Integral $\int^\infty_0 \frac{\cos(\omega x)}{\cosh(ax)-\cos\beta}\text{d}x$ Maybe you can help me, understanding an arising ambiguity:
Consider the integral, which is on page 30 integral (6) of the Bateman Project( see link below)
$$\int^\infty_0 \frac{\cos(\omega x)}{\cosh(ax)+\cos(\beta)}\text{d}x=\frac{\pi}{a\sin(\beta)}\frac{\sinh(\frac{\beta \omega}{a})}{\sinh(\frac{\pi \omega}{a})}$$
which holds for $\text{Re}(a)\pi>\text{Im}(a^*\beta) $.
Say I want to calculate the integral with a minus in the denominatorfor real $a, \beta$ (hence the above restriction on the parameters is always satisfied), which I have naively done by
$\int^\infty_0 \frac{\cos(\omega x)}{\cosh(ax)-\cos(\beta)}\text{d}x=\int^\infty_0 \frac{\cos(\omega x)}{\cosh(ax)+\cos(\beta\pm\pi)}\text{d}x=
\frac{\pi}{a\sin(\beta\pm \pi)}\frac{\sinh(\frac{(\beta\pm\pi) \omega}{a})}{\sinh(\frac{\pi \omega}{a})}=-\frac{\pi}{a\sin(\beta)}\frac{\sinh(\frac{(\beta\pm\pi) \omega}{a})}{\sinh(\frac{\pi \omega}{a})}$,
then I get the ambiguity of the choosen sign infront of the $\pi$ in the $\sinh(\frac{(\beta\pm\pi) \omega}{a})$.
Can anyone explain to me what it is the correct way (sign of $\pi$) to solve this integral with the minus in the denominator?
Link for formula: https://authors.library.caltech.edu/43489/1/Volume%201.pdf
 A: Consider the following series (Gradshteyn, Ryzhik 7th edition, 1.461):
$$ \sum_{k=0}^{\infty} e^{-kt} \sin(kz) = \frac{1}{2} \frac{\sin z}{\cosh t -\cos z} \quad t>0$$
Hence if $a>0$
\begin{align*} I =& \int_{0}^{\infty} \frac{\cos(wx)}{\cosh(ax)-\cos(\beta)} dx \\
= &  2\int_{0}^{\infty} \frac{1}{2} \frac{\sin(\beta)}{\cosh(ax) -\cos(\beta)} \frac{\cos(wx)}{\sin(\beta)}dx\\ 
=&  2\int_{0}^{\infty} \sum_{k=0}^{\infty} e^{-kax} \sin(k\beta) \frac{\cos(wx)}{\sin(\beta)}dx\\
=&  \frac{2}{\sin(\beta)}\sum_{k=0}^{\infty} \sin(k\beta)\int_{0}^{\infty}  e^{-kax}\cos(wx) dx\\ \end{align*}
This last integral is the Laplace transform of the [cosine function]1.
$$\mathcal{L} \left\{\cos(t) \right\} = \int_{0}^{\infty} e^{-st}\cos (at) dt = \frac{s}{s^2+a^2} \quad \Re(s) > |\Im (a)|$$
\begin{align*}
I =& \frac{2}{\sin(\beta)}\sum_{k=0}^{\infty} \sin(k\beta) \int_{0}^{\infty}  e^{-kax}\cos(wx) dx\\
=& \frac{2}{\sin(\beta)}\sum_{k=0}^{\infty} \frac{\sin(k\beta)ka}{k^2a^2+w^2}\\
=& \frac{2}{a\sin(\beta)}\sum_{k=0}^{\infty} \frac{\sin(k\beta)k}{k^2+\left(\frac{w}{a}\right)^2}
\end{align*}
This last series is the following Fourier series for the hyperbolic sine (Gradshteyn, Ryzhik 7th edition, 1.445):
$$\sum_{k=1}^{\infty} \frac{\sin(kx)k}{k^2+\alpha^2} = \frac{\pi}{2} \frac{\sinh\left(\alpha(\pi-x)\right)}{\sinh(\alpha\pi)} \quad 0<x< 2\pi $$
Hence
$$ \boxed{\displaystyle I = \int_{0}^{\infty} \frac{\cos(wx)}{\cosh(ax)-\cos(\beta)} dx = \frac{\pi\sinh\left(\frac{w}{a}(\pi-\beta)\right)}{a\sinh\left(\frac{w\pi}{a}\right)\sin(\beta)} \atop \quad a>0,\; 0<\beta<2\pi }$$
where the cases $w=0$ and $\beta =\pi$ can be calculated as limiting cases.
For the case with a plus in the denominator
$$ J =  \int_{0}^{\infty} \frac{\cos(wx)}{\cosh(ax)+\cos(\bar{\beta})} dx $$
You need the following series
$$ \sum_{k=0}^{\infty} (-1)^{k-1} e^{-kt} \sin(kz) = \frac{1}{2} \frac{\sin z}{\cosh t +\cos z} \quad t>0$$
