6

Show that $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2\left(\frac{a\pi}{2}\right) $$

I think we can solve it by contour integration, but how? and it would be better if there were two ways, the first with real analysis and the second with complex analysis.

giobrach
  • 7,490
mnsh
  • 5,875

4 Answers4

12

We of course assume that $a$ is real here. For a complex analysis, consider the integral

$$\oint_C dz \frac{\sin{a z}}{\sinh{z}}$$

where $C$ is the following contour:

rect cont

Because there are no poles in the interior of the contour, this integral is zero by Cauchy's theorem. On the other hand, the contour integral is equal to

$$\int_{\epsilon}^R dx \frac{\sin{a x}}{\sinh{x}} + i \int_0^{\pi} dy \frac{\sin{a(R+i y)}}{\sinh{(R+i y)}} + \\ \int_R^{\epsilon} dx \frac{\sin{a(x+i \pi)}}{\sinh{(x+i \pi)}} + i \epsilon \int_0^{-\pi/2} d\phi \, e^{i \phi} \frac{\sin{a(i \pi + \epsilon e^{i \phi})}}{\sinh{(i \pi + \epsilon e^{i \phi})}} + \\i \int_{\pi-\epsilon}^{\epsilon} dy \frac{\sin{i a y}}{\sinh{i y}} + i \epsilon \int_{\pi/2}^0 d\phi \, e^{i \phi} \frac{\sin{a \epsilon e^{i \phi}}}{\sinh{\epsilon e^{i \phi}}} $$

We take the limits as $R \to \infty$ and $\epsilon \to 0$; in these limits, the second and sixth integrals vanish. We also need only consider the real part of the above expression, as the integral of interest is real; this then eliminates the fifth integral from consideration altogether, and we need only examine the real part of the third. By expanding the trig functions using, e.g., addition theorems, we get the following relation:

$$\left (1+\cosh{\pi a}\right) \int_0^{\infty} dx \frac{\sin{a x}}{\sinh{x}} - \frac{\pi}{2} \sinh{\pi a} = 0$$

or

$$\int_0^{\infty} dx \frac{\sin{a x}}{\sinh{x}} = \frac{\pi}{2} \tanh{\frac{\pi a}{2}}$$

Of course, this is not the integral desired. Nevertheless, assuming everything is absolutely convergent (which it is here), we may differentiate both sides with respect to $a$ and reverse order of derivative and integral to get

$$\int_0^{\infty} dx \frac{x \cos{a x}}{\sinh{x}} = \frac{\pi^2}{4} \text{sech}^2{\frac{\pi a}{2}}$$

as was to be shown.

Ron Gordon
  • 138,521
  • here is my question: how did you know that the contour of (C) my integral is the picture you put ? – mnsh Jul 29 '13 at 15:52
  • 1
    @hmedan.mnsh: Well, this type of contour is common when you are integrating ratios of trig/hyperbolic functions of different frequencies. It has the advantage of simplifying the evaluation of the integral over each piece of the contour when you know that much of the integral is not going to vanish. It also avoids resorting to summing infinite series as oen did (although that technique worked here). – Ron Gordon Jul 29 '13 at 15:55
  • I've two questions: the fourth integral tends to zero as well? and e.g., means for example? thanks. – Charles Seife Jun 07 '16 at 07:54
  • @CharlesSeife: No. Expand out the numerator and denominator and take the limit as $\epsilon \to 0$. You will find that the $\epsilon$s cancel and you are left with the $\sinh{\pi a}$ term. I do not know what you mean by "means for example." – Ron Gordon Jun 07 '16 at 12:45
  • Oh! got it : ) now, when I expand in the third integral the numerator $\sin(a(x+i\pi)$ I can't achieve your result, but the denominator $\sinh (x+i\pi)=-\sinh x$ does. Thanks. – Charles Seife Jun 07 '16 at 14:31
8

Here's another complex analytic approach.

