How to do the following integral: ∫01xa−1log⁡(x)dx where a≥0 ?

tearstreakdl

tearstreakdl

Answered

2022-01-03

How to do the following integral:
01xa1log(x)dx
where a0 ?

Answer & Explanation

Jenny Sheppard

Jenny Sheppard

Expert

2022-01-04Added 35 answers

Call your integral I(a). Then
I(a)=01xadx=1a+1
I(a)=01xadx=1a+1
as long as a0. Now you need to solve the differential equation
I(a)=1a+1
This is a very easy differential equation to solve, and the solution is
I(a)=log(a+1)+C
where C is some constant. Now we ask, what is that constant? Notice that
I(0)=0111logxdx=0
so we need
I(0)=log(1)+C=0,
or rather C=0. So we conclude that
01xa1logxdx=log(a+1),
as you suggested.
Edward Patten

Edward Patten

Expert

2022-01-05Added 38 answers

We can utilize
01xtdt=x1log(x)
combined with the substitution x1a, to get
01xa1log(x)dx=01x1log(x)x1a1dx
=0101x1a1xtdtdx
=0111a+tdt
=log(1a+1)log(1a)
=log(1+a)
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

We have xa1=ealog(x)1. Hence, the integral is
I=01xa1log(x)dx=01(k=1aklogk(x)k!)dxlog(x)
=k=1akk!01logk1(x)dx
=k=1akk!(1)k1(k1)!=k=1(1)k1akk=log(1+a)

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