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Question Number 104780 by Aziztisffola last updated on 23/Jul/20

Commented by Dwaipayan Shikari last updated on 23/Jul/20

(π^2 /6) Γ(s).ζ(s)=∫_0 ^∞ (x^(s−1) /(e^x +1))dx Γ(2).ζ(2)=∫_0 ^∞ (x^(2−1) /(e^x +1))dx (2−1)!ζ(2)=∫_0 ^∞ (x/(e^x +1))dx ζ(2)=Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) so ∫_0 ^∞ (x/(e^x +1))=(π^2 /6)

$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Gamma\left({s}\right).\zeta\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$\Gamma\left(\mathrm{2}\right).\zeta\left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}−\mathrm{1}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$\left(\mathrm{2}−\mathrm{1}\right)!\zeta\left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}}{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$\zeta\left(\mathrm{2}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}}{{e}^{{x}} +\mathrm{1}}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Commented by Aziztisffola last updated on 23/Jul/20

Thanks sir , I found Γ(2).ζ(2)=(π^2 /6)

$$\mathrm{Thanks}\:\mathrm{sir}\:,\:\mathrm{I}\:\mathrm{found}\:\Gamma\left(\mathrm{2}\right).\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Answered by abdomsup last updated on 23/Jul/20

∫_0 ^∞ (x/(e^x −1))dx =∫_0 ^∞ ((xe^(−x) )/(1−e^(−x) ))dx =∫_0 ^∞ xe^(−x) (Σ_(n==0) ^∞ e^(−nx) )dx =Σ_(n=0) ^∞ ∫_0 ^∞ x e^(−(n+1)x) dx =_((n+1)x=t) Σ_(n=0) ^∞ ∫_0 ^∞ (t/(n+1))e^(−t) (dt/(n+1)) =Σ_(n=0) ^∞ (1/((n+1)^2 )) ∫_0 ^∞ t e^(−t) dt =Σ_(n=1) ^∞ (1/n^2 ).Γ(2) =ξ(2).Γ(2) =(π^2 /6)×1! =(π^2 /6)

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}}{{e}^{{x}} −\mathrm{1}}{dx}\:=\int_{\mathrm{0}} ^{\infty} \frac{{xe}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} \left(\sum_{{n}==\mathrm{0}} ^{\infty} {e}^{−{nx}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} {x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} {dx} \\ $$$$=_{\left({n}+\mathrm{1}\right){x}={t}} \:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \frac{{t}}{{n}+\mathrm{1}}{e}^{−{t}} \frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}} \:{dt} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }.\Gamma\left(\mathrm{2}\right) \\ $$$$=\xi\left(\mathrm{2}\right).\Gamma\left(\mathrm{2}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}×\mathrm{1}!\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by Aziztisffola last updated on 23/Jul/20

Thanks Sir

$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

Commented by mathmax by abdo last updated on 24/Jul/20

you are welcome.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

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