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Question Number 198141 by Erico last updated on 11/Oct/23

$${\underset{0}{\int }}^1\frac{\mathrm{x}(1-\mathrm{x})}{\sin \left(\pi \mathrm{x}\right)}\mathrm{dx}=???$$

Answered by witcher3 last updated on 11/Oct/23

$$\frac{1}{\sin \left(\pi \mathrm{x}\right)}=\frac{{2\mathrm{ie}}^{-\mathrm{i}\pi \mathrm{x}}}{1-{\mathrm{e}}^{-2\mathrm{i}\pi \mathrm{x}}}$$$$=\left(2\mathrm{i}\right)\sum _{\mathrm{n}⩾0}{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{x}(1+2\mathrm{n})}$$$${\mathrm{I}}_{\mathrm{n}}={\int }_0^1\mathrm{x}(1-\mathrm{x}){\mathrm{e}}^{-\mathrm{i}\pi \mathrm{x}(1+2\mathrm{n})}\mathrm{dx}$$$$=\frac{1}{(1+2\mathrm{n})(-\mathrm{i}\pi )}{\int }_0^1(1-2\mathrm{x}){\mathrm{e}}^{-\mathrm{i}\pi \mathrm{x}(1+2\mathrm{n})}\mathrm{dx}$$$$=\frac{1}{{\pi }^2{(1+2\mathrm{n})}^2}\left[(1-2\mathrm{x})\stackrel{-\mathrm{i}\pi \mathrm{x}(1+2\mathrm{n})}{\mathrm{e}}\right]+\frac{2{\pi }^2}{{(1+2\mathrm{n})}^2}{\int }_0^1{\mathrm{e}}^{-\mathrm{i}\pi (1+2\mathrm{n})}\mathrm{dx}$$$$=\frac{2}{(1+2\mathrm{n}){\pi }^2}\left(\frac{2}{\mathrm{i}\pi (1+2\mathrm{n})}\right)=\frac{4}{\mathrm{i}{\pi }^3{(1+2\mathrm{n})}^3}$$$${\int }_0^1\frac{\mathrm{x}(1-\mathrm{x})}{\sin \left(\pi \mathrm{x}\right)}\mathrm{dx}=2\mathrm{i}\sum _{\mathrm{n}⩾0}{\mathrm{I}}_{\mathrm{n}}$$$$=\frac{8}{{\pi }^3}\mathrm{\Sigma }\frac{1}{{(1+2\mathrm{n})}^3}=\frac{7\zeta \left(3\right)}{{\pi }^3}$$$$$$

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