tearstreakdl

2022-01-03

How to do the following integral:

${\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}\left(x\right)}dx$

where$a\ge 0$ ?

where

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.

as long as

This is a very easy differential equation to solve, and the solution is

where C is some constant. Now we ask, what is that constant? Notice that

so we need

or rather

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

${\int}_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}\left(x\right)}dx={\int}_{0}^{1}\frac{x-1}{\mathrm{log}\left(x\right)}{x}^{\frac{1}{a}-1}dx$

$={\int}_{0}^{1}{\int}_{0}^{1}{x}^{\frac{1}{a}-1}{x}^{t}dtdx$

$={\int}_{0}^{1}\frac{1}{\frac{1}{a}+t}dt$

$=\mathrm{log}(\frac{1}{a}+1)-\mathrm{log}\left(\frac{1}{a}\right)$

$=\mathrm{log}(1+a)$

combined with the substitution

Vasquez

Expert2022-01-09Added 669 answers

We have

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