How to prove that $-\frac{\pi}{4}\int_{-\infty}^\infty \frac{\psi\left(\tfrac12+ip\right)+\psi\left(\tfrac12-ip\right)+2\gamma}{\cosh^2\pi p}dp=1$?

Question

The question is motivated by the study of spectral functions of a certain operator arising in a physics problem. The integral in the title is the first from a sequence of integrals $$ I_n=-\frac{\pi}{4}\int_{-\infty}^\infty \frac{\psi\left(\tfrac12+ip\right)+\psi\left(\tfrac12-ip\right)+2\gamma}{\cosh^2\pi p}p^{2n}dp\tag{$\spadesuit$}$$ that seem to be given by rational numbers: $$I_0\stackrel{?}{=}1 ,\qquad I_1\stackrel{?}{=}\frac{1}{36}, \qquad I_2\stackrel{?}{=}-\frac{19}{3600}, \qquad \ldots $$ Simpler integrals $J_n=\displaystyle\int_{-\infty}^\infty \frac{p^{2n}dp}{\cosh^2\pi p}$ can be repackaged into a generating function $$G(x)=\frac{\pi}{2}\displaystyle \int_{-\infty}^\infty \frac{e^{2\pi p x}dp}{\cosh^2\pi p}=\frac{\pi x}{\sin\pi x},\qquad -1<\Re x<1$$ that can be easily computed by residues. Perhaps a similar trick combined with the recurrence relation $\psi(z+1)=\psi(z)+\frac1z$ or/and the series representation $\psi(z)+\gamma=\sum\limits_{n=0}^\infty\left(\frac{1}{n+1}-\frac{1}{n+z}\right)$ can be applied to ($\spadesuit$) but I was not successful with this approach so far, which is why I decided to challenge the non-artificial intelligence.

Answer 1

Denote $\text{Li}_2$ as dilogarithm $$-\frac{\pi}{4}\int_{-\infty}^\infty \frac{\psi\left(\tfrac12+ip\right)+\psi\left(\tfrac12-ip\right)+2\gamma}{\cosh^2\pi p}e^{iap} dp = \frac{e^{a/2} \left(a^2+4 \text{Li}_2\left(1-e^{-a}\right)\right)}{4 \left(e^a-1\right)} \qquad a\in \mathbb{R}$$ RHS is analytic at $a=0$, comparing coefficients $(I_0,\cdots, I_5) = $ $$(1,\frac{1}{36},-\frac{19}{3600},-\frac{1831}{141120},-\frac{1257}{44800},-\frac{451483}{5575680})$$

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Proof sketch: when the factor $e^{iap}, a>0$ is inserted, integral around big semicircle in upper half plane vanishes, integrand has triple poles at $i(\mathbb{Z}+1/2)$. Summing over residues, LHS is $$\sum_{n\geq 0} \frac{1}{4} e^{-a (n+\frac{1}{2})} \left(a^2-4 a H_n+4 (\frac{\pi ^2}{6}-H_n^{(2)})\right)$$ this sum can be done with $\sum_{n\geq 0} e^{-an} H_n^{(k)} = \frac{\text{Li}_k\left(e^{-a}\right)}{1-e^{-a}}$. A little issue is that $\text{Li}_2\left(e^{-a}\right)$ is not analytic at $a=0$, to overcome this, simply use $\text{Li}_2(x) + \text{Li}_2(1-x) = \pi^2/6 - \log x \log(1-x)$.

Answer 2

The following approach uses a rectangular contour in the upper half-plane of fixed height to evaluate $I_{0}$.

Let's first rewrite $I_{0}$ as

$$ I_{0}= -\frac{\pi}{4}\int_{-\infty}^\infty \frac{2 \, \psi\left(\frac{1}{2}-ip\right)+2\gamma}{\cosh^2\pi p}\, \mathrm dp \stackrel{?}{=}1 . $$

Since $$\int_{-\infty}^{\infty}\frac{1}{\cosh^{2} (\pi p)} \, \mathrm dp = \frac{2}{\pi}, $$ the goal is to show that $$\int_{-\infty}^{\infty} \frac{\psi \left(\frac{1}{2}- ip \right)}{\cosh^{2}(\pi p)} \, \mathrm dp = - \frac{2}{\pi} \left(\gamma +1 \right).$$

Let's integrate the function $$f(z) = \frac{z \, \psi \left(\frac{1}{2}- iz \right)}{\cosh^{2}(\pi z)} $$ around a rectangular contour with vertices at $z= \pm R$, $z= \pm R + i$.

There is a double pole inside the contour at $z= \frac{i}{2}$.

