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Question Number 129763 by Eric002 last updated on 18/Jan/21

$$prove$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{1+e^{x^2}}dx=\sqrt{\pi }(1-\sqrt{2})\xi \left(\frac{1}{2}\right)$$

Answered by Dwaipayan Shikari last updated on 18/Jan/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+e^{x^2}}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{nx}^2}dx$$$$=\sqrt{\pi }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{\sqrt{n}}=\sqrt{\pi }\eta \left(\frac{1}{2}\right)=\sqrt{\pi }\zeta \left(\frac{1}{2}\right)(1-\frac{1}{2^{-\frac{1}{2}}})$$$$=\sqrt{\pi }\zeta \left(\frac{1}{2}\right)(1-\sqrt{2})$$

Commented by Eric002 last updated on 18/Jan/21

$$welldone$$

Answered by mnjuly1970 last updated on 18/Jan/21

$$\varphi =2{\int }_0^{\mathrm{\infty }}\frac{1}{1+e^{x^2}}dx\stackrel{x^2=t}{=}{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{t}(1+e^t)}$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{2}-1}}{1+e^t}=\mathrm{\Gamma }\left(s\right).\eta \left(s\right)$$$$where\mathrm{\Gamma }and\eta areEulergammaand$$$$Drichletetafunctionsrespectively.$$$$\therefore \varphi =\mathrm{\Gamma }\left(\frac{1}{2}\right)\eta \left(\frac{1}{2}\right)$$$$=\mathrm{\Gamma }\left(\frac{1}{2}\right)(1-2^{1-\frac{1}{2}})\zeta \left(\frac{1}{2}\right)$$$$=\sqrt{\pi }(1-\sqrt{2})\zeta \left(\frac{1}{2}\right)...$$$$$$$$$$

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