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Question Number 60534 by aliesam last updated on 21/May/19

Commented by maxmathsup by imad last updated on 22/May/19

$$wehave{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+e^{x^2}}=2{\int }_0^{+\mathrm{\infty }}\frac{e^{-x^2}}{1+e^{-x^2}}dx$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-{nx}^2}\right)dx=2\sum _{=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)x^2}dx$$$${=}_{\sqrt{n+1}x=u}2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{\mathit{n}}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{\mathit{d}\mathit{u}}{\sqrt{\mathit{n}+1}}=2\frac{\sqrt{\pi }}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\sqrt{n+1}}$$$$=\sqrt{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\sqrt{n+1}}wehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\sqrt{n+1}}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathit{n}-1}}{\sqrt{\mathit{n}}}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{2n}}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2n+1}}=-\frac{1}{\sqrt{2}}\xi \left(\frac{1}{2}\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2n+1}}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{2n}}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2n+1}}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2n+1}}=\xi \left(\frac{1}{2}\right)-\frac{1}{\sqrt{2}}\xi \left(\frac{1}{2}\right)$$$$=(1-\frac{1}{\sqrt{2}})\xi \left(\frac{1}{2}\right)\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\sqrt{n+1}}=-\frac{1}{\sqrt{2}}\xi \left(\frac{1}{2}\right)+(1-\frac{1}{\sqrt{2}})\xi \left(\frac{1}{2}\right)$$$$=(1-\sqrt{2})\xi \left(\frac{1}{2}\right)\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+e^{x^2}}=\sqrt{\pi }(1-\sqrt{2})\xi \left(\frac{1}{2}\right)$$$$theequalityisproved.$$

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