$$\int_0^{\infty}\frac{e^{-a x^2(x^2-\pi^2)}\cos(2\pi a x^3)}{\cosh x}dx=\frac{\pi}{2}e^{-\pi^4 a/16}.$$ Note the unusual appearance of $x^1,x^2,x^3,x^4$.

  • $\begingroup$ What's the question? Do you want a proof of this? $\endgroup$ – Cameron Williams Sep 7 '13 at 19:39
  • $\begingroup$ Where does it come from? Maple and WA can't find the right-hand side on my machine even for some fixed $a$. $\endgroup$ – Jean-Sébastien Sep 7 '13 at 19:43
  • $\begingroup$ From the looks of it, this is one of those nasty oscillatory integrals, which are handled best in the complex plane using the method of stationary phase / saddle point method. $\endgroup$ – rajb245 Sep 7 '13 at 20:28
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    $\begingroup$ @rajb245: the OP is asserting an exact result (as illustrated below). Why would you use techniques to provide asymptotic approximations? $\endgroup$ – Ron Gordon Sep 8 '13 at 11:09
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    $\begingroup$ Since you are new here, may I call your attention to the "About" link. Especially the segment on accepting answers. As you might consider doing for the incredible answer below. $\endgroup$ – user12802 Sep 9 '13 at 16:32

EDIT @Random Variable found an error in sign in the exponent of the integrand. This is corrected, and does not affect the result. Many thanks to RV for carefully reading through.

To evaluate this integral, consider the following integral in the complex plane:

$$\oint_C \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2}$$

where $C=C_1+C_2+C_3+C_4+C_5+C_6$ as illustrated below:

enter image description here

$$\int_{C_1} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = \int_{i \pi/2}^{R+i \pi/2} \frac{dx}{\sinh{x}} e^{-a (x^2+\pi^2/4)^2}$$ $$\int_{C_2} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = i\int_{\pi/2}^{-\pi/2} \frac{dy}{\sinh{(R+iy)}} e^{-a [(R+i y)^2+\pi^2/4]^2} $$ $$\int_{C_3} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = \int_{R-i \pi/2}^{-i \pi/2} \frac{dx}{\sinh{x}} e^{-a (x^2+\pi^2/4)^2}$$ $$\int_{C_4} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = \int_{-\pi/2}^{-\epsilon} \frac{dy}{\sin{y}} e^{-a (y^2-\pi^2/4)^2} $$ $$\int_{C_5} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = i \epsilon\int_{-\pi/2}^{\pi/2} \frac{d\phi \, e^{i \phi}}{\sinh{(\epsilon e^{i \phi}})} e^{-a [\epsilon^2 e^{i 2 \phi}+\pi^2/4]^2} $$ $$\int_{C_6} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2} = \int_{\epsilon}^{\pi/2} \frac{dy}{\sin{y}} e^{-a (y^2-\pi^2/4)^2} $$

We intend to take $R \to \infty$ and $\epsilon \to 0$. In this case, it should be clear that the integral over $C_2$ will vanish in the former limit. Also, the integrals over $C_4$ and $C_6$ cancel each other out as the sum is an integral over an odd integrand over a symmetric interval.

We are then left with the integrals over $C_1$, $C_3$, and $C_5$. The sum of the first two integrals, over $C_1$ and $C_3$, is

$$\underbrace{\int_{i \pi/2}^{\infty+i \pi/2} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2}}_{z=x+i \pi/2} + \underbrace{\int_{\infty-i \pi/2}^{-i \pi/2} \frac{dz}{\sinh{z}} e^{-a (z^2+\pi^2/4)^2}}_{z=x-i \pi/2} = \\ -i \int_0^{\infty} \frac{dx}{\cosh{x}} e^{-a x^2 (x+i \pi)^2} + i \int_{\infty}^0 \frac{dx}{\cosh{x}} e^{-a x^2 (x-i \pi)^2}$$

which is

$$-i \int_0^{\infty} \frac{dx}{\cosh{x}} e^{-a x^2 (x^2-\pi^2)} \left [ e^{i 2 \pi a x^3} + e^{-i 2 \pi a x^3}\right] = -i 2 \int_0^{\infty} \frac{dx}{\cosh{x}} e^{-a x^2 (x^2-\pi^2)} \cos{2 \pi a x^3} $$

The integral over $C_5$ is, in the limit as $\epsilon \to 0$, is

$$i \epsilon\int_{-\pi/2}^{\pi/2} \frac{d\phi \, e^{i \phi}}{\sinh{(\epsilon e^{i \phi}})} e^{-a [\epsilon^2 e^{i 2 \phi}+\pi^2/4]^2} \approx i \epsilon\int_{-\pi/2}^{\pi/2} \frac{d\phi \, e^{i \phi}}{\epsilon e^{i \phi}} e^{-a (\pi^2/4)^2} = i \pi e^{-a \pi^4/16}$$

By Cauchy's Theorem, the sum of the integrals over these contours is zero. Thus we get

$$-i 2 \int_0^{\infty} \frac{dx}{\cosh{x}} e^{-a x^2 (x^2-\pi^2)} \cos{2 \pi a x^3} + i \pi e^{-a \pi^4/16}=0$$

The result follows.


Apparently I made one edit too many and it was pointed out that the solution, clean as it is now, has become a deux ex machina. Fair enough: I will put back the motivation behind the sleight of hand.

Basically, when I looked at the integral, it screamed "complete the square in the exponentials." So, I rewrote the integral as the real part of

$$\int_0^{\infty} \frac{dx}{\cosh{x}} e^{-a x^2 (x+i \pi)^2}$$

The problem is that, while tempting to make a substitution like $u=x (x+i \pi)$, you end up having to follow a curve in the complex plane. Better, I thought, to work with something more symmetric if I am going to work in the complex plane anyway. Thus I let $x=y-i \pi/2$ so that the above integral becomes

$$-i \int_{i \pi/2}^{\infty+i \pi/2} \frac{dy}{\sinh{y}} e^{-a (y^2+\pi^2/4)^2}$$

And to me it became clear that the contour $C$ above would be useful once I did this. The rest is, well, above.

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    $\begingroup$ +1. You are like a machine for doing integrals. (I mean this as a compliment.) $\endgroup$ – Potato Sep 8 '13 at 1:03
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    $\begingroup$ that's very good answer , integral machine . (+1) $\endgroup$ – what'sup Sep 8 '13 at 1:16
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    $\begingroup$ I guess a consensus has been reached. $\endgroup$ – Ron Gordon Sep 8 '13 at 1:21
  • $\begingroup$ I think your evaluation was better before you edited it. The way it's written now gives no insight into how you picked $f(z)$ and the contour. $\endgroup$ – Random Variable Sep 9 '13 at 18:06
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    $\begingroup$ RV is probably being ridiculed on AoPS because there are certain regulars who have huge egos and do not like it if someone as cognizant as RV points out a mistake. $\endgroup$ – Cody Oct 12 '13 at 15:38

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