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

  • $\begingroup$ My -1 for the missing differential. $\endgroup$
    – user65203
    Jun 17, 2021 at 11:51
  • $\begingroup$ 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. $\endgroup$
    – user65203
    Jun 17, 2021 at 11:54
  • $\begingroup$ Sry I have corrected the mistakes! $\endgroup$
    – SGGS
    Jun 17, 2021 at 12:22
  • 1
    $\begingroup$ There must be additional restriction on $\beta$ because $\cos\beta$ remains the same for infinitely many values of $\beta$, but that gives different $\sinh\frac{\beta\omega}{a}$. $\endgroup$
    – user65203
    Jun 17, 2021 at 12:40

1 Answer 1


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):

$$\sum_{k=1}^{\infty} \frac{\sin(kx)k}{k^2+\alpha^2} = \frac{\pi}{2} \frac{\sinh\left(\alpha(\pi-x)\right)}{\sinh(\alpha\pi)} \quad 0<x< 2\pi $$


$$ \boxed{\displaystyle I = \int_{0}^{\infty} \frac{\cos(wx)}{\cosh(ax)-\cos(\beta)} dx = \frac{\pi\sinh\left(\frac{w}{a}(\pi-\beta)\right)}{a\sinh\left(\frac{w\pi}{a}\right)\sin(\beta)} \atop \quad a>0,\; 0<\beta<2\pi }$$

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

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

  • 1
    $\begingroup$ Thank you very much for that detailed presentation of the solution!!! $\endgroup$
    – SGGS
    Mar 21, 2022 at 14:30

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