Convergent of integral of $1/x^x$

Question

I need to prove if this integral converges (on interval $[1,+\infty))$:

$$\int _1^{\infty }\frac{1}{x^x}\;dx$$

Has anybody any idea how to do it?

thank you. :)

Answer 1

Note that $$\frac{1}{x^x}\leq\frac{1}{x^2}\mbox{ for }x\geq 2,\mbox{ and }\frac{1}{x^x}\leq 1\mbox{ for }1\leq x\leq 2.$$ This implies that for any sufficiently large $N$ $$\int _1^{N}\frac{1}{x^x}dx=\int _1^{2}\frac{1}{x^x}dx+\int _2^{N}\frac{1}{x^x}dx \leq\int _1^{2}dx+\int _2^{N}\frac{1}{x^2}dx =1+\frac{1}{2}-\frac{1}{N}.$$ Letting $N\to\infty$, we get $$\int _1^{N}\frac{1}{x^x}dx\leq\frac{3}{2}.$$