Question-213662

Question Number 213662 by Spillover last updated on 13/Nov/24

Answered by mathmax last updated on 14/Nov/24

I=∫_0 ^∞ (x^2 /(1+e^x ))dx ⇒I=∫_0 ^∞ ((e^(−x) x^2 )/(1+e^(−x) ))dx =∫_0 ^∞ x^2 e^(−x) Σ_(n=0) ^∞ (−1)^n e^(−nx) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ x^2 e^(−(n+1)x) dx (n+1)x=t =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ (t^2 /((n+1)^2 ))e^(−t) (dt/(n+1)) =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^3 ))∫_0 ^∞ t^(3−1) e^(−t) dt =Γ(3).Σ_(n=0) ^∞ (((−1)^n )/((n+1)^3 )) we have Σ_(n=0) ^∞ (((−1)^n )/((n+1)^3 ))=Σ_(n=1) ^∞ (((−1)^(n−1) )/n^3 ) =eta(3)=(1−2^(1−3) )ζ(3)=(1−(1/4))ζ(3) =(3/4)ζ(3) Γ(3)=2!=2 ⇒I=(3/2)ζ(3)

$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{e}^{{x}} }{dx}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {x}^{\mathrm{2}} }{\mathrm{1}+{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}} \sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {e}^{−{nx}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−\left({n}+\mathrm{1}\right){x}} {dx}\:\:\:\left({n}+\mathrm{1}\right){x}={t} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{e}^{−{t}} \frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{3}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$=\Gamma\left(\mathrm{3}\right).\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} } \\ $$$$={eta}\left(\mathrm{3}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{3}} \right)\zeta\left(\mathrm{3}\right)=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\zeta\left(\mathrm{3}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)\:\:\:\:\:\Gamma\left(\mathrm{3}\right)=\mathrm{2}!=\mathrm{2}\:\Rightarrow{I}=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$

Commented by Spillover last updated on 16/Nov/24

$${great}\:{solution} \\ $$

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