A definite integral with hyperbolic cosines

I have a way of showing this without contour integration in the complex plane. There is a bit of a trick involved and, frankly, Mathematica misleads. It should be noted that the condition $|a|+|b| < \pi$ is needed for the integral to converge. Basically, rewrite the $\cosh$'s as exponentials:

$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= 2 \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{1+e^{-2 \pi x}} e^{-\pi x} \\ &= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\infty} dx \: (e^{a x}+e^{-a x}) (e^{b x}+e^{-b x}) e^{-(2 k+1) \pi x} \\ \end{align} $$

Evaluating the integrals, we get

$$= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a+b)} + \frac{1}{(2 k+1)\pi +(a+b)}\right ] $$ $$ + \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a-b)} + \frac{1}{(2 k+1)\pi +(a-b)} \right ] $$

Here I note that $a+b$ and $a-b$ should not be some multiple of $\pi$, so that the above sums behave properly.

To get the sums into a somewhat familiar form, I rearrange them a bit to get

$$= \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a+b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a+b}{2 \pi}\right )}\right ] $$ $$ + \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a-b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a-b}{2 \pi}\right )}\right ] $$

Now, here is the interesting part (at least to me). Let

$$f(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} $$

This looks like it should be a trig function of some sort. It is not; rather, it is something called a Hurwitz-Lerch transcendent, which does not look like it will be much help. That said, it almost looks like a trig function, so I instead considered the following:

$$\begin{align} f(z) + f(1-z) &= \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1-z}\\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{z-(k+1)}\\ &= \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{z+k} \\ &= \frac{\pi}{\sin{\pi z}}\\ \end{align}$$

This is very helpful, because we have precisely this functional form above, e.g.,

$$\frac{1}{2} - \frac{a+b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a+b}{2 \pi} \right ) $$ $$\frac{1}{2} - \frac{a-b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a-b}{2 \pi} \right ) $$

So we get for the integral:

$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= \frac{1}{4} \left [ \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a+b}{2} \right )}} + \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a-b}{2} \right )}} \right ] \\ &= \frac{1}{4} \left [ \frac{1}{\cos{\left ( \frac{a+b}{2} \right )}} + \frac{1}{\cos{\left ( \frac{a-b}{2} \right )}} \right ] \\ &= \frac{\cos{\frac{a}{2}} \cos{\frac{b}{2}}}{\cos{a} + \cos{b}} \end{align}$$

QED

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

$\ds{\int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over \cosh\pars{\pi x}}\,\dd x ={\cos\pars{a/2}\cos\pars{b/2} \over \cos\pars{a} + \cos\pars{b}}:\ {\large ?} \,,\quad \verts{a} + \verts{b} < \pi}$

\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}\ =\ \overbrace{\half\int_{-\infty}^{\infty} {\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x} ^{\dsc{\expo{\pi x}\equiv t\ \imp\ x={1 \over \pi}\,\ln\pars{t}}} \\[5mm]&=\half\int_{0}^{\infty}{\bracks{\pars{t^{\alpha} + t^{-\alpha}}/2} \bracks{\pars{t^{\beta} + t^{-\beta}}/2}\over \bracks{\pars{t^{2} + 1}/\pars{2t}}}\,{\dd t \over \pi t} \end{align}

where $\ds{\alpha \equiv {\verts{a} \over \pi}}$ and $\ds{\beta \equiv {\verts{b} \over \pi}}$

Then, \begin{align} &\color{#66f}{\large% \int_{0}^{\infty}\!\!\!{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x} \\[5mm]&={1 \over 4\pi}\int_{0}^{\infty}\!\!\! {t^{\alpha + \beta} \over t^{2} + 1}\,\dd t +{1 \over 4\pi}\int_{0}^{\infty}\!\!\!{t^{\alpha - \beta} \over t^{2} + 1}\,\dd t +{1 \over 4\pi}\int_{0}^{\infty}\!\!{t^{-\alpha + \beta} \over t^{2} + 1}\,\dd t +{1 \over 4\pi}\int_{0}^{\infty}\!\!{t^{-\alpha - \beta} \over t^{2} + 1}\,\dd t \,\,\,\,\,\pars{1} \end{align}

