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Question Number 127017 by mnjuly1970 last updated on 26/Dec/20
...NICECALCULUS...provethat::∫0∞(x2ln(πx)ππx)dx=1(πln(π))3[(3−2(γ+ln(ln(π)))]
Answered by mindispower last updated on 26/Dec/20
=∫0∞x2ln(πx)e−πxln(π)dxu=πln(π)x⇒dx=duπln(π)⇔∫0∞u2π2ln2(π)ln(uln(π))e−u.duπln(π)=1π3ln3(π)[∫0∞u2ln(u)e−udu−ln(lnπ)∫0∞u2e−u]=1π3ln3(π){Γ′(3)−ln(ln(π).Γ(3)}=Γ(3).Ψ(3)π3ln3(π)−1π3ln3(π)2ln(lnπ)=2(12+1−γ).1π3ln3(π)−2ln(lnπ)π3ln3(π)=3−2γπ3ln3(π)−2ln(lnπ)π3ln3(π)=1(πlnπ)3[(3−2(γ+ln(lnπ))]
Commented by mnjuly1970 last updated on 26/Dec/20
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Answered by Dwaipayan Shikari last updated on 26/Dec/20
∫0∞x2log(πx)ππxdx=1π3∫0∞u2log(u)πuduπx=u=1π3∫0∞e−ulog(π)u2log(u)duulog(π)=t=1(πlog(π))3∫0∞e−tt2log(t)−∫0∞e−tt2log(log(π))=1(πlog(π))3(Γ′(3)−Γ(3)log(log(π)))Γ′(3)=Γ(3)ψ(3)=2(−γ+∑∞1n−1n+2)=−2γ+3=1(πlog(π))3(−2γ+3−2log(log(π))
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