The problem is : Evaluate the integral $$\int_{0}^{\infty} \frac{1}{(1+x^2)\cosh{(ax)}}dx$$

I have tried expand $\frac{1}{\cosh{ax}}$ and give the result in the following way:

First, note that $$\frac{1}{\cosh{(ax)}}=\frac{2e^{-ax}}{e^{-2ax}+1}=\sum_{n=0}^{\infty}2(-1)^n e^{-(2n+1)ax}$$ Secondly, we consider $f(a)=\int_{0}^{\infty} \frac{e^{ax}}{1+x^2}dx$

Some calculation results in $f''(a)+f(a)=\int_{0}^{\infty}e^{ax}dx=-\frac{1}{a}$

We substitute $f(a)=u(a)e^{ia}$ into former result and thus $ (u'(a) e^{2ia})'=-\frac{e^{ia}}{a}.$

Let $E(a)=\int_{0}^{a} \frac{e^{it}}{t}dt=\mbox{Ei}(ia)$ where $\mbox{Ei}(x)$ is the Exponential integral then $$u'(a)= -e^{-2ia} E(a)+c_1 e^{-2ia}.$$

Hence \begin{align*}u(a) &=\frac{1}{2i} e^{-2ia}E(a) - \frac{1}{2i}\int_{0}^{a} \frac{e^{-it}}{t}dt-\frac{1}{2i}c_1 e^{-2ia} +c_2 \\ &=\frac{1}{2i} e^{-2ia}E(a) -\frac{1}{2i}E(-a)-\frac{1}{2i}c_1 e^{-2ia} +c_2\end{align*} We conclude that $$ f(a)=\frac{e^{-ia} \mbox{Ei}(ia)-e^{ia}\mbox{Ei}(-ia)}{2i}+c_1 e^{-ia}+c_2 e^{ia}$$

But I got stuck here, I cannot figure out $c_1$ as well as $c_2$. Also, even $c_1$ and $c_2$ are known, I cannot use the summation to get result for the original question.

Is there other way to tackle this problem? Or can I modify my method to make it feasible to get the desired result? Thanks for your attention!

  • $\begingroup$ Maybe you can try some contour integral. $\endgroup$ – smiley06 Jun 4 '13 at 14:27
  • $\begingroup$ @smiley06 The contour integral was a bit difficult since the $\cosh{x}$ $\endgroup$ – Golbez Jun 4 '13 at 14:30
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    $\begingroup$ Watch the sign in the exponent of the exponential in the denominator in $\;\frac1{\cosh ax}\;$ : it must be positive...and you're missing a factor of two. $\endgroup$ – DonAntonio Jun 4 '13 at 14:40
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    $\begingroup$ @DonAntonio Thanks for your correction! $\endgroup$ – Golbez Jun 4 '13 at 15:29
  • $\begingroup$ $\large a \in {\mathbb R}$ ?. $\endgroup$ – Felix Marin Jul 6 '14 at 18:11

This integral may be evaluated using residue theory. Consider the integral

$$\oint_C \frac{dz}{(1+z^2) \cosh{a z}}$$

where $C$ is a semicircle of radius $R$ in the upper half plane. As $R \to \infty$, the integral about the semicircle vanishes, and we are left with the original integral equaling $i 2 \pi$ time the sum of the residues of the poles of the integrand within $C$. In this case, the poles within $C$ lie at $z_n = i (n+1/2) \pi/a$ for all $n \in \mathbb{N} \cup \{0\}$, and at $z_+ = i$. Evaluating the residues at these poles (which may be accomplished when the integrand is of the form $p(z)/q(z)$ using the formula $p(z_0)/q'(z_0)$ for a pole at $z=z_0$), we find that

$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} = \frac{\pi}{\cos{a}} - \frac{2 \pi}{a} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1/2)^2 \pi^2/a^2 - 1}$$

The sum unfortunately takes the form of a pair of Lerch transcendents

$$\begin{align}\frac{2 \pi}{a} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1/2)^2 \pi^2/a^2 - 1} &= \frac{\pi}{a} \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{(n+1/2)\pi/a-1}-\frac{1}{(n+1/2)\pi/a+1} \right)\\&= \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}-\frac{1}{n+\frac12+\frac{a}{\pi}} \right) \\ &= \Phi\left(-1,1,\frac12-\frac{a}{\pi}\right)-\Phi\left(-1,1,\frac12+\frac{a}{\pi}\right)\end{align}$$


$$\int_0^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} = \frac{\pi}{2\cos{a}} - \frac12 \left[\Phi\left(-1,1,\frac12-\frac{a}{\pi}\right)-\Phi\left(-1,1,\frac12+\frac{a}{\pi}\right) \right ]$$

It should be noted that $a \ne (k+1/2) \pi$ for some $k \in \mathbb{Z}$.


