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

Question

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

Jenny Sheppard · Accepted Answer

Call your integral I(a). ThenI′(a)=∫01xadx=1a+1I′(a)=∫01xadx=1a+1as long as a≥0. Now you need to solve the differential equationI′(a)=1a+1This is a very easy differential equation to solve, and the solution isI(a)=log⁡(a+1)+Cwhere C is some constant. Now we ask, what is that constant? Notice thatI(0)=∫011−1log⁡xdx=0so we needI(0)=log⁡(1)+C=0,or rather C=0. So we conclude that∫01xa−1log⁡xdx=log⁡(a+1),as you suggested.

Edward Patten · Accepted Answer

We can utilize  ∫01xtdt=x−1log⁡(x)  combined with the substitution ↦x1a, to get  ∫01xa−1log⁡(x)dx=∫01x−1log⁡(x)x1a−1dx  =∫01∫01x1a−1xtdtdx  =∫0111a+tdt  =log⁡(1a+1)−log⁡(1a)  =log⁡(1+a)

Vasquez · Accepted Answer

We have xa−1=ealog⁡(x)−1. Hence, the integral is I=∫01xa−1log⁡(x)dx=∫01(∑k=1∞aklogk⁡(x)k!)dxlog⁡(x) =∑k=1∞akk!∫01logk−1⁡(x)dx =∑k=1∞akk!(−1)k−1(k−1)!=∑k=1∞(−1)k−1akk=log⁡(1+a)

How to do the following integral: \int_0^1\frac{x^a-1}{\log(x)}dx where a\geq0 ?

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New Questions in Calculus 1