Let $f_\tau(t)=\frac {\sin t \tau}{e^{t/2}-e^{-t/2}}$ and show that $I :=\int_{-\infty}^\infty f_\tau(t)\,dt = \pi \tanh \pi \tau$.

My attempt:

Consider the integral $\int_R f_\tau(t)\,dt$ over a large rectangle $R$ over the complex numbers going from $-K$ to $K$ to $K+4\pi i$ to $-K+4\pi i$ to $-K$ and let $K\rightarrow \infty$.

Since $f_\tau(t)$ has poles at $0, 2\pi i, 4\pi i$, we have that $$\int_R f_\tau(t)\,dt = \frac 1 2(\operatorname{res}_{0}f_\tau(t) + \operatorname{res}_{4\pi i}f_\tau(t)) + \operatorname{res}_{2\pi i}f_\tau(t).$$

Since the integrals over the vertical sides go to zero,

$$I - \int \frac {e^{-4\pi \tau}e^{it\tau}-e^{4\pi\tau}e^{-it\tau}}{e^{t/2}-e^{-t/2}}\,dt=\frac 1 2(\operatorname{res}_{0}f_\tau(t) + \operatorname{res}_{4\pi i}f_\tau(t)) + \operatorname{res}_{2\pi i}f_\tau(t).$$

Moreover,

$$\operatorname{res}_{0}f_\tau(t) = \lim_{t\rightarrow 0}\,(t-0)\frac{\sin t\tau}{e^{t/2}-e^{-t/2}} = 0.$$

If I am on the right path, I'd like to ask for a hint for the calculation of the other two residues and the integral over the top of the rectangle.

  • 2
    This is an indefinite integral, but the result given in your formula does not depend on $t$. So something is not right. – Hans Engler Sep 9 '16 at 12:13
  • 1
    Are you considering the definite integral $\int_{- \infty}^{+\infty}$? Othrwise this doesn't make sense. – Crostul Sep 9 '16 at 12:20
  • $e^{t/2}-e^{t/2}=0$... ^L^ – tired Sep 9 '16 at 12:41
  • 1
    @HansEngler, Did I clarify the problem you pointed out? – Rodrigo Sep 10 '16 at 9:48
  • 1
    @AméricoTavares, yes! – Rodrigo Sep 10 '16 at 12:39
up vote 1 down vote accepted

\begin{align} I &= \int_{-\infty}^{+\infty}dt\,\frac{\sin(t\tau)}{e^{t/2}-e^{-t/2}}\\ &= \int_{-\infty}^{+\infty}dt\,\frac{1}{2i}\frac{e^{it\tau}}{\sinh(t/2)} \end{align}

The function $$ g_{\tau}(t) = \frac{1}{2i}\frac{e^{it\tau}}{\sinh(t/2)} $$ has poles at $t_n = 2\pi i n$ for all integers $n$. The residue at $t = t_n$ can be calculated as: \begin{align} Res\{2\pi i n\} &=\lim_{t\rightarrow 2\pi i n} \frac{1}{2i} \frac{(t - 2\pi i n)\,e^{i t \tau}}{2 \sinh(t/2)}\\ &= \lim_{t\rightarrow 2\pi i n} \frac{1}{2i}\frac{(t - 2\pi i n)\,e^{-2\pi n \tau}}{\sinh(\pi i n) + \frac{1}{2}\cosh(\pi i n)(t - 2\pi i n)\,+\,...}\\ &= \lim_{t\rightarrow 2\pi i n} \frac{1}{2i}\frac{(t - 2\pi i n)\,e^{-2\pi n \tau}}{0 + \frac{1}{2}{(-1)}^n(t - 2\pi i n)\,+\,...}\\ &=\frac{1}{i} {(-1)}^ne^{-2\pi n \tau} \end{align} In the second line above, I have taken a Taylor series around $t = 2\pi i n$ to the necessary number of terms on the bottom.

For $\tau>0$ take a contour enclosing the upper half complex plane. This contour encloses all the poles with $n>0$, and passes through the $n=0$ pole at $t = 0$. The $t=0$ pole contributes half its residue to the integral. The value of the integral is thus: \begin{align} I &= 2\pi i \left(\sum_{n = 1}^{\infty} Res\{2\pi i n\}\;+\; \frac{1}{2}Res\{0\}\right)\\ &= 2\pi i \left(\sum_{n = 1}^{\infty} \frac{1}{i} {(-1)}^n e^{-2\pi n \tau} + \frac{1}{2}\frac{1}{i}\right)\\ &= \pi + 2\pi\sum_{n = 1}^{\infty} {(-1)}^n e^{-2\pi n \tau}\\ &= \pi - 2\pi \frac{e^{-2\pi n \tau}}{1+e^{-2\pi n \tau}}\\ &= \pi \tanh(\pi \tau) \end{align}

For $\tau<0$ the same procedure - except with a contour enclosing the lower half of the complex plane - yields the same result.

