KAIST POW 2012-22 : Simple integral

Simple integral

 Compute $\int_0^1 \frac{x^k -1}{\ln x} dx$

Proof. Define a function $f : \mathbb{R}\to \mathbb{R}\cup \{ \pm \infty \}$ by $f(k)=\int_0^1 \frac{x^k -1}{ \ln x} dx$. Since $f(0)=0< \infty$ and the integral support $[0,1]$ is compact, $f(k)$ is finite and differentiable near 0. In this neighborhood, and $k>-1$, $f'(k)=\int_0^1 x^k dx =\frac{1}{k+1}$. This implies that for $k>-1$, $f(k)$ is finite and $$ \int_0^1 \frac{x^k -1}{ \ln x} dx=\ln (k+1)\textrm{ for $k>-1$}$$
 For $k< -1$, use $\lim_{k\to {-1^+}} f(k) =-\infty$. Clearly $f(k)$ is decreasing since $x^k$ is decreasing for $x \in [0,1]$. Therefore, $$ \int_0^1 \frac{x^k -1}{ \ln x} dx=-\infty \textrm{ for $k\leq -1$}$$