Asymptotics of $\int_0^{\infty} e^{-kx}(\sin(kx)+\cos(kx)) \frac{x^{4}}{\sinh(x^2)} dx$ at large $k$

Question

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?

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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 $n$.

Answer 1

In this answer we try to find an asymptotic estimate for the integral

$$ I(k)=\int_0^{\infty}e^{-kx}(\sin(kx)+\cos(kx))\frac{x^4}{\sinh(x^2)} $$

as $k\rightarrow +\infty$

Simple trigonometric considerations bring the problem in the more conveniant form

$$ I(k)=\lim_{R \rightarrow\infty}I_R(k)=\lim_{R \rightarrow\infty}\sqrt{2}\Im\left[e^{i\frac{\pi}{4}}\int_0^{R}\underbrace{\frac{x^{4}}{\sinh(x^2)}e^{-kx+ikx}dx}_{f(x)}\right] $$

The OP already showed that the boundary at $x=0$ will not contribute to the integral, so a more sophisticad approach is needed

To proceed we determine a path of steepest descent which is simply $x=te^{i\frac{\pi}{4}}$ in the first quadrant.

Deforming the integral into a triangle with the steepest descent path as hypotenuse (with small semicircles at $x=\sqrt{in}$ to avoid the singularities of $\sinh(x^2)^{-1}$) gives us according to Mr. Cauchy

$$ I_R(k)=\\I_{R e^{i\frac\pi4}}(k)+\sqrt{2}\Im\left[i\int_0^{R}f(R+iy)dy\right]+\sqrt{2} \Im\left[i\pi\sum_{R\geq n\geq 1}\text{Res}(f(x),x={\sqrt{i\pi n}})\right] $$

here the first integral on the right is interpreted as a principal value integral. Therefore, taking limits $R\rightarrow \infty$ we observe $\lim_{R\rightarrow\infty}I_{Re^{i\frac\pi4}}(k)=0$ because $\Im(e^{i\pi/4}f(x))=0$ on this ray. Furthermore the contribution from the opposite leg of the triangle goes as $f(Re^{i\alpha})\sim 2 e^{-R^2}$ and can be safely dismissed.

$$ I(k)=\sqrt{2}\pi\Im\left[i\sum_{n\geq 1}\text{Res}(f(x),x={\sqrt{i\pi n}})\right]\sim\sqrt{2}\pi\Im\left[i\text{Res}(f(x),x={\sqrt{i\pi}})\right] $$

Because residues with bigger $n$ are exponentially surpressed.

The remaining residue is easily calculated according to the residue formula for a simple pole.

Doing the algebra we end up with

$$ I(k)\sim\frac{\pi^{5/2}}{\sqrt{2}}e^{-\sqrt{2\pi}k}+\mathcal{O}(e^{-\sqrt{4\pi}k}) $$

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Path of integration

original path of integration (green), deformed path of integration (blue) and singularities (red) in the complex $x$-plane

Answer 2

Actually for large $k$ the integral seems to go to zero as we can't say anything about the $sin(kx)$ and $cos(kx)$ terms except that they are bounded so : $$ |e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|\leq2e^{-kx}\frac{x^4}{sinh(x^2)}$$

Then you can show that $\frac{x^4}{sinh(x^2)}$ is bounded as it goes to zero continuously when $x\rightarrow\infty$. $$ 2e^{-kx}\frac{x^4}{sinh(x^2)}\leq Me^{-kx}$$ Finally you can split the integral between an indefinite integral and an integral on a compact : $$\int_1^\infty|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx\leq M\int_1^\infty e^{-kx}dx=M\frac{e^{-ka}}{k}$$ The you have : $$\int_1^\infty|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx +\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx\leq M\int_1^\infty e^{-kx}dx+\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx=M\frac{e^{-k}}{k}\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx$$

Answer 3

User tired has already given a good answer. In this answer, we consider a slight variation of tired's method.

I) Define the meromorphic functions

$$ g(z)~:= \frac{z^4}{\sinh (z^2)}~=~z^4\sum_{p\in\mathbb{Z}} \frac{(-1)^p}{z^2-i\pi p}, \tag{1} $$

$$ f(z)~:=~\exp\left( -k z (1\!-\!i) -\frac{i\pi}{4}\right)g(z), \qquad k~>~0,\tag{2} $$

and the integral along the positive $x$-axis

$$ J~:=~\int_0^{\infty}\! \mathrm{d}z~f(z),\tag{3} $$

for later convenience. Notice that the integral along the positive $y$-axis

$$ \int_{i\infty}^0 \mathrm{d}z~f(z)~=~\bar{J}\tag{4} $$

happens to be the complex conjugate!

II) OP's integral is

$$ \frac{I}{\sqrt{2}} ,\tag{5}$$

where

$$ I ~:=~\sqrt{2} \int_0^{\infty}\! \mathrm{d}z~ e^{-kz}( \sin kz + \cos kz) g(z)~=~2\int_0^{\infty}\! \mathrm{d}z~ e^{-kz}\cos\left(k z -\frac{\pi}{4}\right)g(z)$$ $$~\stackrel{(2)+(3)}{=}~2~{\rm Re}~ J ~=~ \bar{J}+J ~\stackrel{(3)+(4)}{=}~ \left( \int_{i\infty}^0+ \int_0^{\infty} \right) \! \mathrm{d}z~f(z) $$ $$~=~ 2\pi i \sum_{n\in\mathbb{N}} {\rm Res}\left(f, \sqrt{i \pi n } \right) ~=~-\pi^{\frac{5}{2}}\sum_{n\in\mathbb{N}}(-1)^n n^{\frac{3}{2}} e^{-k\sqrt{2\pi n}} .\tag{6} $$

In the third-last equality of eq. (6), we see that the integration contour runs from $z=i\infty$ to $z=\infty$. This explains why OP didn't get any contributtion from the origin $z=0$. In the second-last equality of eq. (6), we closed the contour, and used the residue theorem. Note that formula (6) is an exact (as opposed to an approximate) result.