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.
On
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
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 $\frac{3}{4}=\frac{1}{2}H_2$: $$\sum_{n\geq 1}\frac{1}{n(n+2)}=\frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+2}\right)=\frac{H_2}{2}=\frac{3}{4}.$$