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 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.
