How do I show $\lim_{\epsilon\to0}\int_\epsilon^1x^n(\ln(x))^mdx=\frac{-m}{n+1}\lim_{\epsilon\to0}\int_\epsilon^1x^n(\ln(x))^{m-1}dx$

Question

I have an assignemt where I'm to prove $$\int_0^1x^{-x}dx=\sum_{k=1}^\infty k^{-k}$$

And one of the steps towards doing this is showing that

$$\lim_{\epsilon\to0}\int_\epsilon^1x^n(\ln(x))^mdx=\frac{-m}{n+1}\lim_{\epsilon\to0}\int_\epsilon^1x^n(\ln(x))^{m-1}dx$$

$n,m\in\mathbb{N}\setminus\{0\}$

And I really am not sure where to start with this. In the previous steps I have shown that

$$\int_0^1e^{-x\cdot\ln(x)}dx=\int_0^1x^{-x}dx=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1(x\cdot\ln(x))^ndx$$

And that

$$\int_0^1(x\cdot\ln(x))^ndx=\lim_{\epsilon\to0}\int_\epsilon^1x^n(ln(x))^ndx$$

But I'm not sure how any of that helps me here..

Thanks in advance

Answer 1

You're overcomplicating it. Do integration by parts with $dv=x^n dx$ and $u={(\ln{x})}^m$: $$I= \lim_{\varepsilon \to 0} \frac{1}{n+1} x^{n+1} {(\ln{x})}^m \big \rvert_{\varepsilon}^1-\frac{m}{n+1} \int_{\varepsilon}^1 x^{n+1} \frac{{(\ln{x})}^{m-1}}{x} \; dx$$ $$I=\lim_{\varepsilon \to 0}-\frac{m}{n+1} \int_{\varepsilon}^1 x^{n} {(\ln{x})}^{m-1} \; dx$$