Prove the convergence of $\int_1^{\infty} x^{-x}\,dx$

Question

I am trying to prove that the following integral converges: $$\int_1^{\infty} x^{-x}\,dx$$ I have a proof, but I was wondering if there were any improvements that could be made, or comments. Here it is.

First, break the integral up into $$\int_1^{e} x^{-x}\,dx+\int_{e} ^{\infty} x^{-x}\,dx$$ The first integral poses no problem. Then, note that $$x^{-x}=e^{-x\log x}$$ Thus, for $x\in[e,+\infty)$, $$x\ge e \implies x\log x\ge x \implies \log x^x\ge x \implies -x \ge -\log x^x \\ \implies e^{-x\log x}\le e^{-x}$$

This means that, for $x\in[e, \infty]$, $e^{-x\log x} \le e^{-x}$, or $$\int_e^{\infty} e^{-x \log x}\,dx \le \int_e^{\infty} e^{-x}\,dx=e^{-e}$$

Thus, by the comparison test, since $$\int_1^{a} x^{-x}\,dx<\infty$$ and $$\int_a^{\infty} x^{-x} \le e^{-e}<\infty$$ Their sum $$\int_1^e x^{-x}\,dx+\int_e^{\infty} x^{-x}\,dx=\int_1^{\infty} x^{-x}\,dx <\infty$$ is convergent, and our integral is thus convergent.

Answer 1

Maybe limit comparison test?

\begin{align*} \lim_{x\rightarrow\infty}\dfrac{e^{-x\log x}}{e^{-x}}=\lim_{x\rightarrow\infty}e^{x(1-\log x)}=0 \end{align*} since $1-\log x\rightarrow-\infty$ as $x\rightarrow\infty$.

So $e^{-x\log x}\leq e^{-x}$ for large $x>0$, and there is no singularity for $x^{-x}$ on $[1,M]$, so the whole integral converges.

Answer 2

You can completely skip breaking up the integral by using the integral test, which says that

$\int_1^{\infty} x^{-x}\;dx$ converges or diverges with $\sum_{n=1}^{\infty} n^{-n}$

Given that $n!\leq n^n$, $$S_N = \sum_{n=1}^{N}\frac{1}{n^n} \leq \sum_{n=1}^N\frac{1}{n!} \leq \sum_{n=1}^{\infty}\frac{1}{n!} = e-1$$

$S_N$ is monotonically increasing and bounded above, hence it converges. So $\int_1^{\infty} x^{-x}\;dx$ converges.

Answer 3

Splitting the integral is probably easiest, but you can simplify your argument by noticing that for $x \geq 2$, we have $x^x \geq x^2$ and hence $\dfrac{1}{x^x} \leq \dfrac{1}{x^2}$. Hence,

$$ \int_1^{\infty} \frac{1}{x^x}dx = \int_1^2 \frac{1}{x^x} + \int_2^{\infty} \frac{1}{x^x}dx \leq \int_1^2 \frac{1}{x^x} + \int_2^{\infty} \frac{1}{x^2}dx $$

You can check $\displaystyle \int_2^{\infty}\frac{1}{x^2} dx = \frac{1}{2}$. Thus your integral converges.