Log integral to the power $k$

Question

For what $\text{k}$ is this integral convergent?

$$\int_0^e\ln^k(x)\space\text{d}x$$

Mathematica gives $\Re(k)>-1$, but why?

And I came up with:

$$\int_0^e\ln^k(x)\space\text{d}x=e\left\{(-1)^kk!+\sum_{p=0}^{k-1}\frac{k!(-1)^p}{(k-p)!}\right\}$$

For which $k$ is the equality valid?

Answer 1

If $k\leq 0,\;$ it converges since

$\lim_{x\to 0^+} (\ln(x))^k \in \mathbb R$

if $k>0$

$$\lim_{x\to 0^+}\sqrt{x}(\ln(x))^k=0$$

$\implies$

$|\sqrt{x}(\ln(x))^k|\leq 1$ for $x$ near $0$

$\implies |(\ln(x))^k|\leq \frac{1}{\sqrt{x}}$

$\implies \int_0 (\ln(x))^kdx $ converges by comparison test.