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Question Number 134651 by 0731619177 last updated on 06/Mar/21

Answered by Ar Brandon last updated on 06/Mar/21

$$\mathrm{sinx}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}\Rightarrow \sin \pi \mathrm{x}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\left(\pi \mathrm{x}\right)}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}$$$$\mathcal{I}={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{x}(1-{\mathrm{x}}^2)}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\left(\pi \mathrm{x}\right)}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\pi }^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{\mathrm{x}(1-{\mathrm{x}}^2)}\mathrm{dx}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\pi }^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{2\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{2\mathrm{k}}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\pi }^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\left[\frac{{\mathrm{x}}^{2\mathrm{n}+2\mathrm{k}+1}}{(2\mathrm{n}+2\mathrm{k}+1)}\right]}_0^{\mathrm{\infty }}$$

Commented by 0731619177 last updated on 06/Mar/21

$$tanks$$

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