Question Number 200395 by sonukgindia last updated on 18/Nov/23
Answered by witcher3 last updated on 18/Nov/23
$$\mathrm{I}_{\mathrm{10}} \rightarrow\mathrm{I}_{\mathrm{9}} \\ $$$$\mathrm{I}_{\mathrm{10}} =\frac{\mathrm{1}}{\mathrm{m}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{m}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{m}}\beta\left(\frac{\mathrm{1}}{\mathrm{m}};\mathrm{1}−\frac{\mathrm{1}}{\mathrm{m}}\right)=\frac{\pi}{\mathrm{msin}\left(\frac{\pi}{\mathrm{m}}\right)} \\ $$$$\mathrm{I}_{\mathrm{9}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2n}} }=\mathrm{2I}_{\mathrm{10}} \left(\mathrm{2n}\right)=\frac{\pi}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$