Finding the integral $\int_0^1 \frac{x^a - 1}{\log x} dx$ How to do the following integral:
$$\int_{0}^1 \dfrac{x^a-1}{\log(x)}dx$$
where $a \geq 0$?
I was asked this question by a friend, and couldn't think of any substitution that works. Plugging in a=2,3, etc in Wolfram, I get values like $\log(a+1)$, which may be the right answer (for general $a$). Is there a simple way to calculate this integral?
 A: Call your integral $I(a)$. Then
$$
I'(a) = \int_0^1 x^a dx = \frac{1}{a+1}
$$
as long as $a \geq 0$. Now you need to solve the differential equation
$$ I'(a) = \frac{1}{a + 1}.$$
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
$$ I(0) = \int_0^1 \frac{1 - 1}{\log x} dx = 0,$$
so we need
$$ I(0) = \log(1) + C = 0,$$
or rather $C = 0$. So we conclude that
$$
\int_0^1 \frac{x^a - 1}{\log x} dx = \log(a + 1),
$$
as you suggested. $\diamondsuit$
A: We can utilize
$$
\int_0^1x^t\,\mathrm{d}t=\frac{x-1}{\log(x)}
$$
combined with the substitution $x\mapsto x^{1/a}$, to get
$$
\begin{align}
\int_0^1\frac{x^a-1}{\log(x)}\,\mathrm{d}x
&=\int_0^1\frac{x-1}{\log(x)}x^{\frac1a-1}\,\mathrm{d}x\\
&=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\frac1{\frac1a+t}\,\mathrm{d}t\\
&=\log\left(\frac1a+1\right)-\log\left(\frac1a\right)\\[9pt]
&=\log(1+a)
\end{align}
$$
A: A couple of possibilities:


*

*Change variables to $x=e^{-y}$, $dx = -e^{-y} \, dy$, so the integral becomes
$$ \int_0^{\infty} \frac{e^{-y}-e^{-(1+a)y}}{y} \, dy. $$
This is a Frullani integral, so the value is
$$ (1-0)\log{(1+a)/1} = \log{(1+a)}. $$

*Differentiate under the integral sign. I won't go through this one since someone else got there first.

*Do the substitution from 1., but integrate by parts:
$$ I(a) = \left[ (e^{-y}-e^{-(1+a)y})\log{y} \right] + \int_0^{\infty} e^{-y}\log{y} \, dy - (1+a)\int_0^{\infty} e^{-(1+a)y} \log{y} \, dy, $$
The first term is zero since the bracket is $O(y)$ at $0$. Change variables to $u=(1+a)y$ in the last integral. It turns into
$$ -\int_0^{\infty} e^{-u} (\log{u}-\log{(1+a)} \, dy = -\int_0^{\infty} e^{-y}\log{y} \, dy + \log{(1+a)} $$, and then the remaining integrals cancel and you get the result.
A: We have $x^a-1 = e^{a\log(x)}-1$. Hence, the integral is
\begin{align}
I & = \int_0^1 \dfrac{x^a-1}{\log(x)}dx = \int_0^1 \left(\sum_{k=1}^{\infty} \dfrac{a^k \log^k(x)}{k!}\right)\dfrac{dx}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} \int_0^1 \log^{k-1}(x)dx\\
& = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}a^k}k = \log(1+a)
\end{align}
