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Question-200395




Question Number 200395 by sonukgindia last updated on 18/Nov/23
Answered by witcher3 last updated on 18/Nov/23
I_(10) →I_9   I_(10) =(1/m)∫_0 ^∞ (t^((1/m)−1) /(1+t))dt  =(1/m)β((1/m);1−(1/m))=(π/(msin((π/m))))  I_9 =2∫_0 ^∞ (dx/(1+x^(2n) ))=2I_(10) (2n)=(π/(nsin((π/(2n)))))
$$\mathrm{I}_{\mathrm{10}} \rightarrow\mathrm{I}_{\mathrm{9}} \\ $$$$\mathrm{I}_{\mathrm{10}} =\frac{\mathrm{1}}{\mathrm{m}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{m}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{m}}\beta\left(\frac{\mathrm{1}}{\mathrm{m}};\mathrm{1}−\frac{\mathrm{1}}{\mathrm{m}}\right)=\frac{\pi}{\mathrm{msin}\left(\frac{\pi}{\mathrm{m}}\right)} \\ $$$$\mathrm{I}_{\mathrm{9}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2n}} }=\mathrm{2I}_{\mathrm{10}} \left(\mathrm{2n}\right)=\frac{\pi}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$

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