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Question Number 131546 by mathmax by abdo last updated on 05/Feb/21
prove that ∫_0 ^∞ e^(−x) ln(x)dx =−γ
provethat∫0∞e−xln(x)dx=−γ
Answered by Dwaipayan Shikari last updated on 06/Feb/21
I(a)=∫_0 ^∞ e^(−x) x^a dx=Γ(a+1) I′(a)=∫_0 ^∞ e^(−x) x^a log(x)dx=Γ′(a+1) I′(0)=∫_0 ^∞ e^(−x) log(x)dx=Γ′(1)=Γ(1)ψ(1)=−γ
I(a)=∫0∞e−xxadx=Γ(a+1)I′(a)=∫0∞e−xxalog(x)dx=Γ′(a+1)I′(0)=∫0∞e−xlog(x)dx=Γ′(1)=Γ(1)ψ(1)=−γ

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