# 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?

• My -1 for the missing differential.
– user65203
Jun 17, 2021 at 11:51
• As the given value of the integral is in terms of all variables but $\omega$, on concludes that integration is on $\omega$ and this does not make sense.
– user65203
Jun 17, 2021 at 11:54
• Sry I have corrected the mistakes!
– SGGS
Jun 17, 2021 at 12:22
• There must be additional restriction on $\beta$ because $\cos\beta$ remains the same for infinitely many values of $\beta$, but that gives different $\sinh\frac{\beta\omega}{a}$.
– user65203
Jun 17, 2021 at 12:40

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

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

$$\sum_{k=0}^{\infty} (-1)^{k-1} e^{-kt} \sin(kz) = \frac{1}{2} \frac{\sin z}{\cosh t +\cos z} \quad t>0$$

• Thank you very much for that detailed presentation of the solution!!!
– SGGS
Mar 21, 2022 at 14:30