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Question Number 155133 by amin96 last updated on 25/Sep/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^3}=?$$

Answered by mnjuly1970 last updated on 26/Sep/21

$$\frac{{\pi }^3}{32}$$

Answered by qaz last updated on 26/Sep/21

$$\mathrm{S}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}$$$$=\frac{1}{4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}.......{\mathrm{e}}^{\mathrm{x}}\to \tan \theta$$$$=\frac{1}{4}{\int }_0^{\pi /2}{\ln }^2\tan \theta \mathrm{d}\theta$$$${\int }_0^{\pi /2}{\tan }^{2\mathrm{k}-1}\theta \mathrm{d}\theta =\frac{\pi }{2}\csc \left(\mathrm{k}\pi \right)$$$$\Rightarrow {\int }_0^{\pi /2}{\tan }^{2\mathrm{k}-1}\theta \cdot {\ln }^2\tan \theta \mathrm{d}\theta =\frac{{\pi }^3}{8}(\csc \left(\mathrm{k}\pi \right)\cot \left(\mathrm{k}\pi \right)+{\csc }^3\left(\mathrm{k}\pi \right))$$$$\mathrm{let}\mathrm{k}=\frac{1}{2}\mathrm{we}\mathrm{get}\mathrm{S}=\frac{{\pi }^3}{32}$$

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