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How to prove $$\int_0^{\infty } \frac{\sin (x)}{\sqrt{x} \left(\cos ^2(x)+1\right)} \, dx= \sqrt{2\pi} \sum _{k=1}^{\infty } \frac{(-1)^{k-1} \left(\sqrt{2}-1\right)^{2 k-1}}{\sqrt{2 k-1}}$$ Any help will be appreciated.


Update: Using @uniquesailor's hint the problem is solved. Indeed, set $b=3-2 \sqrt{2}$ and make use of $\frac{1}{\cos ^2(x)+1}=\frac{2}{3 \left(\frac{1}{3} \cos (2 x)+1\right)}$, one may broke the integrand into Fourier series based on his Poisson type formula. Then, upon using Fresnel's result $\int_0^{\infty } \frac{\sin (x)}{\sqrt{x}} \, dx=\sqrt{\frac{\pi }{2}}$ and the trigonometric identity $2 \sin (x) \cos (2 n x)=\sin ((2 n+1) x)-\sin ((2 n-1) x)$, the integral is transformed to RHS after rearranging. According to Benidict RHS is also equivalent to $-i \sqrt{2\pi} \chi_{\frac{1}{2}}(i (\sqrt{2}-1))$.

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It can be proved by applying this relationship:

$$2\sum_{n=0}^{\infty} \beta^n cos(nt)-1=\frac{1-\beta^2}{1+\beta^2}\frac{1}{1-\frac{2\beta}{1+\beta^2}cos(t)}$$

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  • $\begingroup$ Good use of Fourier series. I've completed the details. $\endgroup$
    – Iridescent
    Feb 27 '20 at 4:34
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Not quite complete, but this looks promising....

$$ \Phi(z,s,a) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-at}}{1-ze^{-t}}\;dt $$ $$ \chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu} = 2^{-\nu} z \Phi(z^2,\nu,\frac{1}{2}) $$

For your sum we clearly have something like $\nu = \frac{1}{2}, z = (\sqrt{2}-1)$

$$ \chi_{\frac{1}{2}}((\sqrt{2}-1)) = \sum_{k=0}^\infty \frac{(\sqrt{2}-1)^{2k+1}}{\sqrt{2k+1}} =\frac{2^{-1/2}(\sqrt{2}-1)}{\Gamma(\frac{1}{2})}\int_0^\infty \frac{x^{-\frac{1}{2}}e^{-x/2}}{1-(\sqrt{2}-1)^2e^{-x}}\;dx $$

$$ \chi_{\frac{1}{2}}((\sqrt{2}-1)) = \sum_{k=1}^\infty \frac{(\sqrt{2}-1)^{2k-2}}{\sqrt{2k-2}} =\frac{2^{-1/2}(\sqrt{2}-1)}{\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{x}}\frac{e^{-x/2}}{1-(\sqrt{2}-1)^2e^{-x}}\;dx $$

Might have to rearrange the exp/trig terms? There is a missing factor of $(-1)^{k-1}$, perhaps change $z$ slightly. I haven't checked it.

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