Hyperbolic contour integration

Question

$$ \int_{-\infty}^\infty\frac{x\cos(ux)}{\sinh(x/2)}dx $$

If I'm correct, there is a pole at $z = i\pi$ ?

How can I evaluate this integral?

Answer 1

Note that $$ \int_{-\infty}^\infty\frac{x\cos(ux)}{\sinh(x/2)}dx=2\int_{-\infty}^\infty\frac{x\cos(2ux)}{\sinh(x)}dx.$$ Let $f(z)=\frac{z\cos(2uz)}{\sinh(z }$. Choose the rectangle $\gamma$: $z=-R$ to $z=R$ in $x$-axis, $z=R$ to $z=R+\pi i$ in the line $x=R$, $z=-R+\pi i$ to $z=-\varepsilon+\pi i$ and $z=\varepsilon+\pi$ $z=R+\pi i$ in the line $y=\pi i$, $z=-R+\pi i$ to $z=-R$ in the line $x=-R$, and a small semicircle $C_\varepsilon$ below the pole $z=\pi i$ with radius $\varepsilon$. Then $f(z)$ is analytic inside $\gamma$. Note that he contribution from the short sides of the rectangle is zero since $\sinh$ grows exponentially as $x\to\pm\infty$. Then one has
$$ \int_{-\infty}^\infty f(x)dx - \int_{\varepsilon}^\infty f(x+i\pi)dx -\int_{C_\varepsilon} f(x)dz - \int_{-\infty}^{-\varepsilon} f(x+i\pi)dx =0.$$ Note that $z=\pi i+\varepsilon e^{it}$ on $C_\varepsilon$, $t\in[\pi,2\pi]$. So $$ \int_\pi^{2\pi} \frac{(\varepsilon e^{it}+i\pi)\cos[2u(\varepsilon e^{it}+i\pi)]}{\sinh(\varepsilon e^{it}+i\pi)}i\varepsilon e^{it}dt \to \int_{\pi}^{2\pi} \pi\cosh(2\pi u) dt = \pi^2\cosh(2u\pi)$$ as $\varepsilon\to0$. Here $\sinh(x+i\pi) \approx -x$ near $x=\pi i$ is used.

Thus we have $$ \pi^2\cosh(2u\pi) = \int_{-\infty}^\infty f(x)dx - \lim_{\epsilon\to 0}\left(\int_{-\infty}^{-\epsilon} f(x+i\pi)dx+\int_{\epsilon}^\infty f(x+i\pi)dx\right). $$

Using $$\sinh(x+i\pi) = -\sinh(x),\cos(xi)=\cosh(x),\sin(xi)=i\sinh(x)$$ and the principle integral, it is easy to see \begin{eqnarray*} &&\int_{-\infty}^{-\epsilon} f(x+i\pi)dx+\int_{\epsilon}^\infty f(x+i\pi)dx\\ &=&-\int_{-\infty}^{-\epsilon}\frac{(x+\pi i)\cos[2u(x+\pi i)]}{\sinh(x)}dx-\int_{\epsilon}^\infty \frac{(x+\pi i)\cos[2u(x+\pi i)]}{\sinh(x)}dx\\ &=&-\bigg(\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^\infty\bigg)\frac{(x+\pi i)[\cos(2ux)\cosh(2\pi u)-i\sin(2ux)\sinh(2\pi u)]}{\sinh(x)}dx\\ &=&-\bigg(\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^\infty\bigg)\frac{x\cos(2ux)\cosh(2\pi u)+\pi\sin(2ux)\sinh(2\pi u)}{\sinh(x)}dx\\ &&-i\bigg(\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^\infty\bigg)\frac{\pi \cos(2ux)\cosh(2\pi u)-x\sin(2ux)\sinh(2\pi u)}{\sinh(x)}dx\\ &=&-\bigg(\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^\infty\bigg)\frac{x\cos(2ux)\cosh(2\pi u)}{\sinh(x)}dx\\ &\to&-\cosh(2\pi u)\int_{-\infty}^\infty f(x)dx \end{eqnarray*} as $\varepsilon\to0$. $$ \int_{-\infty}^\infty f(x)dx = \frac{\pi^2\cosh(2\pi u)}{1+\cosh(2\pi u)}.$$ So $$ \int_{-\infty}^\infty\frac{x\cos(ux)}{\sinh(x/2)}dx=\frac{2\pi^2\cosh(2\pi u)}{1+\cosh(2\pi u)}. $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \begin{array}{c}\hline \quad Zeros\ \mbox{of}\ \sinh\pars{x \over 2}\ \mbox{are given by}\ \braces{z_{n} = 2n\pi\ic\ \mid\ n \in \mathbb{Z}}\quad \\[2mm] \mbox{So,}\ \color{red}{\large \pi\ic}\ \mathbf{isn't}\ \mbox{a pole of the integrand.} \\ \hline \end{array} $$

