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.


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

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

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

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

letting a=b=1 results in your posted solution.

  • $\begingroup$ Should there be an equality somewhere in the second line? $\endgroup$
    – Meow
    May 19 '14 at 10:03
  • $\begingroup$ @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. $\endgroup$
    – user85798
    May 19 '14 at 11:13
  • $\begingroup$ I'm not sure which paper it is exactly. He touches on it here, but doesn't solve it to the extent given above. $\endgroup$
    – Meow
    May 19 '14 at 11:26
  • 1
    $\begingroup$ @Alyosha See here $\endgroup$
    – user85798
    May 19 '14 at 11:52
  • $\begingroup$ 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. $\endgroup$
    – Cody
    May 19 '14 at 14:25

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