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Saturday, August 21, 2021

Laplace transform I

Laplace transform related to Harmonic numbers

What is the Laplace transform of exln(1ex)


We repost here the solution of this problem posted by @SrinivasR1729 on Twitter.
0eaxexln(1ex)dx=Ha+1a+1

Proof 0eaxexln(1ex)dx=10(1w)aln(w)dw(w1ex)=10(1w)a(ddt|t=0+wt)dw=ddt|t=0+10wt(1w)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: 0eaxexln(1ex)dx=ddt|t=0+B(t+1,a+1)=limt0+(ψ0(t+1)ψ0(a+t+2))B(t+1,a+1)=(ψ0(1)ψ0(a+2))B(1,a+1)=(γψ0(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 0eaxexln(1ex)dx=(γψ0(a+2))B(1,a+1)=Ha+1a+1 0eaxexln(1ex)dx=Ha+1a+1

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