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