From Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$
In the case of zero $\omega$ and integral starts as 0, how do I prove that using contour integral
$\int_{0}^\infty \frac{\sinh(kx)}{\sinh(x)}dx = \frac{1}{2}\tan{\frac{k}{2}}$, $|k|<1$?
Thanks in advance.
For any $a\in(-1,1)$ and $k\in\mathbb{N}^+$ the identity $$\int_{0}^{+\infty}\sinh(ax)e^{-kx}\,dx = \frac{a}{k^2-a^2} \tag{1}$$ holds trivially, so by expanding $\frac{1}{\sinh x}$ as a geometric series we have that $$ \int_{0}^{+\infty}\frac{\sinh(ax)}{\sinh(x)}\,dx=\sum_{m\geq 0}\frac{2a}{(2m+1)^2-a^2}\tag{2} $$ but the RHS of $(2)$ is a well-known series, which can be obtained by applying $\frac{d}{dz}\log(\ldots)$ to the Weierstrass product for the cosine function, or by just invoking Herglotz' trick. The final outcome is $$\forall a\in(-1,1),\qquad \int_{0}^{+\infty}\frac{\sinh(ax)}{\sinh(x)}\,dx=\frac{\pi}{2}\tan\left(\frac{\pi a}{2}\right).\tag{3} $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sinh\pars{kx} \over \sinh\pars{x}} & = \mrm{sgn}\pars{k}\ \overbrace{\int_{0}^{\infty}{1 - \expo{-2\verts{k}x} \over 1 - \expo{-2x}} \expo{-\pars{1 - \verts{k}}x}\,\dd x} ^{\ds{\mbox{It converges whenever}\ \color{red}{\verts{k} < 1}}} \\[5mm] & \stackrel{\expo{-2x}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\,\mrm{sgn}\pars{k}\int_{0}^{1}{1 - x^{\verts{k}} \over 1 - x}\, x^{-1/2 - \verts{k}/2}\,\dd x \\[5mm] & = {1 \over 2}\,\mrm{sgn}\pars{k}\bracks{\int_{0}^{1}{1 - x^{-1/2 + \verts{k}/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-1/2 - \verts{k}/2} \over 1 - x}\,\dd x} \\[5mm] & = {1 \over 2}\,\mrm{sgn}\pars{k}\bracks{% \Psi\pars{{1 \over 2} + {\verts{k} \over 2}} - \Psi\pars{{1 \over 2} - {\verts{k} \over 2}}} \\[5mm] & = {1 \over 2}\,\,\mrm{sgn}\pars{k}\pi \cot\pars{\pi\bracks{{1 \over 2} - {\verts{k} \over 2}}} = {1 \over 2}\,\mrm{sgn}\pars{k}\tan\pars{{\pi \over 2}\,\verts{k}} \\[5mm] & = \bbx{{1 \over 2}\,\tan\pars{{\pi \over 2}\,k}\,,\qquad \verts{k} < 1} \end{align}
