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

Question

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:

$$a=\cos\phi,\,\phi\in\left[0,\frac{\pi}{2}\right]$$

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?

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 $$\int_{\mathbb R} f(x) \,dx = \frac {2 \pi \beta \cosh \frac {(\pi - \phi) \beta} \alpha} {\alpha^2 \sinh \frac {\pi \beta} \alpha}.$$