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

tearstreakdl

Answered question

2022-01-03

How to do the following integral:
${\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}\left(x\right)}dx$
where $a\ge 0$ ?

Answer & Explanation

Jenny Sheppard

Beginner2022-01-04Added 35 answers

Call your integral $I\left(a\right)$. Then $I}^{\prime}\left(a\right)={\int}_{0}^{1}{x}^{a}dx=\frac{1}{a+1$ $I}^{\prime}\left(a\right)={\int}_{0}^{1}{x}^{a}dx=\frac{1}{a+1$ as long as $a\ge 0$. Now you need to solve the differential equation $I}^{\prime}\left(a\right)=\frac{1}{a+1$ This is a very easy differential equation to solve, and the solution is $I\left(a\right)=\mathrm{log}(a+1)+C$ where C is some constant. Now we ask, what is that constant? Notice that $I\left(0\right)={\int}_{0}^{1}\frac{1-1}{\mathrm{log}x}dx=0$ so we need $I\left(0\right)=\mathrm{log}\left(1\right)+C=0$, or rather $C=0$. So we conclude that ${\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}x}dx=\mathrm{log}(a+1)$, as you suggested.

Edward Patten

Beginner2022-01-05Added 38 answers

We can utilize
$\int}_{0}^{1}{x}^{t}dt=\frac{x-1}{\mathrm{log}\left(x\right)$
combined with the substitution $\mapsto {x}^{\frac{1}{a}}$, to get
${\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}\left(x\right)}dx={\int}_{0}^{1}\frac{x-1}{\mathrm{log}\left(x\right)}{x}^{\frac{1}{a}-1}dx$ $={\int}_{0}^{1}{\int}_{0}^{1}{x}^{\frac{1}{a}-1}{x}^{t}dtdx$ $={\int}_{0}^{1}\frac{1}{\frac{1}{a}+t}dt$ $=\mathrm{log}(\frac{1}{a}+1)-\mathrm{log}\left(\frac{1}{a}\right)$ $=\mathrm{log}(1+a)$

Vasquez

Skilled2022-01-09Added 457 answers

We have ${x}^{a}-1={e}^{a\mathrm{log}(x)}-1$. Hence, the integral is $I={\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}(x)}dx={\int}_{0}^{1}(\sum _{k=1}^{\mathrm{\infty}}\frac{{a}^{k}{\mathrm{log}}^{k}(x)}{k!})\frac{dx}{\mathrm{log}(x)}$ $=\sum _{k=1}^{\mathrm{\infty}}\frac{{a}^{k}}{k!}{\int}_{0}^{1}{\mathrm{log}}^{k-1}(x)dx$ $=\sum _{k=1}^{\mathrm{\infty}}\frac{{a}^{k}}{k!}(-1{)}^{k-1}(k-1)!=\sum _{k=1}^{\mathrm{\infty}}\frac{(-1{)}^{k-1}{a}^{k}}{k}=\mathrm{log}(1+a)$