ガウス積分のような定積分

\begin{align*} \int_{0}^{\infty}\frac{1}{1+e^{x^{2}}}dx & =\int_{0}^{\infty}\frac{e^{-x^{2}}}{1+e^{-x^{2}}}dx\\ & =\int_{0}^{\infty}e^{-x^{2}}\sum_{k-0}^{\infty}\left(-e^{-x^{2}}\right)^{k}dx\\ & =\int_{0}^{\infty}\sum_{k-0}^{\infty}\left(-1\right)^{k}e^{-\left(k+1\right)x^{2}}dx\\ & =\sum_{k-0}^{\infty}\left(-1\right)^{k}\int_{0}^{\infty}e^{-\left(k+1\right)x^{2}}dx\\ & =\sum_{k-0}^{\infty}\left(-1\right)^{k}\int_{0}^{\infty}e^{-\left(k+1\right)x^{2}}dx\\ & =\sum_{k-0}^{\infty}\left(-1\right)^{k}\frac{1}{2}\sqrt{\frac{\pi}{k+1}}\cmt{\because\int_{0}^{\infty}e^{-\left|a\right|x^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{\left|a\right|}}}\\ & =\frac{\sqrt{\pi}}{2}\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{1}{k^{\frac{1}{2}}}\\ & =\frac{\sqrt{\pi}}{2}\eta\left(\frac{1}{2}\right)\\ & =\frac{\sqrt{\pi}}{2}\left(\zeta\left(\frac{1}{2}\right)-\frac{2}{2^{\frac{1}{2}}}\zeta\left(\frac{1}{2}\right)\right)\\ & =\frac{\sqrt{\pi}}{2}\cdot\frac{\sqrt{2}-2}{\sqrt{2}}\zeta\left(\frac{1}{2}\right)\\ & =\frac{\sqrt{\pi}}{2}\cdot\frac{2-2\sqrt{2}}{2}\zeta\left(\frac{1}{2}\right)\\ & =-\frac{\sqrt{2}-1}{2}\sqrt{\pi}\zeta\left(\frac{1}{2}\right) \end{align*}