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

Prove ∫_0 ^( ∞) (x^a /(1+e^x ))dx = (1−2^(−a) )𝛇(a+1)𝚪(a+1)

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

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

I(a)=∫_0 ^∞ (x^a /(1+e^x ))dx=Σ_(n=1) ^∞ (−1)^(n+1) ∫_0 ^∞ e^(−nx) x^a dx =Σ_(n=1) ^∞ (−1)^(n+1) (1/n^(a+1) )Γ(a+1) =Γ(a+1)ζ(a+1)(1−(1/2^(a+1−1) ))=a!ζ(a+1)(1−2^(−a) )

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

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