \end{aligned}$$

where we used that $|v| \ge 1-d$, $M \ge N$ and (6.27). By our choice of $d \le d_0$ with $d_0$ as in (6.25), we get for all large enough M

$$\begin{aligned} |R_1(v)| \le \frac{\exp (CM^{1/3})}{(1+d/2)^M}, \end{aligned}$$

which implies (6.37) in the case $|{\mathsf {I}}{\mathsf {m}}(v)| \in [1-d, 1+ d]$ as long as M is sufficiently large.

Next, suppose that $|{\mathsf {I}}{\mathsf {m}}(v)| \in [1 + d, 3]$. Then from (6.38) we have

$$\begin{aligned} \begin{aligned}&\left| \frac{\Gamma (v)}{\Gamma (v+j)} \right| \le \frac{1}{|{\mathsf {I}}{\mathsf {m}}(v)|^j} \text{ and } \left| \frac{ \Gamma (\theta - v - j)}{\Gamma (\theta - v)} \right| \le \frac{1}{|{\mathsf {I}}{\mathsf {m}}(v)|^j}\\&\quad \left| \prod _{n = 1}^{{\mathsf {r}} } \frac{\Gamma (v - a_n)}{\Gamma (v + j - a_n)} \prod _{m = 1}^{{\mathsf {c}}} \frac{\Gamma (\alpha _m - v - j)}{\Gamma (\alpha _m - v)} \right| \le C. \end{aligned} \end{aligned}$$

(6.39)

The above inequalities imply that (observe $k_v \le 3$)

$$\begin{aligned} \sum _{j = 1}^{k_v }|R_j(v)| \le \frac{\exp (CM^{1/3})}{(1 + d)^{N + M}}, \end{aligned}$$

(6.40)

which implies (6.37) in the case $|{\mathsf {I}}{\mathsf {m}}(v)| \in [1+d, 3]$ as long as M is sufficiently large.

Finally, we suppose that $|{\mathsf {I}}{\mathsf {m}}(v)| \ge 3$. In this case (6.39) imply

$$\begin{aligned} \sum _{j = 1}^{k_v }|R_j(v)| \le \sum _{j = 1}^{k_v} \frac{C \cdot \exp ({A \sigma M^{1/3} j})}{|{\mathsf {I}}{\mathsf {m}}(v)|^{j(M+N)}} \le C \sum _{j = 1}^{k_v}|{\mathsf {I}}{\mathsf {m}}(v)|^{-jM} \le C |{\mathsf {I}}{\mathsf {m}}(v)|^{-M},\nonumber \\ \end{aligned}$$

(6.41)

where in the next to last inequality we used that $|{\mathsf {I}}{\mathsf {m}}(v)|^{N} \ge \exp ({A\sigma M^{1/3}})$ for all large enough M, and in the last inequality we bounded the sum by a geometric series. Equation (6.41) implies (6.37) in the case $|{\mathsf {I}}{\mathsf {m}}(v)| \ge 3$ since

$$\begin{aligned} \log (1 + |v|) \le \log (1 + 2 |{\mathsf {I}}{\mathsf {m}}(v)|) \le \log (3) + \log |{\mathsf {I}}{\mathsf {m}}(v)|\le 2 \log |{\mathsf {I}}{\mathsf {m}}(v)|. \end{aligned}$$

Step 5 In this step we establish (6.24) when $|v - a| \le r$. Observe that by Cauchy’s theorem we may deform the $D_v$ in the definition of $K_u(v,v')$ to the contour $C_{b, \pi /4}$ without affecting the value of the integral. The latter is because $v + 1$ is strictly to the right of $C_{b, \pi /4}$ whenever $|v-z_c| \le r$, by our choice of r. Thus we do not cross any poles in the process of the deformation. The decay estminates necessary to deform the contour $D_v$ in the definition of $K_u(v,v')$ to $C_{b, \pi /4}$ near infinity come from Proposition 6.1 applied to $K = [0,\theta ]$, $T =0$ and $\vec {a}, \vec {\alpha }, u$ as in the statement of the proposition.

Consequently, we have

$$\begin{aligned} K_u(v,v') = \frac{1}{2\pi {\mathsf {i}}} \int _{C_{b, \pi /4}} F(w,v',v)dw, \end{aligned}$$

where $F(w,v',v)$ is as in (6.26). Next, by Lemma 5.4, Lemma 5.6 and (6.27) we have

$$\begin{aligned} | e^{MG(v)} e^{M^{1/3}\sigma x (v-z_c)} | \le \exp ( - (\sqrt{2}/2)|v-z_c|^3 \sigma ^3 M/6 + CM^{2/3}) \le e^{CM^{2/3}},\nonumber \\ \end{aligned}$$

(6.42)

and also by Lemma 5.5, Lemma 5.6 and (6.27) provided that M is sufficiently large we have

$$\begin{aligned} | e^{-MG(w)} e^{-M^{1/3} \sigma (w-z_c) } | \le Ce^{-c M |w| \log (1 + |w|)} \text{ if } |w - b| \ge r/2 \text{ and } w \in C_{b, \pi /4}.\nonumber \\ \end{aligned}$$

(6.43)

One also observes that the inequalities in (6.29) and (6.30) are also satisfied if $|v - a| \le r$ and $|w - b| \ge r/2$, $w \in C_{b, \pi /4}$, which together imply that for all large M

$$\begin{aligned}&\left| \int _{C_{b, \pi /4} \cap B_{r/2}^c(b)} F(w,v',v)dw \right| \le e^{CM^{2/3}} \int _{C_{b, \pi /4} \cap B_{r/2}^c(b)} e^{-c M |w| \log (1 + |w|)}|dw| \nonumber \\&\quad \le Ce^{-c M}, \end{aligned}$$

(6.44)

where |dw| denotes integration with respect to arc-length.

