A Challenge on One Integral Problem I want to show that if
$$I(t)=\int_0^\infty e^{-t^2x}\,\frac{\sinh(2tx)}{\sinh(x)}\,dx$$
then for $t^2\neq1$
$$I(t)=4t\sum_{n=0}^\infty \frac{1}{\left(2n+1+t^2\right)^2-4t^2}$$
Finally, show that
$$\lim_{t\to1}\left[I(t)-\frac{4t}{\left(t^2-1\right)^2}\right]=\frac{3}{4}$$
Could anyone here help me to solve it? Thanks.
 A: For the first identity, it is sufficient to write $\frac{e^{-t^2 x}}{\sinh x}$ as a geometric series and integrate it termwise against $\sinh(2tx)$:
$$\frac{e^{-t^2 x}}{\sinh x}=2\frac{e^{-(t^2+1)x}}{1-e^{-2x}}=2\sum_{n=0}^{+\infty}e^{-(t^2+2n+1)x},$$
$$\int_{0}^{+\infty}e^{-(t^2+2n+1)x}\sinh(2tx)\,dx=\frac{2t}{(t^2+2n+1)^2-4t^2}.$$
For the second point, it is sufficient to notice that:
$$I(t)-\frac{4t}{(1-t^2)^2}=4t\sum_{n\geq 1}\frac{1}{(2n+1+t^2)^2-4t^2}$$
hence the limit of the RHS for $t\to 1$ is given by:
$$\sum_{n\geq 1}\frac{4}{(2n+2)^2-4}=\sum_{n\geq 1}\frac{1}{n(n+2)}$$
that is a well-known telescopic series, converging to $\frac{3}{4}=\frac{1}{2}H_2$:
$$\sum_{n\geq 1}\frac{1}{n(n+2)}=\frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+2}\right)=\frac{H_2}{2}=\frac{3}{4}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
{\rm I}\pars{t}&
=\color{#66f}{\large\int_{0}^{\infty}\expo{-t^{2}x}\,{\sinh\pars{2tx} \over \sinh\pars{x}}\,\dd x}\ =\ 
\overbrace{\int_{0}^{\infty}\expo{-t^{2}x}\,
{\expo{2tx} - \expo{-2tx} \over \expo{x} - \expo{-x}}\,\dd x}
^{\ds{\color{#c00000}{\expo{-x}\equiv y\ \imp\ x = -\ln\pars{y}}}}
\\[5mm]&=\int_{1}^{0}y^{\, t^{2}}\,{y^{\, -2t} - y^{\, 2t} \over 1/y - y}
\,\pars{-\,{\dd y \over y}}\ =\
\overbrace{\int_{0}^{1}{y^{\, t^{2} - 2t} - y^{\,t^{2} + 2t} \over 1 - y^{2}}\,\dd y}
^{\ds{\color{#c00000}{y^{2}\ \mapsto\ y}}}
\\[5mm]&=\int_{0}^{1}{y^{\, t^{2}/2\ -\ t} - y^{\,t^{2}/2\ +\ t} \over 1 - y}
\,\half\,y^{-1/2}\,\dd y
=\half\int_{0}^{1}
{y^{\,\pars{t^{2} - 1}/2\ -\ t} - y^{\,\pars{t^{2} - 1}/2\ +\ t} \over 1 - y}\,\dd y
\\[5mm]&=\half\int_{0}^{1}
{1 - y^{\,\pars{t^{2} - 1}/2 + t} \over 1 - y}\,\dd y
-\half\int_{0}^{1}
{1 - y^{\,\pars{t^{2} - 1}/2 - t} \over 1 - y}\,\dd y
\\[5mm]&=\color{#66f}{\large\half\,\Psi\pars{{t^{2} + 1 \over 2} + t}
-\half\,\Psi\pars{{t^{2} + 1 \over 2} - t}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the Digamma Function and we used the identity
$\ds{\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t = \Psi\pars{z} + \gamma}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant. Then,

\begin{align}
{\rm I}\pars{t}&
=\color{#66f}{\large\int_{0}^{\infty}\expo{-t^{2}x}\,{\sinh\pars{2tx} \over \sinh\pars{x}}\,\dd x}
\\[5mm]&=\color{#66f}{\large\half\,\Psi\pars{\bracks{t + 1}^{2} \over 2}
-\half\,\Psi\pars{\bracks{t - 1}^{2} \over 2}}
\end{align}

\begin{align}&\color{#66f}{\large%
\lim_{t\ \to\ 1}\bracks{{\rm I}\pars{t} - {4t \over \pars{t^{2} - 1}^{2}}}}
=\lim_{t\ \to\ 1}\bracks{\half\,\Psi\pars{\bracks{t + 1}^{2} \over 2}
-\half\,\Psi\pars{\bracks{t - 1}^{2} \over 2} - {4t \over \pars{t^{2} - 1}^{2}}}
\\[5mm]&=\lim_{t\ \to\ 1}\bracks{\half\,\Psi\pars{2}
-\half\,\Psi\pars{\bracks{t - 1}^{2} + 2 \over 2}+\half\,{2 \over \pars{t - 1}^{2}} - {4t \over \pars{t^{2} - 1}^{2}}}
\\[5mm]&={\Psi\pars{2} - \Psi\pars{1} \over 2}
+\lim_{t\ \to\ 1}{\pars{t - 1}^{2} \over \pars{t^{2} - 1}^{2}}
={1 \over 2} + \lim_{t\ \to\ 1}{1 \over \pars{t +  1}^{2}}
=\color{#66f}{\large{3 \over 4}}
\end{align}
