Proof of 3.981.3 Gradshteyn ed.8.
$$\int^{\infty}_{0} \frac{\cos(ax)}{\cosh(\beta x)}dx = \frac{\pi}{2\beta}\operatorname{sech}(\frac{a\pi}{2\beta})$$
I was interested in the derivation (not necessarily rigorously proved) of the above result. I've tried using differentiation under the integral sign using $$I(a) = \int^{\infty}_{0} \frac{\cos(ax)}{\cosh(x)}dx$$ but with no success.
A hint would be highly appreciated.
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $\mathrm{sech}$.
Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue:
$\sum_{k}Re_{z=z_k}\left[\pi \cot(\pi z)f(z)\right] \sim \sum_{k}Re_{z=z_k}\left[\pi \csc(\pi z)f(z)\right]$
Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
– Elianther Mar 25 '19 at 00:29Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as $$I=\int \cos (a x)\, \text{sech}(b x)\,dx$$ write $$(a^2+b^2)\,e^{-bx}\,I=(b+i a)e^{- i a x} \, _2F_1\left(1,\frac{b-i a}{2 b};\frac{3b-ia}{2};-e^{2 b x}\right)+$$ $$(b-i a) e^{ i a x} \, _2F_1\left(1,\frac{b+i a}{2 b};\frac{3b+i a}{2 b};-e^{2 b x}\right)$$ where appear the gaussian hypergeometric functions. Using the limits $$K=\int_0^\infty \cos (a x)\, \text{sech}(b x)\,dx=\frac{\psi \left(\frac{3b-ia}{4 b}\right)+\psi \left(\frac{3b+i a}{4 b}\right)-\psi \left(\frac{b-i a}{4 b}\right)-\psi \left(\frac{b+i a}{4 b}\right)}{4 b}$$ and using the properties of the digamma function $$K=\frac{\pi }{2 b} \text{sech}\left(\frac{\pi a}{2 b}\right)$$
The required integral, after substituting $\ u = bx,$ can be rewritten as:
$$\ I = \frac{1}{b}\cdot \int_0^{\infty} \frac{\cos(Au)}{\cosh u}du$$
Where $\ A = \frac{a}{b}$
Consider the Maclaurin expansion of $\cos(Au)$:
$$\cos(Au) = \sum_{n=0}^{\infty}(-1)^n \cdot \frac{(Au)^{2n}}{(2n)!}$$
Substituting this into the integral and rearranging the terms, we get:
$$ I = \frac{1}{b}\cdot \sum_{n=0}^{\infty}\frac{(-1)^nA^{2n}}{(2n)!} \cdot \int_0^{\infty} \frac{u^{2n}}{\cosh u} du $$
$$ I = 2\cdot \frac{1}{b}\cdot \sum_{n=0}^{\infty}\frac{(-1)^nA^{2n}}{(2n)!} \cdot \int_0^{\infty} \frac{u^{2n}e^u}{1+e^{2u}} du $$
As seen here, the integral inside the sum evaluates to $\Gamma(2m+1) \cdot \beta(2m+1) = (2m)! \cdot \beta(2m+1)$
Where $\beta(n)$ is the Dirichlet Beta function and $\Gamma(n)$ is the Gamma function.
Using this result, the integral evaluates to
$$ I = 2\cdot \frac{1}{b}\cdot \sum_{n=0}^{\infty}(-1)^nA^{2n}\cdot \beta(2n+1) $$
Consider the Taylor expansion of the hyperbolic secant function at $\ x = Az$:
$$\operatorname{sech}(Az) = \sum_{n=0}^{\infty}\frac{E_{2n}}{(2n)!}\cdot A^{2n}z^{2n}$$
Also consider the property of the Dirichlet Beta function:
$$\ (-1)^n \cdot \beta(2n+1) = \frac{E_{2n}}{2\cdot (2n)!} \cdot (\frac{\pi}{2})^{2n+1}$$
Where $\ E_n$ is the nth Euler number.
Substituting the previous identity into the integral, we get:
$$\ I = 2\cdot \frac{1}{b}\cdot \sum_{n=0}^{\infty} \frac{E_{2n}}{2\cdot (2n)!} \cdot A^{2n} (\frac{\pi}{2})^{2n+1}$$
$$ = \frac{1}{b}\cdot \frac{\pi}{2} \cdot \sum_{n=0}^{\infty} \frac{E_{2n}}{(2n)!} \cdot (A\frac{\pi}{2})^{2n}$$
$$ I = \frac{\pi}{2b} \cdot \operatorname{sech}(\frac{A\pi}{2})$$
Putting $\ A = \frac{a}{b}$, we get:
$$ I = \frac{\pi}{2b} \cdot \operatorname{sech}(\frac{a\pi}{2b})$$
Q.E.D.
It is enough to show that $I=\int_0^\infty\frac{\cos Ax}{\cosh x}dx=\frac{\pi}2\operatorname{sech}\tfrac{\pi A}{2}.\tag1$
Let $I_1=\int_0^\infty\frac{e^{Aix}}{e^x+e^{-x}}dx$ and $I_2=\bar{I}_1$, its complex conjugate. Then $I=I_1+I_2.$
By $u=e^x$ sub, we have $I_1=\int_1^\infty \frac{u^{Ai}}{u^2+1}du$ and $I_2=\int_1^\infty \frac{u^{-Ai}}{u^2+1}du= \int_0^1 \frac{u^{Ai}}{u^2+1}du.$
Hence $I=\int_0^\infty \frac{u^{Ai}}{u^2+1}du=\frac1{1+(-1)^{Ai}} \int_{-\infty}^\infty \frac{u^{Ai}}{u^2+1}du=\frac1{1+e^{-\pi A}} \int_{-\infty}^\infty \frac{u^{Ai}}{u^2+1}du.\tag2$
On the other hand, by the residue theorem with the upper semi-circle contour, $\int_{-\infty}^\infty \frac{u^{Ai}}{u^2+1}du=2\pi i\frac{i^{Ai}}{2i}=\pi e^{-\frac{\pi A}2}.\tag3$
From $(2)$ and $(3)$, we obtain $(1)$.