I should note that, in response to @GrahamHesketh's query, the result above may be shown to be equal to a difference between two digamma functions as follows:

$$\int_0^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} = \frac12 \left [ \psi\left(\frac{3}{4}+\frac{a}{2 \pi} \right)-\psi\left(\frac{1}{4}+\frac{a}{2 \pi} \right) \right ]$$

This may be accomplished by noting that

$$\frac{\pi}{\cos{a}} = \sum_{n=-\infty}^{\infty} (-1)^n \frac{1}{n+\frac12-\frac{a}{\pi}} = \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}+\frac{1}{n+\frac12+\frac{a}{\pi}} \right) $$

$$\psi\left(\frac{1}{4}+\frac{a}{2 \pi} \right) = \sum_{n=0}^{\infty}\left (\frac{1}{n+1}- \frac{1}{n+\frac12 \left (\frac12+\frac{a}{\pi}\right)}\right )$$

$$\psi\left(\frac{3}{4}+\frac{a}{2 \pi} \right) = \sum_{n=0}^{\infty} \left ( \frac{1}{n+1}-\frac{1}{n+1-\frac12 \left (\frac12-\frac{a}{\pi}\right)}\right )$$

To establish equality, note that the result I posted above boils down to

$$\frac{\pi}{\cos{a}} - \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}-\frac{1}{n+\frac12+\frac{a}{\pi}} \right) = 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{n+\frac12+\frac{a}{\pi}}$$

Equality between the above sum and the difference between the two $\psi$ terms is established by comparing the summands term by term.

  • $\begingroup$ Not sure if it's of any interest but after some poking maple gives: $$\frac{2 \pi}{a} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1/2)^2 \pi^2/a^2 - 1} ={\frac {\pi }{\cos \left( a \right) }}-\Psi \left( \frac{3}{4}+{\frac {a}{2\pi }}\right) +\Psi \left( \frac{1}{4}+{\frac {a}{2\pi }} \right) $$ with $\Psi$ the digamma function. I don't how to derive it. $\endgroup$ – Graham Hesketh Jun 4 '13 at 16:26

Here is another solution: Let

$$\hat{f}(\xi) = \int_{\Bbb{R}} f(x)e^{-2\pi i \xi x} \, dx$$

denote the Fourier transform of $f$. Then it is well-known that

$$ (\mathrm{sech} \, \pi x)^{\wedge} = \mathrm{sech} \, \pi \xi \quad \text{and} \quad \left( \frac{1}{a^2 + \pi^2 x^2} \right)^{\wedge} = \frac{1}{a} e^{-2a |\xi|}. $$

Also, if both $f$ and $g$ are in $L^2$, then

$$ \int_{\Bbb{R}} \hat{f} g = \int_{\Bbb{R}} f \hat{g}. $$

In particular, plugging $f(x) = \mathrm{sech} \, \pi x$ we have

$$ \int_{\Bbb{R}} \frac{g(x)}{\cosh \pi x} \, dx = \int_{\Bbb{R}} \frac{\hat{g}(x)}{\cosh \pi x} \, dx. $$

This shows that

\begin{align*} \int_{0}^{\infty} \frac{dx}{(x^2 + 1) \cosh a x} &= \frac{\pi a}{2} \int_{-\infty}^{\infty} \frac{dx}{(a^2 + \pi^2 x^2) \cosh \pi x} \\ &= \frac{\pi}{2} \int_{-\infty}^{\infty} \frac{e^{-2a|x|}}{\cosh \pi x} \, dx = \pi \int_{0}^{\infty} \frac{e^{-2a x}}{\cosh \pi x} \, dx \\ &= 2\pi \int_{0}^{\infty} \frac{e^{-(2a+\pi) x} (1 - e^{-2\pi x})}{1 - e^{-4\pi x}} \, dx \\ &= \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{a}{2\pi}+\frac{1}{4}} (1 - t^{\frac{1}{2}})}{1 - t} \, \frac{dt}{t} \qquad (t = e^{-4\pi x}) \\ &= \frac{1}{2} \left[ \psi_{0}\left( \frac{a}{2\pi}+\frac{3}{4} \right) - \psi_{0}\left( \frac{a}{2\pi}+\frac{1}{4} \right) \right], \end{align*}

where we exploited the identity

$$ \psi_{0}(s) = -\gamma + \int_{0}^{1}\frac{t - t^{s}}{1 - t} \, \frac{dt}{t}. $$

