Integral of $\ln(x)\operatorname{sech}(x)$ How can I prove that:
$$\int_{0}^{\infty}\ln(x)\,\operatorname{sech}(x)\,dx=\int_{0}^{\infty}\frac{2\ln(x)}{e^x+e^{-x}}\,dx\\=\pi\ln2+\frac{3}{2}\pi\ln(\pi)-2\pi\ln\!\Gamma(1/4)\approx-0.5208856126\!\dots$$
I haven't really tried much of anything worth mentioning; I've had basically no experience with $\ln\!\Gamma$.
 A: $$
\begin{align}
\int_0^\infty\frac{2\log(x)}{e^x+e^{-x}}\,\mathrm{d}x
&=\frac{\partial}{\partial t}\int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x
\end{align}
$$
$$
\begin{align}
\int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x
&=\int_0^\infty2x^te^{-x}\left(1-e^{-2x}+e^{4x}-\dots\right)\,\mathrm{d}x\\
&=2\Gamma(t+1)\left(1-\frac1{3^{t+1}}+\frac1{5^{t+1}}-\dots\right)\\[6pt]
&=2\Gamma(t+1)\beta(t+1)
\end{align}
$$
where $\beta(s)$ is the Dirichlet beta function.
Now we need to compute $\frac{\mathrm{d}}{\mathrm{d}x}\Gamma(x)\beta(x)$ at $x=1$.
$\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ as shown in this answer.
$\beta(1)=\frac\pi4$ using Gregory's Series. Finally, we need to compute
$$
\beta'(1)=\sum_{k=0}^\infty(-1)^k\frac{\log(2k+3)}{2k+3}
$$
which I am attempting to do in a manner similar to this answer. I will try to finish this in a bit.
A: You can get the value of the integral you're interested in from the integral $$I(a) =\int_{0}^{\infty} \frac{\ln (1+\frac{x^{2}}{a^{2}})}{\cosh x} \, dx, \quad a>0.$$
Notice that $\lim_{a \to \infty} I(a) = 0$.
Differentiating under the integral sign, we get $$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{2a}{(a^{2}+x^{2})\cosh x} \, dx - \frac{2}{a} \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx  \\ &= \int_{0}^{\infty} \frac{2}{(1+u^{2})\cosh (au)} \, du - \frac{\pi}{a}.  \end{align}$$
From the answers to this question, we know that  $$\int_{0}^{\infty} \frac{2}{(1+u^{2}) \cosh (au)} \, du=  \psi\left(\frac{3}{4}+ \frac{a}{2 \pi} \right) - \psi \left(\frac{1}{4} + \frac{a}{2 \pi} \right). $$
Therefore, $$ \begin{align} I(a) &=  2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right) \right] - \pi \ln (a)+ C \\ &= 2 \pi \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] +C. \end{align}$$
Letting $ a \to \infty$, we get $$ 0 = 2 \pi \lim_{a \to \infty}  \log \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] + C.$$
Using Stirling's approximation formula for the gamma function, we see that $$ \frac{\Gamma(x+\frac{1}{2})}{\Gamma(x)} \sim \frac{\sqrt{\frac{2\pi}{x+1/2}} \left(\frac{x+1/2}{e} \right)^{x+1/2}}{\sqrt{\frac{2\pi}{x}} \left(\frac{x}{e} \right)^{x}} =\sqrt{x} \left(1+ \frac{1}{2x} \right)^{x} e^{-1/2} \sim \sqrt{x} .$$
Therefore, $$ \lim_{a \to \infty}  \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] = \lim_{a \to \infty} \ln \ \frac{\sqrt{\frac{1}{4} + \frac{a}{2 \pi}}}{\sqrt{a}} = \lim_{a \to \infty} \ln \sqrt{\frac{1}{4a}+\frac{1}{2 \pi}} = - \frac{\ln (2 \pi)}{2}, $$
which implies $$C= \pi \ln (2 \pi).$$
So we have $$I(a)  = 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)  - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right]  - \pi \ln (a) +  \pi \ln (2 \pi).$$
But since $$ \int_{0}^{\infty} \frac{\ln (a^{2}+x^{2})}{\cosh x} \, dx = I(a) + \ln(a^{2}) \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx =  I(a) + \pi \ln (a),$$ it follows that  $$2 \int_{0}^{\infty} \frac{\ln x}{\cosh x} \, dx = \lim_{a \to 0^{+}}  2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)  - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right]  +  \pi \ln (2 \pi). $$
The final step is to apply the reflection formula for the gamma function.