The problem is reduced to the evaluation of $\ds{{1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t}$, $\ds{\pars{~\mbox{with}\ \verts{\Re\pars{\mu}} < 1~}}$, in the complex plane. For this purpose we use a 'key-hole contour' which takes care of the $\ds{z^{\mu}\mbox{-branch cut}}$: $$ z^{\mu}=\verts{z}^{\mu}\exp\pars{\ic\mu\,{\rm Arg}\pars{z}}\,,\qquad z\not= 0\,,\qquad \verts{\,{\rm Arg}\pars{z}} < \pi $$

\begin{align}&{1 \over 4\pi}\,2\pi\ic\pars{% {\expo{\pi\mu\ic/2} \over 2\ic} + {\expo{-\pi\mu\ic/2} \over -2\ic}} =\dsc{\half\,\ic\sin\pars{\pi\mu \over 2}} \\[5mm]&={1 \over 4\pi} \int_{-\infty}^{0}{\pars{-t}^{\mu}\expo{\pi\mu\ic} \over t^{2} + 1}\,\dd t +{1 \over 4\pi} \int_{0}^{-\infty}{\pars{-t}^{\mu}\expo{-\pi\mu\ic} \over t^{2} + 1}\,\dd t \\[5mm]&=\expo{\pi\mu\ic}\,{1 \over 4\pi} \int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t -\expo{-\pi\mu\ic}\,{1 \over 4\pi} \int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t \\[5mm]&=\dsc{2\ic\sin\pars{\pi\mu}\pars{% {1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t}}\ \imp\ \begin{array}{|c|}\hline\\ \quad {1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t ={1 \over 8}\,\sec\pars{\pi\mu \over 2}\quad \\ \\ \hline \end{array} \end{align}

Then, the expression $\pars{1}$ is reduced to:

\begin{align} &\color{#66f}{\large% \int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x} ={1 \over 4}\,\sec\pars{\verts{a} + \verts{b} \over 2} +{1 \over 4}\,\sec\pars{\verts{a} - \verts{b} \over 2} \\[5mm]&={1 \over 4}\,{\cos\pars{\bracks{\verts{a} - \verts{b}}/2} +\cos\pars{\bracks{\verts{a} + \verts{b}}/2}\over \cos\pars{\bracks{\verts{a} - \verts{b}}/2} \cos\pars{\bracks{\verts{a} + \verts{b}}/2}} \\[5mm]&={1 \over 4}\,{2\cos\pars{a}\cos\pars{b}\over \cos^{2}\pars{a/2}\cos^{2}\pars{b/2} - \sin^{2}\pars{a/2}\sin^{2}\pars{b/2}} \\[5mm]&=\half\,{\cos\pars{a}\cos\pars{b}\over \cos^{2}\pars{a/2}\cos^{2}\pars{b/2}- \bracks{1 - \cos^{2}\pars{a/2}}\bracks{1 - \cos^{2}\pars{b/2}}} \\[5mm]&=\half\,{\cos\pars{a}\cos\pars{b}\over -1 + \cos^{2}\pars{a/2} + \cos^{2}\pars{b/2}} =\half\,{\cos\pars{a}\cos\pars{b}\over -1 + \bracks{1 + \cos\pars{a}}/2 + \bracks{1 + \cos\pars{b}}/2} \end{align}

Finally,

$$ \color{#66f}{\large% \int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x} =\color{#66f}{\large% {\cos\pars{a/2}\cos\pars{b/2} \over \cos\pars{a} + \cos\pars{b}}}\,,\qquad \verts{a} + \verts{b} < \pi $$