Estimate integral $\int\limits_0^{\infty}\operatorname{sech}\left(\varepsilon x\right)\cos\left(kx\right)\,dx,\,$ with $\,k,\varepsilon>0$

Question

$ \newcommand{\sech}{\operatorname{sech}} $ Is there any analytic/asymptotic way to estimate the value of the integral:

$$ \int_{0}^{\infty} \sech\left(\varepsilon x\right)\cdot \cos\left(kx\right)\,dx , \qquad k,\varepsilon > 0, $$ where $\varepsilon $ is a small parameter.

Thank you.

Answer 1

$$\newcommand{\sech}{\operatorname{sech}} \begin{align*}\int_{0}^{+\infty}\frac{\cos(nx)}{\cosh x}\,dx &= 2\sum_{m\geq 0}(-1)^m \int_{0}^{+\infty}\cos(n x)e^{-(2m+1)x}\,dx\\&=2\sum_{m\geq 0}\frac{(-1)^m (2m+1)}{n^2+(2m+1)^2}\\&=\frac{\pi}{2}\,\sech\frac{\pi n}{2}\end{align*}$$ since: $$ \text{Res}\left(\sec\frac{\pi z}{2},z=2m+1\right)=\frac{2}{\pi}(-1)^{m+1} $$ hence it follows that: $$ \int_{0}^{+\infty}\sech(\varepsilon x)\cos(kx)\,dx = \frac{\pi}{2\varepsilon}\,\sech\left(\frac{\pi k}{2\varepsilon}\right)$$ as checked by Claude Leibovici with a CAS. It is well-known that the hyperbolic secant is more or less a fixed point of the Fourier transform.