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Question Number 204569 by pticantor last updated on 21/Feb/24

$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathit{o}\mathit{f}$$$$\mathit{I}={\int }_0^{+\mathrm{\infty }}\mathit{l}\mathit{n}(1+{\mathit{e}}^{-\mathit{x}})\mathit{d}\mathit{x}\mathit{n}\mathit{o}\mathit{w}\mathit{i}\mathit{n}\mathit{g}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{{\mathit{n}}^2}=\frac{{\pi }^2}{6}$$

Answered by witcher3 last updated on 22/Feb/24

$$\mathrm{\forall }\mathrm{x}\in {\mathbb{R}}_{+},{\mathrm{e}}^{-\mathrm{x}}<1$$$$\ln (1+{\mathrm{e}}^{-\mathrm{x}})=\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}$$$${\int }_0^{\mathrm{\infty }}\ln (1+{\mathrm{e}}^{-\mathrm{x}})\mathrm{dx}=\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}\mathrm{dx}$$$$=\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}=\eta \left(2\right)=(1-2^{1-2})\zeta \left(2\right)=\frac{{\pi }^2}{12}$$

Commented by pticantor last updated on 23/Feb/24

$$plshowdoyoumanagetohave(1-2^{1-2})\zeta \left(2\right)?$$

Commented by witcher3 last updated on 23/Feb/24

$$\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}=\sum _{\mathrm{n}⩾0}\frac{1}{{(2\mathrm{n}+1)}^2}-\frac{1}{{4\mathrm{n}}^2}$$$$=\mathrm{\Sigma }(\frac{1}{{\mathrm{n}}^2}-\frac{1}{{\left(2\mathrm{n}\right)}^2}-\frac{1}{{4\mathrm{n}}^2})=\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^2}-\frac{1}{2}\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^2}=\zeta \left(2\right)-\frac{1}{2}\zeta \left(2\right)=(1-2^{1-2})\zeta \left(2\right)$$

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