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    $\begingroup$ Nice! Your solutions are very original and I get a lot out of them. (+1) $\endgroup$ – Ron Gordon Jun 4 '13 at 18:39
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    $\begingroup$ Convolutions are neat. +1, of course. $\endgroup$ – J. M. is a poor mathematician Jun 4 '13 at 19:18
  • $\begingroup$ The 1st of the "well-known" formulas not only makes all the job, but is also very nice. $\endgroup$ – Start wearing purple Jun 4 '13 at 19:47

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{1 \over \pars{1 + x^{2}}\cosh\pars{ax}}\,\dd x:\ {\large ?}}$

Zeros of $\ds{\cosh\pars{ax}}$ are given by $\ds{\pars{\! n + \half\!}\,{\pi \over a}\,\ic.\ n \in {\mathbb Z}.\ \mbox{Similarly,}\ \pm\ic\ \mbox{are zeros of}\ \pars{x^{2} + 1}}$.

\begin{align}&\left.\color{#c00000}{% \int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\cosh\pars{ax}}} \right\vert_{\,a\ >\ 0} =-\,\half\,\Im\int_{-\infty}^{\infty}{\dd x \over \pars{x + \ic}\cosh\pars{ax}} \\[3mm]&=-\,\half\,\Im\pars{% 2\pi\ic\sum_{n = 0}^{\infty}{1 \over \bracks{\pars{n + 1/2}\pi\ic/a + \ic} \braces{a\sinh\pars{\bracks{n + 1/2}\pi\ic}}}} \\[3mm]&=-\,\half\,\Im\braces{% 2\pi\ic\sum_{n = 0}^{\infty}{1 \over \bracks{\pars{n + 1/2}\pi/a + 1}\ic \bracks{a\,\ic\pars{-1}^{n}}}} =\left.\color{#00f}{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + 1/2 + a/\pi}} \right\vert_{\,a\ >\ 0}\quad\pars{1} \end{align}

\begin{align}\left.\color{#00f}{\sum_{n = 0}^{\infty}% {\pars{-1}^{n} \over n + 1/2 + a/\pi}}\right\vert_{\,a\ >\ 0} &=\sum_{n = 0}^{\infty} \pars{{1 \over 2n + 1/2 + a/\pi} - {1 \over 2n + 3/2 + a/\pi}} \\[3mm]&=\sum_{n = 0}^{\infty} {1 \over \pars{2n + 3/2 + a/\pi}\pars{2n + 1/2 + a/\pi}} \\[3mm]&={1 \over 4}\sum_{n = 0}^{\infty} {1 \over \bracks{n + 3/4 + a/\pars{2\pi}}\bracks{n + 1/4 + a/\pars{2\pi}}} \\[3mm]&=\half\bracks{\Psi\pars{{3 \over 4} + {a \over 2\pi}} -\Psi\pars{{1 \over 4} + {a \over 2\pi}}}_{\,a\ >\ 0} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function.

\begin{align} &\color{#66f}{\large% \int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\cosh\pars{ax}} = \half\bracks{% \Psi\pars{{3 \over 4} + {\verts{a} \over 2\pi}} -\Psi\pars{{1 \over 4} + {\verts{a} \over 2\pi}}}} \end{align}

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we have $$\int_{-\infty}^{+\infty}\dfrac{1}{\cosh{a\pi x}}\dfrac{\beta}{x^2+\beta^2}=\psi\left(\dfrac{a\beta}{2}+\dfrac{3}{4}\right)-\psi\left(\dfrac{a\beta}{2}+\dfrac{1}{4}\right)$$

  • $\begingroup$ Can you add a small comment on how to derive this? So far I can only check numerically... (+1) $\endgroup$ – Start wearing purple Jun 4 '13 at 15:10
  • $\begingroup$ I was also wondering how to derive this result. Can you give a detailed proof? $\endgroup$ – Golbez Jun 4 '13 at 15:28
  • $\begingroup$ It's somehow close to the laplace transform of $\frac{1}{\cosh{x}}$. But a bit difficult to check their relationship. $\endgroup$ – Golbez Jun 4 '13 at 15:32

You can try to use the residue theorem for a sequence of contours that go between the poles of $\frac{1}{\cosh(x)}$. Then there is of course left the job of writing the estimates that show the equality between the limit of values of contour integrals and the real integral. Good luck.


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