Test for convergence for improper integral $1/x^x$

Question

I am having trouble determining if this is convergence or divergence

$$\int^1_0 1/x^x dx$$

Answer 1

It suffices to show that

$$\lim_{x \to 0^+} x^{-x} =1,$$ because then we could consider the integral

$$\int_0^1 x^{-x} dx$$

as an integral of a continuous function on a compact set (which has convergent value).

Taking logs, this is the equivalent to

$$\lim_{x \to 0^+} -x \log x =0.$$

This is a standard result, which can be shown using l'Hopital's rule.

Answer 2

Hint: Use the limit comparison test.

Answer 3

There is another way which is based on Caparison Test again. In fact, the following limit says that the improper integral is convergent: $$\lim_{x\to 0^+}~~x^{1/2}\times\frac{1}{x^x}=0<\infty$$