How do I Evaluate the Integral $\int_{-\infty }^{\infty } \frac{1}{e^{x^2}+1} \, dx$?

Question

Like the question states, how do I evaluate the integral $$\int_{-\infty }^{\infty } \frac{1}{e^{x^2}+1} \, dx$$ I know of no methods to evaluate this, but when I plug this into Mathematica and Wolfram Alpha, it returns $$\int_{-\infty }^{\infty } \frac{1}{e^{x^2}+1} \, dx=\left(1-\sqrt{2}\right) \sqrt{\pi } \zeta \left(\frac{1}{2}\right)$$ where $\zeta (x)$ is the Riemann Zeta Function.

Answer 1

The given integral equals $$ 2\int_{0}^{+\infty}\frac{dx}{1+e^{x^2}} = \int_{0}^{+\infty}\frac{dz}{\sqrt{z}(1+e^{z})}=\sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\frac{e^{-nz}}{\sqrt{z}}\,dz $$ or $$ \sqrt{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{\sqrt{n}} = \sqrt{\pi}\,\eta\left(\frac{1}{2}\right)=(1-\sqrt{2})\sqrt{\pi}\,\zeta\left(\frac{1}{2}\right) $$ as claimed by WA.