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.

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.

