Evaluating Integral using Leibnitz's rule

Relevant wiki: Differentiation Under the Integral Sign

In order to evaluate the given integral, we resort to the following device.

Define $\phi(\alpha)=\int_0^1\frac{x^{\alpha}-1}{\ln x}dx; \alpha>0$

then by Leibnitz's rule:

$\phi'(\alpha)=\large \int_0^1 \frac{\partial}{\partial \alpha}\left(\frac{x^{\alpha}-1}{\ln x}\right)= \int_0^1\frac{x^{\alpha}\ln x}{\ln x}dx= \int_0^1x^{\alpha}dx= \frac{1}{\alpha +1}$ .

Integrating with respect to $\alpha$ ,

$\phi(\alpha)= \ln(\alpha + 1) + c$ But since $\phi(0)=0 ,\ and \ c=0$ and so $\phi(\alpha)=\ln(\alpha)+1$ .

Then, the value of our integral is $\boxed{\ln 2}$ .

The applicability of Leibnitz's rule is ok here since we are defining: $F(x,\alpha)=\bigg(\frac{x^\alpha -1}{\ln x}\bigg), \text{where} \ 0<x<1$ ,

$F(0,\alpha)=0, F(1,\alpha)=\alpha$ , then $F(x,\alpha)$ is continuous in both $x$ and $\alpha$ for $0\le x\le 1$ and for all finite $\alpha$ .