TITLE: Improper integral of $\log x \operatorname{sech} x$
QUESTION [7 upvotes]: How to prove the following?
$$ \int_0^\infty \log x \operatorname{sech}x\,dx = \frac{\pi}{2} \log\left( \frac{4\pi^3}{\Gamma(1/4)^4} \right)
$$
I obtained the right side with CAS. It seems like this function has many poles on the imaginary axis so a simple contour integral cannot be used. I also tried the following
$$\begin{align*}
\int_0^\infty \log x \operatorname{sech} x\, dx
&= \int_0^\infty \operatorname{sech} x \left. \frac{\partial}{\partial s} x^s \right|_{s=0} dx \\
&= \left. \frac{\partial}{\partial s} \int_0^\infty x^s \operatorname{sech} x\,dx \right|_{s=0}
\end{align*}$$
However, the last integral is very difficult to evaluate and contains terms of $\zeta$ functions.

REPLY [3 votes]: This direct proof was discovered by Abdulhafeez Ayinde Abdulsalam (see https://arxiv.org/abs/2203.02675).
Let $\displaystyle \Delta\left(a\right) = \int_0^{\infty} \frac{\log\left(x^2 +  a^2\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x$.
Then
\begin{align*}
\Delta\left(a\right) &= \int_0^{\infty} \frac{\log\left(\left|a\right| - ix\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x + \int_0^{\infty} \frac{\log\left(\left|a\right| + ix\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x
\end{align*}
where $i=\sqrt{-1}$.
\begin{align*}
\Delta\left(a\right) &= 2\int_0^{\infty} \frac{\log\left(\left|a\right| - ix\right)}{e^{-2 \pi x} + 1} e^{-\pi x} \, \mathrm{d}x +  2\int_0^{\infty} \frac{\log\left(\left|a\right| + ix\right)}{e^{-2 \pi x} + 1} e^{-\pi x} \, \mathrm{d}x
\\&= -\frac{2}{\pi}\int_0^{\infty} \log\left(\left|a\right| - ix\right) \,\, \mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right) - \frac{2}{\pi}\int_0^{\infty} \log\left(\left|a\right| + ix\right) \,\, \mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right)
\\&=\frac{-2i}{\pi}\int_0^{\infty} \frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right| - ix}\, \mathrm{d}x + \frac{2i}{\pi}\int_0^{\infty} \frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right| + ix} \, \mathrm{d}x + \ln{a}
\end{align*}
Therefore
\begin{align*}
\Delta\left(a\right) - \ln{a} &= \frac{-2i}{\pi}\int_0^{\infty} \arctan\left(e^{-\pi x}\right) \int_0^{\infty} e^{-t\left(\left|a\right| - ix\right)} \,\, \mathrm{d}t \, \mathrm{d}x 
\\&\quad+ \frac{2i}{\pi}\int_0^{\infty} \arctan\left(e^{-\pi x}\right) \int_0^{\infty} e^{-t\left(\left|a\right| + ix\right)} \,\, \mathrm{d}t \, \mathrm{d}x
\\&= \frac{-2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} e^{itx} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t 
\\&\quad+\frac{2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} e^{-itx} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t
\\&= \frac{-2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \left(e^{itx} - e^{-itx}\right) \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t
\end{align*}
By Euler's formula,
$$e^{itx} - e^{-itx} = 2i\sin{\left(tx\right)}.$$
Therefore
\begin{align*}
\Delta\left(a\right) - \ln{a} &= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \sin{\left(tx\right)} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t 
\\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \mathrm{d}\left(\frac{-\cos{\left(tx\right)}}{t}\right) \arctan\left(e^{-\pi x}\right) \,\, \mathrm{d}t
\\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\left( \frac{\pi}{4t} - \frac{\pi}{2t}\int_0^{\infty}\frac{\cos{\left(tx\right)}}{\cosh\left(\pi x\right)} \, \mathrm{d}x\right) \,\, \mathrm{d}t 
\\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\left( \frac{\pi}{4t} - \frac{1}{2t}\int_0^{\infty}\frac{\cos{\left(\frac{tx}{\pi}\right)}}{\cosh\left(x\right)}\, \mathrm{d}x\right) \,\, \mathrm{d}t
\\&= \frac{4}{\pi}\int_0^{\infty} e^{-\left|a\right|t} \left(\frac{\pi}{4t} - \frac{\pi}{4t}\mathrm{sech}\left(\frac{t}{2}\right)\right) \,\, \mathrm{d}t 
\\&= \int_0^{\infty} e^{-\left|a\right|t}\left(\frac{1}{t} - \frac{1}{t}\mathrm{sech}\left(\frac{t}{2}\right)\right) \,\, \mathrm{d}t 
\\&=  \int_0^{\infty} \left(\frac{e^{-\left|a\right|t}}{t} - \frac{2e^{-\left(\left|a\right| + \frac{1}{2}\right)t}}{t\left(1 + e^{-t}\right)} \right)\,\, \mathrm{d}t {\stackrel{\,\,t \rightarrow 2t}{=}} \int_0^{\infty} \left(\frac{e^{-2\left|a\right|t}}{t} - \frac{2e^{-\left(2\left|a\right| + 1\right)t}}{t\left(1 + e^{-2t}\right)} \right)\,\, \mathrm{d}t  
