I would like to solve the following improper integral:

$$\int_{-\infty}^{\infty}\frac{1-a\cosh(\alpha x)}{(\cosh(\alpha x)-a)^2}\cos(\beta x)\,dx$$

where $a$, $\alpha$ and $\beta$ are real constants. The constant $a$ is so that $0<a<1$.

I tried to evaluate this integral by contour integration. Because $0<a<1$, we can write $a$ as:


So, the integrand functions has poles of second order at $z=i(\pm\phi+2k\pi)/\alpha$, $k\in\mathbb{Z}$. Now a contour at the complex plane should be thought. I considered a semicircunference centered at the origin with a radius $R\to \infty$. A problem appears, which is to define $R$ as a sequence $R_n$, so that the arc doesn't intersect the poles. This is not straightforward due to the existence of $\alpha$ in $\frac{2k\pi}{\alpha}$. Is there a simpler way to evaluate this integral?


1 Answer 1


The radii do not have to change continuously, one can take a sequence of contours which lie between the consecutive poles. Let $\alpha > 0, \,\beta > 0$, $$f(x) = \frac {1 - \cos \phi \cosh \alpha x} {(\cosh \alpha x - \cos \phi)^2} e^{i \beta x},$$ then the residues in the upper half-plane are $$\operatorname*{Res}_{x = i (\phi + 2 \pi k)/\alpha} f(x) = \frac \beta {i \alpha^2} e^{(-\phi - 2 \pi k) \beta/\alpha}, \quad k \geq 0, \\ \operatorname*{Res}_{x = i (-\phi + 2 \pi k)/\alpha} f(x) = \frac \beta {i \alpha^2} e^{(\phi - 2 \pi k) \beta/\alpha}, \quad k \geq 1$$ and evaluating the sums gives $$\int_{\mathbb R} f(x) \,dx = \frac {2 \pi \beta \cosh \frac {(\pi - \phi) \beta} \alpha} {\alpha^2 \sinh \frac {\pi \beta} \alpha}.$$

  • $\begingroup$ Shouldn't the residuals be: $$\operatorname*{Res}_{z = i (\pm\phi + 2 \pi k)/\alpha}f(z) =\frac{\beta}{i\alpha^2}\left(1\pm i\frac{\alpha}{\beta}\cot\phi\right)e^{-\frac{\beta}{\alpha}(\pm \phi+2k\pi)}$$ ? $\endgroup$ Feb 25, 2019 at 22:15
  • 1
    $\begingroup$ Let $x_0 = i \phi/\alpha$, then $$\frac 1 {(\cosh \alpha x - \cos \phi)^2} = -\frac {\csc^2 \phi} {\alpha^2 (x - x_0)^2} - \frac { i \cos \phi \,\csc^3 \phi} {\alpha (x - x_0)} + O(1), \\ (1 - \cos \phi \cosh \alpha x) e^{i \beta x} = \\ e^{-\beta \phi/\alpha} \sin^2 \phi + i e^{-\beta \phi/\alpha} \sin \phi \, (\beta \sin \phi - \alpha \cos \phi) (x - x_0) + O(|x - x_0|^2).$$ The residue is $a_{-2} b_1 + a_{-1} b_0$. $\endgroup$
    – Maxim
    Feb 26, 2019 at 0:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .