While working on a physics problem I came across an integral of this form: $$\int_0^{\infty} e^{-kx}(\sin(kx)+\cos(kx)) \frac{x^{4}}{\sinh(x^2)} dx$$
For my purposes the numerical evaluation of this integral is fine, and I have done that, but I started to wonder whether it is possible to derive the large $k$ asymptotics in a simple way. Numerically the asymptotics looks exponential, but can I get a simple formula somehow?
I tried a few things that I could think of, like integration by parts, but to no avail. Maybe some kind of saddle point approximation would work, but I guess you have to start by extending the integral to $\int_{-\infty}^{\infty}$, but I am not sure how. I don't have much experience with this kind of thing, can anyone help?
One approach, that does not work is this:
If we substitute $y=kx we get:
$$\int_0^\infty e^{-y} (\sin y + \cos y) \frac{y^4 / k^4}{\sinh(y^2/k^2) } dy/k = \frac{1}{k^3} \int_0^\infty e^{-y} (\sin y + \cos y) \frac{y^4/k^2}{\sinh (y^2/k^2)} dy$$
Now here, the function $e^{-y}$ makes a cut-off of the integral, so for large $k$ you might think that you can make a Taylor expansion in $\frac{y^4/k^2}{\sinh (y^2/k^2)}$, if you tried you would get as a leading term $c/k^3$ where
$ c = \int_0^\infty e^{-y} (\sin y + \cos y) y^2 dy$
But, this integral is zero, so this term that seems to be the leading asymptotic term is zero. If you try to use the higher order term in the Taylor expansion of $\frac{y^4/k^2}{\sinh (y^2/k^2)}$ you will get contants that are:
$ c = \int_0^\infty e^{-y} (\sin y + \cos y) y^{2+4n} dy$
but all of these are zero. I guess this means that all of these power like terms are zero, so the integral goes to 0 faster than $O(1/k^n)$ for any $n$.