Edited to add:

Here's another method. $$ \frac{\sin(t\tau)}{e^{t/2}-e^{-t/2}} = \frac{e^{-t/2}\sin(t\tau)}{1-e^{-t}} = e^{-t/2}\sin(t\tau)\sum_{n = 0}^\infty e^{-nt} $$ Thus: \begin{align} I &= 2\int_0^{\infty}dt\, \frac{\sin(t\tau)}{e^{t/2}-e^{-t/2}}\\ &=2\sum_{n=0}^{\infty}\int_0^{\infty}dt\,e^{-(n+\frac{1}{2})t} \sin(t\tau)\\ &=2\sum_{n=0}^{\infty} \frac{\tau}{\tau^2+{(n+\frac{1}{2})}^2}\\ &=\sum_{n=-\infty}^{\infty} \frac{\tau}{\tau^2+{(n+\frac{1}{2})}^2} \end{align}

This final sum can be evaluated in various ways, such as by the Poisson summation formula, to yield $I = \pi\tanh(\pi\tau)$.

  • I used $\tanh(\pi \tau) = \frac {1-e^{-2\pi \tau}}{1+e^{-2\pi \tau}} = \frac {e^{2\pi \tau}-1}{e^{2\pi \tau}-1}$ to get that for $\tau>0, \tanh(\pi \tau) =\sum_{n=0}^\infty (-e^{-2\pi\tau})^n + (-e^{-2\pi\tau})^{n+1}$ and for $\tau>0, \tanh \pi \tau = -\sum_{n=0}^\infty(-e^{2\pi\tau})^n+(-e^{2\pi\tau})^{n+1}$. But how do I get that for any $\tau$ $\tanh \pi \tau = \sum_{n=1}^\infty(-e^{2\pi\tau})^n-(-e^{-2\pi\tau})^n$? – Rodrigo Sep 13 '16 at 23:48
  • You're right, it's not quite right, is it? I'll fix it later... – John Barber Sep 14 '16 at 0:58
  • @Rodrigo Fixed, I think. I had been sloppy about convergence. – John Barber Sep 14 '16 at 2:11

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \bbox[#fee,5px,border:2px groove navy]{\quad% \mbox{Hereafter, we evaluate the integral without using the Residue Theorem.}\quad} $$

$\ds{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over \expo{t/2} - \expo{-t/2}}\,\dd t = \pi\tanh\pars{\pi\tau}:\ ?}$.

\begin{align} \color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over \expo{t/2} - \expo{-t/2}}\,\dd t} & = 2\int_{0}^{\infty}{\sin\pars{\tau t}\expo{-t/2} \over 1 - \expo{-t}}\,\dd t = 2\sum_{n = 0}^{\infty}\, \Im\int_{0}^{\infty}\expo{-\pars{n + 1/2 - \tau\,\ic}t}\,\,\,\,\dd t \\[5mm] & = -\ic\sum_{n = 0}^{\infty}\, \pars{{1 \over n + 1/2 - \tau\,\ic} - {1 \over n + 1/2 + \tau\,\ic}}\dd t \\[5mm] & = -\ic\bracks{\Psi\pars{\half + \tau\,\ic} - \Psi\pars{\half - \tau\,\ic}}\qquad \pars{~\Psi:\ Digamma\ Function~} \end{align} With Euler Reflection Formula: $$ \color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over \expo{t/2} - \expo{-t/2}}\,\dd t} = -\ic\braces{\pi\cot\pars{\pi\bracks{\half - \tau\,\ic}}} = \color{#f00}{\pi\tanh\pars{\pi\tau}} $$

For a rectangular-like contour, consider

$$\oint_C dz \frac{\sin{\tau z}}{\sinh{(z/2)}} $$

where $C$ is the rectangle with vertices $-R$, $R$, $R+i 2 \pi$, $-R+i 2 \pi$, modified by a semicircular detour around $z=i 2 \pi$ of radius $\epsilon$. The contour integral is therefore

$$\int_{-R}^R dx \frac{\sin{\tau x}}{\sinh{(x/2)}} + i \int_0^{2 \pi} dy \frac{ \sin{\tau (R+i y)}}{\sinh{((R+i y)/2)}} \\ - PV \int_{-R}^R dx \frac{\sin{\tau (x+i 2 \pi)}}{\sinh{(x/2+i \pi)}} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\sin{\tau (\epsilon e^{i \phi}+ i 2 \pi)}}{\sinh{(\epsilon e^{i \phi}/2+i \pi)}} \\- i \int_0^{2 \pi} dy \frac{ \sin{\tau (-R+i y)}}{i \sinh{((-R+i y)/2)}} $$

In the limit as $R \to \infty$ the second and fifth integrals vanish. The contour integral then becomes in this limit and the limit $\epsilon \to 0$,

$$\left [1+\cosh{(2 \pi \tau)} \right ] \int_{-\infty}^{\infty} dx \frac{\sin{\tau x}}{\sinh{(x/2)}} + i \sinh{(2 \pi \tau)} PV \int_{-\infty}^{\infty} dx \frac{\cos{\tau x}}{\sinh{(x/2)}} - 2 \pi \sinh{2 \pi \tau} $$

The integral over cosine vanishes due to symmetry. Further, by Cauchy's theorem, the contour integral vanishes. Thus,

$$\int_{-\infty}^{\infty} dx \frac{\sin{\tau x}}{\sinh{(x/2)}} = 2 \pi \frac{\sin{2 \pi \tau}}{1+\cos{2 \pi \tau}} = 2 \pi \tanh{\pi \tau}$$

The stated result follows.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.