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I am trying to show that$$\displaystyle \int_0^\infty \frac {\cos {\pi x}} {e^{2\pi \sqrt x} - 1} \mathrm d x = \dfrac {2 - \sqrt 2} {8}$$

I have verified this numerically on Mathematica.

I have tried substituting $u=2\pi\sqrt x$ then using the cosine Maclaurin series and then the $\zeta \left({s}\right) \Gamma \left({s}\right)$ integral formula but this doesn't work because interchanging the sum and the integral isn't valid, and results in a divergent series.

I am guessing it is easy with complex analysis, but I am looking for an elementary way if possible.

Venus
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user85798
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1 Answers1

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This integral is one of Ramanujan's in his Collected Papers where he also shows the connection with the sin case.

Define $$\phi(a,b)=\int_{0}^{\infty}\frac{\cos(\frac{a\pi x}{b})}{e^{2\pi \sqrt{x}}-1}dx$$

if $a$ and $b$ are both odd. In this case, they are both $a=b=1$.

Then, $$\phi(a,b)= \frac{1}{4}\sum_{k=1}^{b}(b-2k)\cos\left(\frac{k^{2}\pi a}{b}\right)-\frac{b}{4a}\sqrt{b/a}\sum_{k=1}^{a}(a-2k)\sin\left(\frac{\pi}{4}+\frac{k^{2}\pi b}{a}\right).$$

Letting $a=b=1$ results in your posted solution.

Gary
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Cody
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  • Should there be an equality somewhere in the second line? – Meow May 19 '14 at 10:03
  • @Alyosha I believe that is supposed to be the solution to the general integral. A proof would be nice though. I'm guessing I won't find a proof in Ramanujan's papers. – user85798 May 19 '14 at 11:13
  • I'm not sure which paper it is exactly. He touches on it here, but doesn't solve it to the extent given above. – Meow May 19 '14 at 11:26
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    @Alyosha See here – user85798 May 19 '14 at 11:52
  • I found it in his Collected Papers and in Notebook 4 around page 297. I can not find them free online, but I do have them in pdf format. A friend sent them to me several years ago. I think Notebook 4 can be found free online in a pdf. But, that link Oliver gave is nice. – Cody May 19 '14 at 14:25
  • One thing he does is makes a sub like $x\to x^{2}$. Then, one obtains $$2\int_{0}^{\infty}\frac{x\cos(\pi x^{2})}{e^{2\pi x}-1}dx$$. Then, he uses the series for cos to get: $$2\int_{0}^{\infty}\frac{x}{e^{2\pi x}-1}\sum_{k=0}^{\infty}\frac{(-1)^{k}(\pi x^{2})^{2k}}{(2k)!}$$. $$2\sum_{k=0}^{\infty}\frac{(-1)^{k}\pi^{2k}}{(2k)!}\int_{0}^{\infty}\frac{x^{4k+1}}{e^{2\pi x}-1}dx$$. The integral can be written in terms of Bernoulli numbers. Of course, the rest of it is in the notebook. I will try to find the link for notebook 4 and post a link to it. – Cody May 19 '14 at 14:32
  • See here:http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20IV.pdf – Cody May 19 '14 at 14:35
  • @cody In my question I say that I tried that but was not allowed to do it because the interchange of the sum and integral isn't valid. That final sum in your previous comment is divergent. It appears Ramanujan got around this, since in his notebook the sum is finite. – user85798 May 19 '14 at 22:14
  • Yeah, sorry I could not be of more help. But, I think that notebook 4 has a solution in it as to what Ramanujan done to resolve it. It is on page 298 around 9.5 and 9.6. Looking at it again, he uses $$2\int_{0}^{\infty}\frac{x\cos(\theta x^{2})}{e^{2\pi x}-1}dx$$ and uses the series I mentioned when $\theta\to 0, ;\ \theta\to \infty$. It looks that way. Things are kind of scattered around. also, down on page 302, they list out examples for various values of theta. – Cody May 20 '14 at 12:34
  • See http://ramanujan.sirinudi.org/Volumes/published/ram12.html – Paramanand Singh Jan 28 '23 at 01:14