How to prove that $-\frac{\pi}{4}\int_{-\infty}^\infty \frac{\psi\left(\tfrac12+ip\right)+\psi\left(\tfrac12-ip\right)+2\gamma}{\cosh^2\pi p}dp=1$? 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.
 A: 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) = $
$$(1,\frac{1}{36},-\frac{19}{3600},-\frac{1831}{141120},-\frac{1257}{44800},-\frac{451483}{5575680})$$

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 $\text{Li}_2(x) + \text{Li}_2(1-x) = \pi^2/6 - \log x \log(1-x)$.
