<s>[INST] 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 $\ln\!\Gamma$.
 [/INST] 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.
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