Integration Questions

Question Number 134291 by Lordose last updated on 02/Mar/21

$$\\$$$$\:\boldsymbol{\mathrm{Prove}}\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{dx}\:=\:\left(\mathrm{1}β\mathrm{2}^{β\mathrm{a}} \right)\boldsymbol{\zeta}\left(\mathrm{a}+\mathrm{1}\right)\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right) \\$$$$\\$$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} }{\mathrm{1}+{e}^{{x}} }{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {e}^{β{nx}} {x}^{{a}} {dx} \\$$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}^{{a}+\mathrm{1}} }\Gamma\left({a}+\mathrm{1}\right) \\$$$$=\Gamma\left({a}+\mathrm{1}\right)\zeta\left({a}+\mathrm{1}\right)\left(\mathrm{1}β\frac{\mathrm{1}}{\mathrm{2}^{{a}+\mathrm{1}β\mathrm{1}} }\right)={a}!\zeta\left({a}+\mathrm{1}\right)\left(\mathrm{1}β\mathrm{2}^{β{a}} \right) \\$$