How to prove a generalized integral identity $$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})}dt=-\frac{1}{4}+\frac{\gamma}{2}
$$
where $\gamma$ = Euler Gamma
$$
\int_{0}^{\infty }\frac{t}{( e^{2\pi t}-1)(1+t^{2}) ^{2}}dt=\frac{\pi^2}{24} -\frac{3}{8}
$$
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(
1+t^{2})^{3}}dt=\frac{\pi^2}{96} +\frac{\zeta(3)}{8} -\frac{7}{32} 
$$
 A: By
$$
\int_{0}^{\infty }\frac{\arctan \frac{t}{z}}{e^{2\pi t}-1}dt=\frac{z-z\ln z}{2}-\frac{1}{4}\ln \frac{2\pi }{z}+\frac{1}{2}\ln \Gamma (z)      [{Re}z>0]
$$
we have
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) }dt=\frac{\ln z}{4}-\frac{1}{4z}-\frac{1}{2}\psi \left(
z\right), 
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{2}}dt=\frac{1}{4z}\psi ^{\prime }\left( z\right) -\frac{%
1}{4z^{2}}-\frac{1}{8z^{3}} ,
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{3}}dt=\frac{1}{16z^{3}}\psi ^{\prime }\left( z\right) -
\frac{1}{16z^{2}}\psi ^{\prime \prime }\left( z\right) -\frac{1}{8z^{4}}-
\frac{3}{32z^{5}} 
$$
and the recurrence formula
$$
I_{p}(z)=\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right)^p }dt=\frac{-1}{2pz}I_{p-1}^{\prime }(z).
$$
For $\psi ^{(k)}(z)$ we have the following identities
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(n+z)=\psi ^{(k)}(z)+k!(-1)^{k}\sum\limits_{l=1}^{n-1}\frac{1}{%
\left( {l+z}\right) ^{k+1}}, \\
\psi ^{(k)}(z-n)=\psi ^{(k)}(z)+k!\sum\limits_{l=1}^{n}\frac{1}{\left( {l-n-z%
}\right) ^{k+1}}.%
\end{array}
\end{equation}
and
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(0)=\left\{
\begin{array}{c}
-\gamma ,k=0 \\
k{{!(-1)^{k+1}}\zeta (k+1)},k>0%
\end{array}%
\right. , \\
\psi ^{(k)}(\frac{1}{2})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
k{{!(-1)^{k+1}}}\left( 2^{k+1}-1\right) {\zeta (k+1)},k>0%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{1}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
-\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{3}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right. 
\end{array}
\end{equation}
