Integral Question x^x

Question

How do I show that $\int_{0}^{1} \frac{1}{x^{x}} = \sum_{k=1}^{\infty} \frac{1}{k^k}$. Also, how do I show that $\int_{0}^{1} \frac{1}{1-x} + \frac{1}{ln(x)} dx = \lim_{n \rightarrow \infty} (\sum_{k=1}^{n} \frac{1}{k}) - ln(k)$

Answer 1

Convert x^-x to the taylor series by $$x^{-x}=e^{-xlnx}$$ and then integrating the series from zero to 1. $$\int_0^1\sum_{n=0}^\infty \frac{(-xlnx)^n}{n!}$$ Hope this has helped and put you in the right direction!