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Here is a seemingly challenging integral some may try their hand at.

$$ \int_{0}^{\infty} {\cos\left(\pi x^{2}\right)\over 1 + 2\cosh\left(\,2\,\pi\,x\,/\,\sqrt{\,3\,}\,\right)}\,{\rm d}x ={\,\sqrt{\,2\,}\, - \,\sqrt{\,6\,}\, + 2\, \over 8} $$

It appears to be rather tough. I think maybe it is one of Ramanujans.

But, some of the clever individuals on SE may come up with a method (Ron, robjohn, etc. ) :)

Maybe residues will work, so I wasn't sure rather or not to tag it under contour integration. Of course, I am not nearly as adept at it as many here at SE, so perhaps some one has an approach....residues or otherwise.

Felix Marin
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Cody
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  • May I ask where you found this result ? It is so beautiful. – Claude Leibovici Feb 05 '14 at 11:25
  • I got to thinking that the technique that robjohn used on $\int_{0}^{\infty}\frac{\sin(\pi x^{2})}{\sinh^{2}(\pi x)}dx$ may be useful in some fashion with this one. If we let $z=x\pm \frac{i}{\sqrt{3}}$ and subtract, then we get $\frac{\sqrt{3}\cosh(\pi x/\sqrt{3})\cos(\pi x^{2})}{\sinh(\sqrt{3}\pi x)}$. or maybe a rectangle with height $\frac{\sqrt{3}i}{2}$. Maybe I am full of it though. I can not locate the integral I mentioned that robjohn had done. It is here somewhere. – Cody Feb 05 '14 at 20:07
  • Here is the problem I was talking about that robjohn done sometime back: http://math.stackexchange.com/questions/61605/tough-integral-involving-sinx2-and-sinh2-x/455751#455751 I thought perhaps with the correct function, we may be able to tackle it this way. I liked robjohn's clever solution to this one in the link. – Cody Feb 06 '14 at 23:48
  • Claude, I got this from Integrals and Series. He is the one who initially posted it on his site. – Cody Feb 07 '14 at 18:55

1 Answers1

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It should be little surprise that this can be attacked via contour integration. As always, the challenge is to find the contour and the function to integrate over that contour. At the risk of presenting a deus ex machina, I am simply going to present these and demonstrate how they provide what we need.

Consider the following integral:

$$\oint_C dz \;f(z) $$

where

$$f(z) = \frac{e^{i \pi z^2}}{\sinh{\left (\sqrt{3} \pi z\right )} \left [ 2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} z\right )}-1\right ]}$$

and $C$ is the rectangle with vertices in the complex plane $\pm R\pm i \sqrt{3}/2$. This contour integral is thus equal to

$$\int_{-R}^R dx \; f\left (x-i \frac{\sqrt{3}}{2}\right ) + i \int_{-\sqrt{3}/2}^{\sqrt{3}/2} dy \, f(R+i y) \\ \int_R^{-R} dx\; f\left (x+i \frac{\sqrt{3}}{2}\right ) +i \int_{\sqrt{3}/2}^{-\sqrt{3}/2} dy \, f(-R+i y)$$

It should be clear that the second and fourth integrals vanish as $R\to\infty$, as these integrals vanish as $\pi e^{-\pi R}$ as $R\to\infty$. In this limit, then, the contour integral is equal to

$$\int_{-\infty}^{\infty} dx \; \left [ f\left (x-i \frac{\sqrt{3}}{2}\right ) - f\left (x+i \frac{\sqrt{3}}{2}\right )\right ] = i 2 e^{-i 3 \pi/4}\int_{-\infty}^{\infty} dx \frac{e^{i \pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1}$$

I chose not to clutter up this space with the algebra involved in producing this last equation. The reader, however, should prove to his/herself that this is indeed correct.

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of $f$ inside $C$. I leave it to the reader to verify that these poles are at $z=0$, $z=\pm i \sqrt{3}/3$, and $z=\pm i\sqrt{3}/6$. The residues at these poles are straightforward to compute because the poles are simple:

$$\operatorname*{Res}_{z=0} f(z) = \frac1{\sqrt{3} \pi}$$ $$\operatorname*{Res}_{z=\pm i \sqrt{3}/3} f(z) = \frac{e^{-i \pi/3}}{2 \sqrt{3} \pi}$$ $$\operatorname*{Res}_{z=\pm i \sqrt{3}/6} f(z) = -\frac{e^{-i \pi/12}}{2 \pi}$$

So the residue theorem states, equivalently, that

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i \pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} &= \frac{e^{i 3 \pi/4}}{\sqrt{3}} + \frac{e^{i 5 \pi/12}}{\sqrt{3}} - e^{i 2 \pi/3}\\ &= \frac1{\sqrt{6}} (-1+i) + \frac1{2 \sqrt{6}} \left [ (\sqrt{3}-1) + i (\sqrt{3}+1)\right ] + \frac12 (1-i \sqrt{3})\end{align}$$

The integral of interest here is equal to $1/2$ the real part of the above, which is

$$\int_0^{\infty} dx \frac{\cos{\pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} = \frac{2+\sqrt{2}-\sqrt{6}}{8} $$

which was to be shown.

ADDENDUM

We may as well reap the reward of the imaginary part we get for free:

$$\int_0^{\infty} dx \frac{\sin{\pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} = \frac{\sqrt{2}+\sqrt{6}-2 \sqrt{3}}{8} $$

Ron Gordon
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  • Excellent RG. I knew you would get it sooner or later. Man, I knew it was there. I was so close in finding the correct f(z). :( I even had the correct rectangle vertices, but failed to get the right f(z). I tried what you used sans the sinh in the denominator. ARG. Very good RG. This is a good one for your blog if you're so inclined. Again....kudos for your contour prowess. :) – Cody Feb 07 '14 at 10:11
  • @Cody: Still, this demonstrates that you are getting it - your thought process is much further along than most people who attempt these problems. – Ron Gordon Feb 07 '14 at 11:10
  • Thanks Ron. But, what I had done was used the same rectangle you did, but I tried $\frac{\sinh(\sqrt{3}\pi x)e^{\pi iz^{2}}}{2cosh(\frac{2\pi}{\sqrt{3}}x)-1}$ to see what would happen. The sinh should have been in the denominator. Then, I gave up because it got late. But, I feel encouraged that I was on the right path. Of course, this came from watching you, robjohn, and others over the past months slash years. – Cody Feb 07 '14 at 19:00
  • @Cody: wow, you were so close! Yeah, I hate to say this (my "deus ex machina" comment), but attacking integrals like this is just a lot of experience, and in the end, just guessing. Really hard to explain how I got there systematically. By now you more than most people know what I mean. – Ron Gordon Feb 07 '14 at 19:08
  • @Cody:

    Another possibly is to consider $$f(z) = \frac{\cos[\pi(z^2+\frac{3}{4})]}{\sinh(\sqrt{3} \pi z)[2 \cosh(\frac{2 \pi}{\sqrt{3}} z)-1]}$$

    Then $$ f(x- \frac{i \sqrt{3}}{2}) - f(x +\frac{i \sqrt{3}}{2}) = \frac{2i \cos (\pi x^{2})}{1 + 2 \cosh (\frac{2 \pi}{\sqrt{3}}x)}$$

     – Random Variable Feb 09 '14 at 06:30
  • Very good, RV. Yep, there is usually more than one way to skin an integral. :) Nice observation. – Cody Feb 09 '14 at 10:10