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Question Number 200395 by sonukgindia last updated on 18/Nov/23

Answered by witcher3 last updated on 18/Nov/23

$${\mathrm{I}}_{10}\to {\mathrm{I}}_9$$$${\mathrm{I}}_{10}=\frac{1}{\mathrm{m}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{m}}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{\mathrm{m}}\beta (\frac{1}{\mathrm{m}};1-\frac{1}{\mathrm{m}})=\frac{\pi }{\mathrm{msin}\left(\frac{\pi }{\mathrm{m}}\right)}$$$${\mathrm{I}}_9=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2\mathrm{n}}}={2\mathrm{I}}_{10}\left(2\mathrm{n}\right)=\frac{\pi }{\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}$$

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