What is the Laplace transform of e−xln(1−e−x)
We repost here the solution of this problem posted by @SrinivasR1729 on Twitter.
∫∞0e−axe−xln(1−e−x)dx=−Ha+1a+1
Proof ∫∞0e−axe−xln(1−e−x)dx=∫10(1−w)aln(w)dw(w↦1−e−x)=∫10(1−w)a(ddt|t=0+wt)dw=ddt|t=0+∫10wt(1−w)adw=ddt|t=0+B(t+1,a+1) Recall that the derivative of the complete beta function: ddaB(a,b)=B(a,b)(ψ(0)(a)−ψ(0)(a+b)) Therefore: ∫∞0e−axe−xln(1−e−x)dx=ddt|t=0+B(t+1,a+1)=limt→0+(ψ0(t+1)−ψ0(a+t+2))B(t+1,a+1)=(ψ0(1)−ψ0(a+2))B(1,a+1)=(−γ−ψ0(a+2))B(1,a+1) where we used the fact that ψ(0)(1)=−γ, the Euler constant.
Recall that a hamonic number for real and complex values can be expressed with the following relation: Hx=ψ(0)(x+1)+γ Also note that B(1,a+1)=Γ(1)Γ(a+1)Γ(a+2)=Γ(a+1)Γ(a+1)(a+1)=1a+1 Hence ∫∞0e−axe−xln(1−e−x)dx=(γ−ψ0(a+2))B(1,a+1)=−Ha+1a+1 ∫∞0e−axe−xln(1−e−x)dx=−Ha+1a+1
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