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An alternative: \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{x\cos\pars{ux} \over \sinh\pars{x/2}}\,\dd x} = 2\int_{0}^{\infty} {x\pars{\expo{\ic\verts{u}x} + \expo{-\ic\verts{u}x}}/2 \over \pars{\expo{x/2} - \expo{-x/2}}/2}\,\dd x \\[5mm] = &\ 2\int_{0}^{\infty} {x\bracks{\expo{-\pars{1/2 - \ic\verts{u}}x} + \expo{-\pars{1/2 + \ic\verts{u}}x}} \over 1 - \expo{-x}}\,\dd x \\[5mm] \stackrel{t\ =\ \expo{-x}}{=}\,\,\,& 2\int_{1}^{0}{\bracks{-\ln\pars{t}} \pars{t^{1/2 - \ic\verts{u}} + t^{1/2 + \ic\verts{u}}} \over 1 - t}\,\pars{-\,{\dd t \over t}} \\[5mm] = & -2\,\int_{0}^{1}{\ln\pars{t} \pars{t^{-1/2 - \ic\verts{u}} + t^{-1/2 + \ic\verts{u}}} \over 1 - t}\,\dd t \\[5mm] = &\ -2\,\lim_{\nu \to 0}\partiald{}{\nu}\int_{0}^{1}{\pars{t^{\nu} - 1} \pars{t^{-1/2 - \ic\verts{u}} + t^{-1/2 + \ic\verts{u}}} \over 1 - t}\,\dd t \\[5mm] = &\ 2\,\lim_{\nu \to 0}\partiald{}{\nu}\bracks{% \int_{0}^{1}{1 - t^{\nu - 1/2 - \ic\verts{u}} \over 1 - t}\,\dd t + \int_{0}^{1}{1 - t^{\nu - 1/2 + \ic\verts{u}} \over 1 - t}\,\dd t} \\[5mm] = &\ 2\,\lim_{\nu \to 0}\partiald{}{\nu}\bracks{% \Psi\pars{\nu + {1 \over 2} - \ic\verts{\mu}} + \Psi\pars{\nu + {1 \over 2} + \ic\verts{\mu}}} \end{align}

where $\ds{\Psi}$ is the Digamma Function.

Then, \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{x\cos\pars{ux} \over \sinh\pars{x/2}}\,\dd x} = 2\bracks{\Psi'\pars{{1 \over 2} - \ic\verts{\mu}} + \Psi'\pars{{1 \over 2} + \ic\verts{\mu}}} \\[5mm] = &\ \bbx{2\pi^{2}\,\mrm{sech}^{2}\pars{\pi u}} \quad \pars{\begin{array}{l} \mbox{In the last line, I used the} \\[2mm] Euler\ Reflection\ Formula \end{array}} \end{align}

Answer 3

We have $$\frac {\sin u x} {\sinh(x/2)} = \frac {i (e^{-i u x} - e^{i u x})} {e^{x/2} - e^{-x/2}}, \\ \operatorname*{Res}_{x = 2 \pi i k} \frac 1 {e^{x/2} - e^{-x/2}} = (-1)^k.$$ Deform the contour of integration to $(-\infty, -\epsilon] \cup C_\epsilon \cup [\epsilon, \infty)$, where $C_\epsilon$ is a small half-circle in the lower half-plane centered at the origin. For $u > 0$, evaluate the integral containing $e^{i u x}$ ($e^{-i u x}$) by closing the contour into the upper (lower) half-plane: $$I(u) = \int_{\mathbb R} \frac {\sin u x} {\sinh(x/2)} dx = 2 \pi i \left( -i - 2 i \sum_{k \geq 1} (-1)^k e^{-2 \pi u k} \right) = \\ 2 \pi - \frac {4 \pi} {1 + e^{2 \pi u}} = 2 \pi \tanh \pi u.$$ The same holds for $u < 0$ because the integrand is an odd function of $u$. Then $$\int_{\mathbb R} \frac {x \cos u x} {\sinh(x/2)} dx = I'(u).$$