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

Commented byDwaipayan 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)

Commented byAziztisffola last updated on 23/Jul/20

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

Commented byAziztisffola last updated on 23/Jul/20

Thanks Sir

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

you are welcome.