Evaluating $\int_{0}^{\infty}\frac{1}{x}\big (\frac{\sinh ax}{\sinh x}-ae^{-2x}\big )dx$ Some time ago, stumbled out of an integral:
$$\int_{0}^{\infty}\frac{1}x{}\left (\frac{\sinh ax}{\sinh x}-ae^{-2x}\right )dx=\ln\frac{\pi\cos\frac{a\pi}{2}}{\Gamma^2(\frac{a+1}{2})};\left | a \right |<1$$  
I have no idea where to start?
 A: To be specific, let $I(a)$ denote the integral and differentiate the integral with respect to $a$. Then by referring to the contents and notations in my blog posting,
\begin{align*}
I'(a)
&= \int_{0}^{\infty} \left( \frac{x \cosh ax}{\sinh x} - e^{-2x} \right) \, \frac{dx}{x} \\
&= \int_{0}^{\infty} \left( \frac{x e^{-(1-a)x} + x e^{-(1+a)x}}{1 - e^{-2x}} - e^{-2x} \right) \, \frac{dx}{x} \\
&= \int_{0}^{\infty} \left( \frac{1}{2} \frac{x e^{-(1-a)x/2} + x e^{-(1+a)x/2}}{1 - e^{-x}} - e^{-x} \right) \, \frac{dx}{x} \qquad (2x \mapsto x) \\
&= \frac{1}{2} \left\{ \mathrm{ctr} \, \left( \frac{x e^{-(1-a) x/2}}{1 - e^{-x}} \right) + \mathrm{ctr} \, \left( \frac{x e^{-(1+a) x/2}}{1 - e^{-x}} \right)  \right\} - \mathrm{ctr} \, (e^{-x}) \\
&= - \frac{1}{2} \left\{ \psi_{0}\left(\frac{1-a}{2}\right) + \psi_{0}\left(\frac{1+a}{2}\right) \right\}.
\end{align*}
Since $I(0) = 0$, we can retrieve $I(a)$ by a simple integration:
$$ I(a) = \int_{0}^{a} I'(t) \, dt = \log\Gamma\left(\frac{1-a}{2} \right) - \log\Gamma\left(\frac{1+a}{2} \right). $$
Now the rest follows by the Euler's reflection formula.
A: Differentiate wrt to a to get a tabulated integral and then reintegrate to find $$(-1/2)\int_0^a[\psi((1+x)/2)+\psi((1-x)/2)]dx=\log \left( \Gamma\left(\frac{1-a}{2}\right)\Gamma\left(\frac{1+a}{2} \right)\right)$$
