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I have to find Fourier Sine Transform for the following function :

$$ f(x) = (\sinh(\pi x))^{-1}.$$

However, I'm having trouble solving the integral, which is

$$ \int_0^\infty \frac{\sin \alpha x}{\sinh\pi x}\ \mathsf dx.$$

I've already tried to turn trigonometric functions into exponents, switch to complex numbers, but unsuccessfully. Could you please give me some tips on how to approach solving this integral? Thank you in adnvance.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation} \mbox{Note that}\quad \int_{0}^{\infty}{\sin\pars{\alpha x} \over \sinh\pars{\pi x}}\,\dd x = {1 \over \pi}\int_{0}^{\infty}{\sin\pars{\nu x} \over \sinh\pars{x}} \,\dd x\,,\qquad\nu \equiv {\alpha \over \pi}\label{1}\tag{1} \end{equation}

Then, \begin{align} \int_{0}^{\infty}{\sin\pars{\nu x} \over \sinh\pars{x}}\,\dd x & = \Im\int_{0}^{\infty}{\expo{\ic\nu x} - 1 \over \pars{\expo{x} - \expo{-x}}/2} \,\dd x = 2\,\Im\int_{0}^{\infty}{\expo{-\pars{1 - \ic\nu}x} - \expo{-x} \over 1 - \expo{-2x}}\,\dd x \\[5mm] & \stackrel{t\ =\ \expo{{\large -2x}}}{=}\,\,\, 2\,\Im\int_{1}^{0}{t^{\pars{1 - \ic\nu}/2} - t^{1/2} \over 1 - t}\,\pars{-\,{1 \over 2t}}\,\dd t = \Im\int_{0}^{1}{t^{\pars{-1 - \ic\nu}/2} - t^{-1/2} \over 1 - t}\,\dd t \\[5mm] & = -\,\Im\int_{0}^{1}{1 - t^{\pars{-1 - \ic\nu}/2} \over 1 - t}\,\dd t \\[5mm] & = -\,{1 \over 2\ic}\bracks{\Psi\pars{{1 \over 2} - {\nu \over 2}\,\ic} - \Psi\pars{{1 \over 2} + {\nu \over 2}\,\ic}} \qquad\pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = -\,{1 \over 2\ic} \braces{\pi\cot\pars{\pi\bracks{{1 \over 2} + {\nu \over 2}\,\ic}}} \qquad\pars{~Euler\ Reflection\ Formula~} \\[5mm] & = {\pi \over 2\ic}\tan\pars{{\pi\nu \over 2}\,\ic} = {\pi \over 2}\tanh\pars{\pi\nu \over 2} \end{align}

With \eqref{1}:

$$ \bbx{\int_{0}^{\infty}{\sin\pars{\alpha x} \over \sinh\pars{\pi x}}\,\dd x = {1 \over 2}\,\tanh\pars{\alpha \over 2}} $$

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We can express this integral as a series: \begin{align} I(\alpha)&=\int_0^\infty \frac{\sin \alpha x}{\sinh\pi x}\,dx\\ &=2\int_0^\infty \frac{e^{-\pi x}\sin \alpha x}{1-e^{-2\pi x}}\,dx\\ &=2\int_0^\infty \sum_{n=0}^\infty e^{-(2n+1)\pi x}\sin \alpha x\,dx\\ &=2 \sum_{n=0}^\infty\int_0^\infty e^{-(2n+1)\pi x}\sin \alpha x\,dx\\ &=\sum_{n=0}^\infty\frac{2\alpha}{\alpha^2+(2n+1)^2\pi^2} \end{align} The obtained series is classically evaluated. See the answers here for example. Finally, $$I(\alpha)=\frac 12 \tanh \frac\alpha 2 $$

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