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

Question

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!

Answer 1

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

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

Therefore

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

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

ADDENDUM

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.

Answer 2

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

Answer 3

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.

Answer 4

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

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