And the integral vanishes on the left and right sides of the contour as $R \to \infty$ since $\cosh^{2}(\pi z)$ grows exponentially as $\Re(z) \to \pm \infty$.

We therefore have $$ \begin{align} \int_{-\infty}^{\infty} \frac{x \, \psi \left(\frac{1}{2}- ix \right)}{\cosh^{2}(\pi x)} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{(x+i) \, \psi \left(\frac{3}{2}- ix \right)}{\cosh^{2}(\pi x)} \, \mathrm dx &= 2 \pi i \operatorname{Res}[f(z), i/2] \\ &= 2 \pi i \left(- \frac{\psi(1) + \frac{1}{2} \, \psi^{(1)}(1)}{\pi^{2}} \right) \\ &= i \left(\frac{2 \gamma}{\pi}- \frac{\pi}{6} \right). \end{align}$$

Using the functional equation $\psi(z+1) = \psi(z) + \frac{1}{z} $, the left side of the above equation can be rewritten as $$\small 2 \int_{-\infty}^{\infty} \frac{x}{\left( 4x^{2}+1 \right) \cosh^{2}(\pi x)} \, \mathrm dx - i\int_{-\infty}^{\infty} \frac{\mathrm dx }{\left(4x^{2}+1\right)\cosh^{2}(\pi x)} \, - i \int_{-\infty}^{\infty} \frac{\mathrm dx}{\cosh^{2}(\pi x)} - i \int_{0}^{\infty} \frac{\psi \left(\frac{1}{2}-ix \right)}{\cosh^{2}(\pi x)} \, \mathrm dx .$$

Since the value of first integral is zero, we have $$\int_{-\infty}^{\infty} \frac{\psi \left(\frac{1}{2}-ix \right)}{\cosh^{2}(\pi x)} \, \mathrm dx = \frac{\pi}{6} - \frac{2 \gamma}{\pi} - \frac{2}{\pi} - \int_{-\infty}^{\infty} \frac{\mathrm dx}{\left(4x^{2}+1 \right)\cosh^{2}(\pi x)} .$$

To evaluate the integral $$\int_{-\infty}^{\infty} \frac{\mathrm dx}{\left(4x^{2}+1 \right)\cosh^{2}(\pi x)} $$ we can use Plancherel's theorem.

Since the Fourier transform of $\frac{1}{\cosh^{2} (\pi x)}$ is $\frac{\omega}{\sqrt{2} \pi^{3/2} \sinh(\frac{\omega}{2})} $ and the Fourier transform of $\frac{1}{4x^{2}+1} $ is $\frac{1}{2} \sqrt{\frac{\pi}{2}}e^{-|\omega|/2} $, we have

$$ \begin{align} \int_{-\infty}^{\infty} \frac{\mathrm dx}{\left(4x^{2}+1 \right)\cosh^{2}(\pi x)} &= \frac{1}{4\pi} \int_{-\infty}^{\infty}\frac{xe^{-|x|/2}}{\sinh(\frac{x}{2})} \, \mathrm dx \\ &= \frac{1}{2 \pi} \int_{0}^{\infty} \frac{x e^{-x/2}}{\sinh(\frac{x}{2})} \, \mathrm dx \\ &= \frac{1}{\pi} \int_{0}^{\infty} \frac{x e^{-x}}{1-e^{-x}} \, \mathrm dx \\ &= \frac{1}{\pi} \int_{0}^{\infty} x e^{-x} \sum_{n=0}^{\infty} e^{-nx} \, \mathrm dx \\ &= \frac{1}{\pi} \sum_{n=0}^{\infty} \int_{0}^{\infty} x e^{-(n+1)x} \, \mathrm dx \\ &= \frac{1}{\pi} \sum_{n=0}^{\infty} \frac{1}{\left( n+ 1\right)^{2}} \\ &= \frac{\pi}{6}. \end{align}$$

Therefore, $$\int_{-\infty}^{\infty} \frac{\psi \left(\frac{1}{2}- ip \right)}{\cosh^{2}(\pi p)} \, \mathrm dp = \frac{\pi}{6} - \frac{2 \gamma}{\pi} - \frac{2}{\pi} -\frac{\pi}{6} = - \frac{2}{\pi} \left( \gamma +1\right) , $$ which is what we wanted to show.

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In general, for $a>0$, we have

$$ \int_{-\infty}^{\infty} \frac{\psi \left(a- ip \right)}{\cosh^{2}(\pi p)} \, \mathrm dp = \frac{1}{\pi} \left(2 \psi\left(a+ \frac{1}{2} \right)+ (2a-1) \psi^{(1)}\left(a+ \frac{1}{2} \right) - 2 \right).$$