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Question Number 147411 by mnjuly1970 last updated on 20/Jul/21

Answered by mnjuly1970 last updated on 20/Jul/21

$$:\stackrel{\sqrt{x}:=y}{=}{\int }_0^{\mathrm{\infty }}\frac{2ydy}{1+e^y}=2{\int }_0^{\mathrm{\infty }}\frac{ydy}{1+e^y}$$$$:=2\eta \left(2\right)\mathrm{\Gamma }\left(2\right)=2.\frac{{\pi }^2}{12}=\frac{{\pi }^2}{6}$$$$$$$$$$

Commented by tabata last updated on 20/Jul/21

$$sirwhatsthemean\left(\eta \right)$$

Commented by mnjuly1970 last updated on 20/Jul/21

$$\eta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n^s}etafunction$$$$\eta \left(s\right)=(1-2^{1-s})\zeta \left(s\right)$$$$\eta \left(2\right)=(1-2^{1-2})\zeta \left(2\right)=\frac{1}{2}\zeta \left(2\right)=\frac{{\pi }^2}{12}$$$$\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}zetafunction...$$

Commented by Olaf_Thorendsen last updated on 20/Jul/21

$$\mathrm{the}\mathrm{Dirichlet}\mathrm{eta}\mathrm{function}$$

Answered by Olaf_Thorendsen last updated on 20/Jul/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{dx}{1+e^{\sqrt{x}}}$$$$\mathrm{I}\left(x\right)=\int \frac{dx}{1+e^{\sqrt{x}}}$$$$$$$$\mathrm{I}\left(x\right)=\int (1-\frac{e^{\sqrt{x}}}{1+e^{\sqrt{x}}})dx$$$$\mathrm{I}\left(x\right)=\int (1-2\sqrt{x}\frac{\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}}{1+e^{\sqrt{x}}})dx$$$$\mathrm{I}\left(x\right)=x-2\sqrt{x}\ln (1+{\mathrm{e}}^{\sqrt{x}})+\int \frac{\ln (1+e^{\sqrt{x}})}{\sqrt{x}}dx$$$$\mathrm{I}\left(x\right)=x-2\sqrt{x}\ln (1+{\mathrm{e}}^{\sqrt{x}})-{2\mathrm{Li}}_2(-e^{\sqrt{x}})$$$${\mathrm{Li}}_2(-1)=\eta \left(2\right)=(1-2^{1-2})\zeta \left(2\right)=\frac{1}{2}\zeta \left(2\right)$$$$\mathrm{Li}(-z)=\frac{1}{2}{\mathrm{Li}}_2\left(z^2\right)-{\mathrm{Li}}_2\left(z\right)$$$$\Rightarrow \mathrm{Li}(-e^{\sqrt{x}})=\frac{1}{2}{\mathrm{Li}}_2\left(e^{2\sqrt{x}}\right)-{\mathrm{Li}}_2\left(\sqrt{x}\right)$$$${\mathrm{Li}}_s\left(e^{\mu }\right)\underset{\mathrm{\infty }}{\sim }-\frac{{\mu }^2}{\mathrm{\Gamma }(s+1)}$$$$\Rightarrow {\mathrm{Li}}_2(-e^{\sqrt{x}})\underset{\mathrm{\infty }}{\sim }-\frac{1}{2}.\frac{4x}{2}+\frac{x}{2}=-\frac{x}{2}$$$$\mathrm{I}\left(x\right)\underset{\mathrm{\infty }}{\sim }x-2\sqrt{x}\times \sqrt{x-}2(-\frac{x}{2})\to 0$$$$\mathrm{I}\left(x\right)\underset{0}{\sim }-{2\mathrm{Li}}_2(-1)=-2\times \frac{1}{2}\zeta \left(2\right)=-\zeta \left(2\right)$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{dx}{1+e^{\sqrt{x}}}={\mathrm{I}}_{\mathrm{\infty }}-\mathrm{I}\left(0\right)=0-(-\zeta \left(2\right))$$$$\mathrm{I}=\zeta \left(2\right)=\frac{{\pi }^2}{6}$$

Commented by Tawa11 last updated on 03/Aug/21

$$\mathrm{Great}$$

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