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

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.

I tried to evaluate this integral by contour integration. Because $$0, we can write $$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?

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 $$\int_{\mathbb R} f(x) \,dx = \frac {2 \pi \beta \cosh \frac {(\pi - \phi) \beta} \alpha} {\alpha^2 \sinh \frac {\pi \beta} \alpha}.$$
• Shouldn't the residuals be: $$\operatorname*{Res}_{z = i (\pm\phi + 2 \pi k)/\alpha}f(z) =\frac{\beta}{i\alpha^2}\left(1\pm i\frac{\alpha}{\beta}\cot\phi\right)e^{-\frac{\beta}{\alpha}(\pm \phi+2k\pi)}$$ ? – Élio Pereira Feb 25 '19 at 22:15
• 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 $a_{-2} b_1 + a_{-1} b_0$. – Maxim Feb 26 '19 at 0:26