Since $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx \quad\textrm{and}\quad \int_{0}^{\infty}\frac{\sin ax}{\sinh x}dx$$ are uniformly convergent, differentiation and integration commute. Also, the integrals are even. We find $$\begin{eqnarray*} \int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx &=& \frac{\partial}{\partial a} \int_{0}^{\infty}\frac{\sin ax}{\sinh x}dx \\ &=& \frac{1}{2} \frac{\partial}{\partial a} \int_{-\infty}^{\infty} \frac{\sin ax}{\sinh x}dx \\ &=& \frac{1}{2}\frac{\partial}{\partial a} \mathrm{Im} \int_{-\infty}^{\infty}\frac{e^{iax}}{\sinh x}dx. \end{eqnarray*}$$ We close the contour for the last integral along a semicircular contour in the upper half plane. Here we assume, with little loss of generality, that $\mathrm{Re}(a)>0$. (Due to the good behavior of $e^{iax}$, the integral along this contour is zero.) We pick up the residues in the upper half plane occurring at $x=in\pi$, where $n=1,2,\ldots$. Thus, $$\begin{eqnarray*} \int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx &=& \frac{1}{2}\frac{\partial}{\partial a} \mathrm{Im} \,2\pi i \sum_{n=1}^\infty \mathrm{Res}_{x_n=i n\pi} \frac{e^{iax_n}}{\sinh x_n} \\ &=& \frac{1}{2}\frac{\partial}{\partial a} \mathrm{Im} \,2\pi i \sum_{n=1}^\infty \frac{e^{-n\pi a}}{\cosh i n\pi} \\ &=& \pi \frac{\partial}{\partial a} \sum_{n=1}^\infty (-1)^n e^{-n\pi a} \\ &=& -\pi \frac{\partial}{\partial a} \frac{1}{1+e^{a\pi}} \\ &=& \frac{\pi^2 e^{a\pi}}{(1+e^{a\pi})^2} \\ &=& \frac{\pi^2}{4}\mathrm{sech}^2\frac{a\pi}{2}. \end{eqnarray*}$$

user26872
  • 19,465
1

I'm going to use the approach of Ron Gordon and try to find the integral $\int_0^{\infty}\dfrac{\sin ax}{\sinh x} dx$, but with real methods. First of all, we have that, $$\dfrac 1{\sinh x}=2\sum_{k=2n+1,n\in\mathbb{N}}exp(-kx)$$ So, $$\int_0^{\infty}\dfrac{\sin ax}{\sinh x} dx=2\int_0^{\infty}\sin ax \sum_{k=0}^\infty \exp(-(2k+1)x)dx$$$$=2a\sum\dfrac{1}{a^2+(2k+1)^2}=\pi \coth \pi a-\frac\pi2 \coth\frac{\pi a}{2}$$

Differentiating with respect to a we get the answer.(The sum is obtained by taking the infinite product for $\sinh x$, taking logs and differentiating.)

Ishan Banerjee
  • 6,459
0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax} \over \sinh\pars{x}}\,\dd x = {\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}}:\ {\Large ?}.}$

\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax} \over \sinh\pars{x}}\,\dd x} = \totald{}{a}\int_{0}^{\infty}{\sin\pars{ax} \over \sinh\pars{x}}\,\dd x \\[5mm] = &\ 2\,\totald{}{a}\Im\int_{0}^{\infty} {\expo{\ic ax} - 1 \over \expo{x} - \expo{-x}}\,\dd x \\[5mm] = &\ 2\,\totald{}{a}\Im\int_{0}^{\infty} {\expo{-\pars{1 - \ic a}x} - \expo{-x} \over 1 - \expo{-2x}} \,\dd x \\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,& \totald{}{a}\Im\int_{0}^{\infty} {\expo{-\pars{1/2 - \ic a/2}x}\,\,\, - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x \\[5mm] = &\ \totald{}{a}\Im\left[\int_{0}^{\infty} {\expo{-x} - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x \right. \\[2mm] &\ \phantom{\totald{}{a}\Im\left[\right.} \left.-\int_{0}^{\infty} {\expo{-x} \expo{-\pars{1/2 - \ic a/2}x} \over 1 - \expo{-x}} \,\dd x\right] \\[5mm] = &\ \totald{}{a}\Im\bracks{\Psi\pars{1 \over 2} - \Psi\pars{{1 \over 2} - {a \over 2}\,\ic}} \\[5mm] = &\ \totald{}{a}\bracks{-\Psi\pars{1/2 - a\ic/2} + \Psi\pars{1/2 + a\ic/2} \over 2\ic} \\[5mm] = &\ -\,{\ic \over 2} \totald{}{a}\bracks{\pi\cot\pars{\pi\bracks{{1 \over 2} - {a \over 2}\,\ic}}} \\[5mm] = &\ -\,{\pi\,\ic \over 2} \totald{}{a}\bracks{\tan\pars{{\pi a \over 2}\,\ic}} = {\pi \over 2} \totald{}{a}\bracks{\tanh\pars{\pi a \over 2}} \\[5mm] = &\ \bbx{{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}} \\ & \end{align}
Felix Marin
  • 89,464