 tearstreakdl

2022-01-03

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

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
$I\left(a\right)=\mathrm{log}\left(a+1\right)+C$
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
${\int }_{0}^{1}\frac{{x}^{a}-1}{\mathrm{log}x}dx=\mathrm{log}\left(a+1\right)$,
as you suggested. Edward Patten

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

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

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