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 $-i \sqrt{2\pi} \chi_{\frac{1}{2}}(i (\sqrt{2}-1))$.
Best Answer
It can be proved by applying this relationship:
$$2\sum_{n=0}^{\infty} \beta^n cos(nt)-1=\frac{1-\beta^2}{1+\beta^2}\frac{1}{1-\frac{2\beta}{1+\beta^2}cos(t)}$$