3
$\begingroup$

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 $n$.

$\endgroup$
  • 1
    $\begingroup$ By the change of variable $x\to kx$, the integral is $$k^{-3/2}\int_0^{\infty} e^{-x}(\sin(x)+\cos(x))x^{1/2} \frac{x/k}{\sinh(x/k)} dx\sim ck^{-3/2}$$ when $k\to\infty$, with $$c=\int_0^{\infty} e^{-x}(\sin(x)+\cos(x))x^{1/2} dx$$ and it seems that $$c=\frac{\sqrt{\pi(2+\sqrt{2})}}{4\sqrt[4]{2}}$$ $\endgroup$ – Did Oct 12 '16 at 11:10
  • $\begingroup$ Sorry, I wrote in the integral incorrectly. That is very silly of me. :( I actually tried this kind of thing, and it does not work out, because, with the change of variable kx=y we have: $\int e^{-y} (\sin y + \cos y) \frac{y^4 / k^4}{\sinh(y^2/k^2) } dy/k$ we would get $c/k^3$, but $c=\int_0^\infty e^-x (\cos x + \sin x) x^2 = 0$ $\endgroup$ – evilcman Oct 12 '16 at 11:23
  • $\begingroup$ the method sould work anyways...have a look at this plot wolframalpha.com/input/… $\endgroup$ – tired Oct 12 '16 at 11:30
  • $\begingroup$ the agreement in the relevant region is impressive $\endgroup$ – tired Oct 12 '16 at 11:31
  • $\begingroup$ The agreement of the integrands is very good, but the integral in that case gives 0. That means that the leading asymptotic behavior of the integral comes from the tiny difference. If you go further in the Taylor expansion, you get the same thing. $\endgroup$ – evilcman Oct 12 '16 at 11:35
3
$\begingroup$

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

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


Path of integration

enter image description here

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Where do you close the contour? $\endgroup$ – evilcman Oct 18 '16 at 15:41
  • $\begingroup$ i deform the original contour in green into the new contour which is blue in my picture $\endgroup$ – tired Oct 18 '16 at 17:23
  • $\begingroup$ I don't get it. Usually to get a sum of residues you need to close a contour with the poles inside. Here you don't seem to have a closed contour. $\endgroup$ – evilcman Oct 18 '16 at 18:06
  • $\begingroup$ i can deform my contour as i wish as long as i collect the effects of all singularities that i cross. in this case this results in the half residues an the principal value integral $\endgroup$ – tired Oct 18 '16 at 18:14
  • 1
    $\begingroup$ Oh. I get it. You use that f is holomorphic inside the triangle in you picture, so if you had a closed contour which would be the green line connected with the blue in the opposite direction you would get zero. So instead you evaluate explicitly the blue integral. $\endgroup$ – evilcman Oct 18 '16 at 18:22
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Correction to the answer (v2): The word contributtion should be contribution. $\endgroup$ – Qmechanic Oct 25 '16 at 19:47
-1
$\begingroup$

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 : $$\int_1^\infty|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx +\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx\leq M\int_1^\infty e^{-kx}dx+\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx=M\frac{e^{-k}}{k}\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This proves that the integral converges for every $k>0$ and that its value is $O(1/k)$ when $k\to\infty$ -- which are not the OP's question, I am afraid. $\endgroup$ – Did Oct 12 '16 at 11:47
  • $\begingroup$ I am sorry the previous edit was false as I only shown that the integral was bounded. $\endgroup$ – H.C. Lefevre Oct 12 '16 at 11:53
  • $\begingroup$ In the expanded version of your answer, why do you "forget" the integral on $(0,a)$ in the last computation? $\endgroup$ – Did Oct 12 '16 at 12:42
  • $\begingroup$ Actually it is bounded for a fixed value of $a$ but you have to show that it goes to zero too ... there is a flaw in my attempt $\endgroup$ – H.C. Lefevre Oct 12 '16 at 13:10

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.