On the other hand, if $v \in B_r(a)$ and $w \in B_{r/2}(b)$ we can perform the change of variables ${\tilde{v}} = (v - z_c)M^{1/3} \sigma$, ${\tilde{w}} = (w - z_c) M^{1/3} \sigma$ and apply Lemma 5.4 and Lemma 5.6 to obtain

$$\begin{aligned}&\left| \int _{C_{b, \pi /4} \cap B_{r/2}(b)} F(w,v',v)dw \right| \\&\quad \le C \int _{C_{\mu + \rho , \pi /4}} \left| \frac{1}{{\tilde{w}}M^{-1/3} \sigma ^{-1} + z_c - v'} \right| e^{c \log (1 + |{\tilde{v}}|) + c \log (1 + |{\tilde{w}}|)} \\&\qquad \times \frac{ \pi \cdot \exp \left( {-(\sqrt{2}/2)^3|{\tilde{v}}|^3/12 - (\sqrt{2}/2)^3|{\tilde{w}}|^3/12 + |{\tilde{w}}| A+ |{\tilde{v}}|A }\right) }{M^{1/3}\sigma | \sin (\pi M^{-1/3} \sigma ^{-1}({\tilde{v}} - {\tilde{w}})) |} |dw|. \end{aligned}$$

We remark that in deriving the last inequality we bounded the double product of gamma functions in the definition of $F(w,v',v)$ by $C e^{c \log (1 + |{\tilde{v}}|) + c \log (1 + |{\tilde{w}}|)}$. To see why such a bound holds we can use (6.30), and note that the right side of the top line is O(1) for $v \in B_r(a)$ and $w \in B_r(b)$ since $r \le 1/2$ by assumption, while the rational functions on the second line can be controlled by $Ce^{c \log (1 + |{\tilde{v}}|) + c \log (1 + |{\tilde{w}}|)}$. We also mention that the bound on

$$\begin{aligned} | \exp (M[G(z_c + {\tilde{v}} \sigma ^{-1} M^{-1/3}) - G(z_c + {\tilde{w}} \sigma ^{-1} M^{-1/3})])| \end{aligned}$$

via

$$\begin{aligned} C \exp \left( -(\sqrt{2}/2)^3|{\tilde{v}}|^3/12 - (\sqrt{2}/2)^3|{\tilde{w}}|^3/12 \right) \end{aligned}$$

can be obtained using only Lemma 5.4 in the case $|{\tilde{w}} - b| \le M^{1/3 - 1/10}$ and $|{\tilde{v}} - a| \le M^{1/3 - 1/10}$. When $M^{1/3 - 1/10} \le |{\tilde{w}} - b| \le (r/2) M^{1/3}$ and $M^{1/3 - 1/10} \le |{\tilde{v}} - a| \le r M^{1/3}$ one needs to further invoke Lemma 5.6. The way this estimate is established is very similar to what we earlier did in Step 4 of the proof of Theorem 3.2 in Sect. 3.3 so we omit the details.

From Lemma 3.13 we know that

$$\begin{aligned} \left| \frac{\pi M^{-1/3}\sigma ^{-1} }{\sin (\pi M^{-1/3} \sigma ^{-1}({\tilde{v}} - {\tilde{w}})) } \right| \le C, \end{aligned}$$

and also we know by definition of our contours that

$$\begin{aligned} \left| \frac{1}{{\tilde{w}}M^{-1/3} \sigma ^{-1} + z_c - v'} \right| \le CM^{1/3}. \end{aligned}$$

Combining the last few estimates we see that

$$\begin{aligned}&\left| \int _{C_{b, \pi /4} \cap B_{r/2}(b)} F(w,v',v)dw \right| \le CM^{1/3} \int _{C_{\mu + \rho , \pi /4}} e^{c \log (1 + |{\tilde{v}}|) + c \log (1 + |{\tilde{w}}|)} \cdot \nonumber \\&\quad e^{-(\sqrt{2}/2)^3|{\tilde{v}}|^3/12 - (\sqrt{2}/2)^3|{\tilde{w}}|^3/12+ |{\tilde{w}}| A + |{\tilde{v}}| A} |d{\tilde{w}}| \le CM^{1/3}, \end{aligned}$$

(6.45)

where the last inequality used the fact that $C_{\mu + \rho , \pi /4}$ does not depend on M and the integral is finite by the cube in the exponential. Equations (6.44) and (6.45) imply (6.24) when $|v - z_c| \le r$. This suffices for the proof. $\square$