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Question Number 127017 by mnjuly1970 last updated on 26/Dec/20

               ...NICE     CALCULUS...    prove that ::  ∫_0 ^( ∞)  (((x^2 ln(πx))/π^(πx) ))dx    =(1/((πln(π))^3 ))[(3−2(γ+ln(ln(π)))]

...NICECALCULUS...provethat::0(x2ln(πx)ππx)dx=1(πln(π))3[(32(γ+ln(ln(π)))]

Answered by mindispower last updated on 26/Dec/20

=∫_0 ^∞ x^2 ln(πx)e^(−πxln(π)) dx  u=πln(π)x  ⇒dx=(du/(πln(π)))  ⇔∫_0 ^∞ (u^2 /(π^2 ln^2 (π)))ln((u/(ln(π))))e^(−u) .(du/(πln(π)))  =(1/(π^3 ln^3 (π)))[∫_0 ^∞ u^2 ln(u)e^(−u) du−ln(lnπ)∫_0 ^∞ u^2 e^(−u) ]  =(1/(π^3 ln^3 (π))){Γ′(3)−ln(ln(π).Γ(3)}  =((Γ(3).Ψ(3))/(π^3 ln^3 (π)))−(1/(π^3 ln^3 (π)))2ln(lnπ)  =2((1/2)+1−γ).(1/(π^3 ln^3 (π)))−((2ln(lnπ))/(π^3 ln^3 (π)))  =((3−2γ)/(π^3 ln^3 (π)))−((2ln(lnπ))/(π^3 ln^3 (π)))=(1/((πlnπ)^3 ))[(3−2(γ+ln(lnπ))]

=0x2ln(πx)eπxln(π)dxu=πln(π)xdx=duπln(π)0u2π2ln2(π)ln(uln(π))eu.duπln(π)=1π3ln3(π)[0u2ln(u)euduln(lnπ)0u2eu]=1π3ln3(π){Γ(3)ln(ln(π).Γ(3)}=Γ(3).Ψ(3)π3ln3(π)1π3ln3(π)2ln(lnπ)=2(12+1γ).1π3ln3(π)2ln(lnπ)π3ln3(π)=32γπ3ln3(π)2ln(lnπ)π3ln3(π)=1(πlnπ)3[(32(γ+ln(lnπ))]

Commented by mnjuly1970 last updated on 26/Dec/20

peace be upon you    sir power ..mercey..

peacebeuponyousirpower..mercey..

Answered by Dwaipayan Shikari last updated on 26/Dec/20

∫_0 ^∞ ((x^2 log(πx))/π^(πx) )dx=(1/π^3 )∫_0 ^∞ ((u^2 log(u))/π^u )du       πx=u  =(1/π^3 )∫_0 ^∞ e^(−ulog(π)) u^2 log(u)du                 ulog(π)=t  =(1/((πlog(π))^3 ))∫_0 ^∞ e^(−t) t^2 log(t)−∫_0 ^∞ e^(−t) t^2 log(log(π))  =(1/((πlog(π))^3 ))(Γ′(3)−Γ(3)log(log(π)))  Γ′(3)=Γ(3)ψ(3)=2(−γ+Σ^∞ (1/n)−(1/(n+2)))=−2γ+3  =(1/((πlog(π))^3 ))(−2γ+3−2log(log(π))

0x2log(πx)ππxdx=1π30u2log(u)πuduπx=u=1π30eulog(π)u2log(u)duulog(π)=t=1(πlog(π))30ett2log(t)0ett2log(log(π))=1(πlog(π))3(Γ(3)Γ(3)log(log(π)))Γ(3)=Γ(3)ψ(3)=2(γ+1n1n+2)=2γ+3=1(πlog(π))3(2γ+32log(log(π))

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