2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Shouldn't the residuals be: $$\operatorname*{Res}_{z = i (\pm\phi + 2 \pi k)/\alpha}f(z) =\frac{\beta}{i\alpha^2}\left(1\pm i\frac{\alpha}{\beta}\cot\phi\right)e^{-\frac{\beta}{\alpha}(\pm \phi+2k\pi)}$$ ? $\endgroup$ – Élio Pereira Feb 25 '19 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 '19 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.