\\&{\stackrel{z \rightarrow e^{-t}}{=}} -\int_0^{1} \left(z^{2\left|a\right|} - \frac{2z^{2\left|a\right|+1}}{1 + z^2} \right)\,\, \frac{\mathrm{d}z}{z\ln{z}} = -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1}{z} - \frac{2}{1 + z^2}\right)\,\, \mathrm{d}z 
\\&= -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1 + z^2 - 2z}{z\left(1 + z^2\right)}\right)\,\, \mathrm{d}z = -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{\left(1 - z\right)^2}{z\left(1 + z^2\right)}\right)\,\, \mathrm{d}z 
\\&= \int_0^{1} \frac{z^{2\left|a\right| - 1}\left(1 - z\right)}{1 + z^2}\int_0^1 z^p \mathrm{d}p\,\, \mathrm{d}z = \int_0^{1}\int_0^1 \frac{z^{2\left|a\right| + p - 1}\left(1 - z\right)}{1 + z^2} \mathrm{d}p\,\, \mathrm{d}z  
\\&= \int_0^{1}\int_0^1 \sum_{k=0}^{\infty} \left(-1\right)^k z^{2\left|a\right| + p + 2k - 1}\left(1 - z\right) \mathrm{d}z\,\, \mathrm{d}p \\&= \int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \int_0^1 z^{2\left|a\right| + p + 2k - 1}\left(1 - z\right) \mathrm{d}z\,\, \mathrm{d}p 
\\&= \int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \left(\frac{1}{2\left|a\right| + p + 2k} - \frac{1}{2\left|a\right| + p + 2k + 1}\right)\, \mathrm{d}p 
\\&= \frac{1}{2}\int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \left(\frac{1}{k + \frac{2\left|a\right| + p}{2}} - \frac{1}{k + \frac{2\left|a\right| + p + 1}{2}}\right)\, \mathrm{d}p 
\\&= -\frac{1}{4}\int_0^{1} \left(\psi_0\left(\frac{2\left|a\right| + p}{4}\right) - \psi_0\left(\frac{2\left|a\right| + p}{4} + \frac{1}{2}\right) \right.
\\&\qquad\qquad\qquad\left.- \psi_0\left(\frac{2\left|a\right| + p + 1}{4}\right) + \psi_0\left(\frac{2\left|a\right| + p + 1}{4} + \frac{1}{2}\right)\right)  \mathrm{d}p
\\&= -\log\left(\frac{\Gamma\left(\frac{2\left|a\right| + p}{4}\right)\Gamma\left(\frac{2\left|a\right| + p+ 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + p}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + p + 1}{4}\right)}\right)\biggr\vert_0^1 
\\&= - \log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)\Gamma\left(\frac{2\left|a\right| + 2}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right) 
\\&= - \log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2\Gamma\left(\frac{2\left|a\right| + 2}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2 \Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)}{\Gamma\left(\frac{2\left|a\right|}{4} + \frac{1}{2}\right)}\right) 
\\&=  -\log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2\Gamma\left(\frac{\left|a\right|}{2} + 1 \right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2 \Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{\left|a\right|}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) 
\\&= -\log\left(\frac{\left|a\right|\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2}{2\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2}\right) = 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right).
\end{align*}
Hence
\begin{align}
\Delta\left(a\right) &= 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right) + \ln{\left|a\right|} 
\\&= 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right).\tag{1}\label{rzq}
\end{align}
Taking the limit of \eqref{rzq} as $a \to 0$, we have
$$\int_0^{\infty} \frac{\log\left(x\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x = \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right),$$
which implies
$$ \int_0^{\infty} \ln{x}\mathrm{sech}\left(\pi x\right)\mathrm{d}x= \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
$$ \frac{1}{\pi}\int_0^{\infty} \ln\left(\frac{x}{\pi}\right)\mathrm{sech}\left(x\right)\mathrm{d}x = \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
$$\int_0^{\infty} \ln\left(\frac{x}{\pi}\right)\mathrm{sech}\left(x\right)\mathrm{d}x = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
$$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x - \ln{\pi}\int_0^{\infty} \mathrm{sech}\left(x\right)\mathrm{d}x = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
$$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x - \frac{\pi}{2}\ln{\pi} = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$
$$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x= \pi\log\left(\frac{\sqrt{2}\pi^{\frac{1}{2}}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right) = \pi\log\left(\frac{2\pi^{\frac{3}{2}}}{\Gamma\left(\frac{1}{4}\right)^2}\right).$$