Euler Integral + Euler Series

Note that the integral is $I \; = \; \int_0^\infty \frac{\cos \sqrt{x}}{e^{2\pi \sqrt{x}} - 1}\,dx \; = \; \sum_{n=1}^\infty \int_0^\infty \cos\sqrt{x} e^{-2\pi n \sqrt{x}}\,dx \; = \; 2\sum_{n=1}^\infty \int_0^\infty u\cos u e^{-2\pi n u}\,du$ Note that

$\int_0^\infty u \cos u e^{-pu}\,du \; = \; -\frac{d}{dp}\int_0^\infty \cos u e^{-pu}\,du \; =\; -\frac{d}{dp}\frac{p}{p^2+1} \; =\; \frac{p^2 - 1}{(p^2+ 1)^2}$

for $p > 0$ , and hence $I \; =\; 2\sum_{n=1}^\infty \frac{4\pi^2 n^2 - 1}{(4 \pi^2 n^2 + 1)^2} \; =\; 1 - \tfrac14\mathrm{cosech}^2\tfrac12$ using some standard series identities, which means that $I = \boxed{0.079326405792207681055}$ .

$\displaystyle\int_0^∞ \dfrac{\cos(\sqrt{x})}{e^{2π\sqrt{x}}-1} dx =^{\textrm {substituting} {x=u^2} }\int_0^∞ 2u\cos(u)\sum_{n=1}^∞ e^{-2πnu} du$

Considering $\cos(u)= \dfrac{e^{iu} +e^{-iu}}{2}$ , the integral becomes

$\displaystyle \sum_{n=1}^∞ \int_0^∞ u e^{-u(2πn-i)} + ue^{-u(2πn+i)} du$

$\displaystyle= \sum_{n=1}^∞ \dfrac{1}{(2πn-i)^2} +\dfrac{1}{(2πn+i)^2}=\dfrac{1}{4} \sum_{n=1}^∞ \dfrac{1}{(πn-i/2)^2} +\dfrac{1}{(πn+i/2)^2}$ Now $\sum_{n=-∞ }^∞ \dfrac{1}{(πn+x)^2} = \csc^2(x)$

Here it is $\displaystyle \dfrac{1}{4} \sum_{n=-∞}^∞ \dfrac{1}{(πn+\frac{i}{2})^2} +1 = 1-\dfrac{e}{(e-1)^2}$ See this question