If we had started with the integral $\int_{0}^{\infty} \frac{\ln(a^{2}+x^{2})}{\cosh x} \, dx $, we wouldn't have had a known initial condition.
A: Just a partial answer for now.
We have to compute:
$$ I = \int_{0}^{+\infty}\frac{2\log x}{e^{x}+e^{-x}}\,dx = 2\frac{\partial}{\partial\alpha}\left.\left(\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz\,\right)\right|_{\alpha=0^+}$$
Since:
$$\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz=\int_{1}^{+\infty}\frac{\log^\alpha t}{t^2+1}\,dt=\int_{0}^{1}\frac{(-\log t)^\alpha}{1+t^2}\,dt$$
and
$$\int_{0}^{1}(-\log t)^\alpha\, t^{2n}\,dt = \frac{\Gamma(\alpha+1)}{(2n+1)^{\alpha+1}},$$
it follows that:
$$\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz = \Gamma(\alpha+1)\cdot\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{\alpha+1}}=\Gamma(\alpha+1)\cdot\beta(\alpha+1).\tag{1}$$
Now we "just" need to differentiate the RHS of $(1)$ with respect to $\alpha$ and take the limit as $\alpha\to 0^+$. With the aid of Mathematica (see here for the relation between the Dirichlet beta function and the Hurwitz zeta function) for computing the Taylor series of the $\Gamma$ and $\beta$ functions I got:
$$I=\frac{1}{2} \left(-\pi  (\gamma+\log 4)-\text{StieltjesGamma}\left[1,\frac{1}{4}\right]+\text{StieltjesGamma}\left[1,\frac{3}{4}\right]\right).$$
A: If several change of variables are made, it can be shown that this integral is equivalent to the famous Vardi integral, $\displaystyle \int_{\pi/4}^{\pi/2}\ln(\ln(\tan(x)))dx=\frac{\pi}{2}\ln\left(\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right)$, which has been done on the site. Well, it's twice the Vardi integral. 
Vardi's Integral: $\int_{\pi/4}^{\pi/2} \ln (\ln(\tan x))dx $
Begin with $\int_{\pi/4}^{\pi/2}\ln(\ln(\tan(x)))dx$ and let $1/t=\tan(x)$.
It then becomes:
$$\int_{0}^{1}\frac{\ln(\ln(1/t))}{t^{2}+1}dt$$
Let $u=1/t$ and it becomes:
$$\int_{1}^{\infty}\frac{\ln(\ln(u)))}{u^{2}+1}du$$
Now, let $w=\ln(u)$ and it becomes:
$$1/2\int_{0}^{\infty}\frac{\ln(w)}{\cosh(w)}dw$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x
    =\pi\ln\pars{2} + {3 \over 2}\,\pi\ln\pars{\pi}
    -2\pi\ln\pars{\Gamma\pars{1 \over 4}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x}
=\ \overbrace{2\int_{0}^{\infty}{\expo{x}\ln\pars{x} \over \expo{2x} + 1}\,\dd x}
^{\ds{\mbox{Set}\ \expo{x} \equiv t\ \imp\ x = \ln\pars{t}}}\ =\
\overbrace{2\int_{1}^{\infty}{t\ln\pars{\ln\pars{t}} \over t^{2} + 1}\,{\dd t \over t}}
^{\ds{t\ \mapsto\ {1 \over t}}}
\\[3mm]&=2\int_{1}^{0}{\ln\pars{\ln\pars{1/t}} \over 1/t^{2} + 1}\,
\pars{-\,{\dd t \over t^{2}}}
\end{align}

Then,
$$\begin{array}{|c|}\hline\\
\quad\color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x}
=2\int_{0}^{1}{\ln\pars{\ln\pars{1/t}} \over 1 + t^{2}}\,\dd t\quad
\\ \\ \hline
\end{array}
$$

I already evaluated this integral.
  The result is:
  $$
\color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x}
=\pi\ln\pars{\Gamma^{2}\pars{3/4} \over \root{\pi}}
$$

Also,
$$
\Gamma\pars{3 \over 4} = {\pi \over \Gamma\pars{1/4}\sin\pars{\pi/4}}
={\root{2}\pi \over \Gamma\pars{1/4}}
$$
such that

\begin{align}&\color{#66f}{\large\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x}
=\pi\ln\pars{{2\pi^{2} \over \Gamma^{2}\pars{1/4}}\,{1 \over \root{\pi}}}
=-\pi\ln\pars{\Gamma^{2}\pars{1/4} \over 2\pi^{3/2}}
\\[3mm]&=\color{#66f}{\large\pi\ln\pars{2} + {3 \over 2}\,\pi\ln\pars{\pi}
-2\pi\ln\pars{\Gamma\pars{1 \over 4}}}
\end{align}

