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I came across the interesting family of integrals that was studied by Ramanujan and was one of subjects of his first letter to Hardy.

$$ \phi(n):=\int_{0}^{\infty} \frac{\cos n x}{e^{2 \pi \sqrt{x}}-1} d x $$

He offers a functional equation

$$ \int_{0}^{\infty} \frac{\sin n x}{e^{2 \pi \sqrt{x}}-1} d x=\phi(n)-\frac{1}{2 n}+\phi\left(\frac{\pi^{2}}{n}\right) \sqrt{\frac{2 \pi^{3}}{n^{3}}} $$

And give some special cases,

$$ \phi(0)=\frac{1}{12} ; \quad \phi\left(\frac{\pi}{2}\right)=\frac{1}{4 \pi} ; \quad \phi(\pi)=\frac{2-\sqrt{2}}{8} ; \quad \phi(2 \pi)=\frac{1}{16} $$

$$ \begin{gathered} \phi\left(\frac{2 \pi}{5}\right)=\frac{8-3 \sqrt{5}}{16} ; \quad \phi\left(\frac{\pi}{5}\right)=\frac{6+\sqrt{5}}{4}-\frac{5 \sqrt{10}}{8} \\ \phi\left(\frac{2 \pi}{3}\right)=\frac{1}{3}-\sqrt{3}\left(\frac{3}{16}-\frac{1}{8 \pi}\right) \end{gathered} $$

The particular case $\phi(0)=\frac{1}{12}$ is easy to proof:

$$ \begin{aligned} \phi(0)&=\int_{0}^{\infty} \frac{1}{e^{2 \pi \sqrt{x}}-1} d x \qquad (x \mapsto x^2)\\ &=2 \int_{0}^{\infty} \frac{x}{e^{2 \pi x}-1} d x \qquad (2\pi x \mapsto x)\\ &=\frac{1}{2 \pi^2} \int_{0}^{\infty} \frac{x}{e^{ x}-1} d x\\ &=\frac{1}{2 \pi^2} \zeta(2)\\ &=\frac{1}{12} \qquad \blacksquare \end{aligned} $$

I am interesting in computing $\phi(\pi)=\frac{2-\sqrt{2}}{8}$

$$ \phi(n)=\int_{0}^{\infty} \frac{\cos n x}{e^{2 \pi \sqrt{x}}-1} d x $$

Letting $x \mapsto x^2$ we obtain

$$ \begin{aligned} \phi(n)&=2\int_{0}^{\infty} \frac{x\cos n x^2}{e^{2 \pi x}-1} d x \qquad (2 \pi x \mapsto x)\\ &=\frac{1}{2 \pi^2}\int_{0}^{\infty} \frac{x\cos \frac{n x^2}{4 \pi^2} }{e^{ x}-1} d x \end{aligned} $$

Recall the Bernoulli polynomials generating function

$$\sum_{k=0}^{\infty} \frac{B_{k} \, x^k}{k!} = \frac{x}{e^{x} - 1}$$

Then, letting $ n \rightarrow \pi$ we get

$$ \begin{aligned} \phi(\pi)&=\frac{1}{2 \pi^2}\sum_{k=0}^{\infty} \frac{B_{k} }{k!}\int_{0}^{\infty} x^k\cos \frac{ x^2}{4 \pi} d x \qquad (\frac{x^2}{4 \pi}=w)\\ &=\frac{1}{2 \pi^2}\sum_{k=0}^{\infty} \frac{B_{k} }{k!}(2 \sqrt{\pi})^{k+1}\int_{0}^{\infty} w^{\frac{k-1}{2}}\cos w d w\\ &=\frac{1}{2 \pi^2}\sum_{k=0}^{\infty} \frac{B_{k} }{k!}(2 \sqrt{\pi})^{k+1}\Gamma \left(\frac{k+1}{2} \right)\cos\left(\frac{(k+1) \pi}{4} \right) \end{aligned} $$

Which does not seem very promising!

Any idea how to proceed?

Ricardo770
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  • Find a function $f_n(x)$ which gives $$-\frac{i}{2\sqrt{x}}\left(f_n(i\sqrt{x})-f_n(-i\sqrt{x})\right)=\cos(nx),$$ then you will have $$-\frac{i}{2}\int_0^\infty\frac{f_n(i\sqrt{x})-f_n(-i\sqrt{x})}{e^{2\pi\sqrt{x}}-1}\frac{dx}{\sqrt x}=\phi(n)=\frac12f_n(0)+\sum_{k\ge1}f_n(k).$$ This is of course due to Ramanujan Summation. – clathratus Sep 13 '21 at 01:50
  • the aforementioned $f(x)$ in my other comment must be holomorphic for $\Re(x)\ge0$ and $|f|$ must be bounded by some $C/|x|^{1+\epsilon}$ for positive $C, \epsilon$ – clathratus Sep 13 '21 at 02:02
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    Why don't you instead look at the proof by Ramanujan? Check http://ramanujan.sirinudi.org/Volumes/published/ram12.html – Paramanand Singh Sep 14 '21 at 16:54
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    And contrary to the popular perception, Ramanujan provided proofs for many of his results and I have found his technique far more economical and easy to understand than the modern approaches. – Paramanand Singh Sep 14 '21 at 16:57
  • @ParamanandSingh thank you very much for the link to Ramanujan´s paper. I didn´t know it. I´ll have a look. – Ricardo770 Sep 14 '21 at 17:10

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