<s>[INST] Evaluation of $\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx$ I want to show that
$$\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx=\frac{\psi^{(1)}(\frac{t}{2\pi})-4\psi^{(1)}(\frac{t}{\pi})}{8\pi}t+\frac14
$$
where $\psi^{(1)}(z)$ is the first derivative of the digamma function, but I've never come across such an integral and don't know what to do. Any help  would be greatly appreciated.
 [/INST] We first modifiy the integral to be calculated, by remarking that, changing $x=-u$,
\begin{align}
I&=\int^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)}\,dx\\
&=-\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{-tu} +1\right)}\,du\\
&=-\int_{-\infty }^{0}\frac{ue^{tu} }{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du\\
&=-\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2}\,du+\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du\\
&=\frac{1}{2}+I^-
\end{align}
where 
\begin{equation}
I^-=\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{equation} 
Then, we have
\begin{align}
I&=I^-+\frac{1}{2}\\
I+I^-&=\int_{-\infty }^{\infty}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{align}
Thus,
\begin{equation}
I=\frac{1}{4}+\frac{1}{2}\int_{-\infty }^{\infty}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{equation} 
This integral is evaluated by the residue method, by closing the contour with the upper half circle. It can easily be verified that the half-circle contribution vanishes. Interior poles lie at 


*

*$u=i$ with residue 
\begin{align}
R_i&=\frac{it\exp(it)}{4(1+\exp(it))^2}\\
&=\frac{it}{16\cos^2\left( t/2 \right)}\\
&=\frac{it}{4}\sum_{n=-\infty}^\infty\frac{1}{\left( t+\pi(2n+1) \right)^2}\\
&=\frac{it}{2}\sum_{n\ge0}\frac{t^2+\pi^2(2n+1)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}
\end{align}
(using the series representation for $\sec^2(z)$).

*$u=i(2n+1)\pi/t$, where $n=0,1,2,\ldots$ with residues 
\begin{align}
R_n&=\frac{-i\pi\left( 2n+1\right)t^2 }{\left(\left( (2n+1)i\pi \right)^2+t^2\right)^2}\\
&=\frac{-i\pi\left( 2n+1\right)t^2 }{\left(t^2-\pi^2(2n+1)^2\right)^2}
\end{align}
Then
\begin{align}
I&=\frac{1}{4}+\frac{1}{2}2i\pi\frac{it}{2}
\sum_{n\ge0}\left[\frac{t^2+\pi^2(2n+1)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}-\frac{2\pi\left( 2n+1\right)t }{\left(t^2-\pi^2(2n+1)^2\right)^2}
\right]\\
&=\frac{1}{4}-\frac{\pi t}{2}\sum_{n\ge0}\frac{\left( t-\pi(2n+1) \right)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}\\
&=\frac{1}{4}-\frac{\pi t}{2}\sum_{n\ge0}\frac{1}{\left( t+\pi(2n+1) \right)^2}
\end{align}
The summation can be written as
\begin{align}
I&=\frac{1}{4}-\frac{ t}{8\pi}\sum_{n\ge0}\frac{1}{\left( \frac{t}{2\pi}+n+\frac{1}{2} \right)^2}\\
&=\frac{1}{4}-\frac{ t}{8\pi}\psi^{(1)}\left( \frac{t}{2\pi}+\frac{1}{2} \right)
\end{align}
From the classical identity
\begin{equation}
\psi\left(2z\right)=\tfrac{1}{2}\left(\psi\left(z\right)+\psi\left(z+\tfrac{1}{2}\right)\right)+\ln 2
\end{equation} 
we obtain, after differentiation
\begin{equation}
\psi^{(1)}\left(z+\tfrac{1}{2}\right)=4\psi^{(1)}\left(2z\right)-\psi^{(1)}\left(z\right)
\end{equation} 
and thus
\begin{equation}
I=\frac{1}{4}-\frac{t}{8\pi}\left[4\psi^{(1)}\left( \frac{t}{\pi}\right)-\psi^{(1)}\left( \frac{t}{2\pi}\right)\right]
\end{equation} 
as expected.

</s>