Prove that $\int^3_0 \frac{1}{x^x}dx$ converges

Question

How do you prove that $\int^3_0 \frac{1}{x^x}dx$ converges? I've learned the integral of $x^x$, but I don't think the question requires me to use it here as we have not been taught the integral of $x^x$ in class.

Answer 1

$\int_{1}^{3}\frac{dx}{x^x}$ is obviously convergent, so the only issue is to show that $\int_{0}^{1}\frac{dx}{x^x}$ is convergent.
On the other hand over $(0,1)$ we have $0\leq -x\log x\leq \sqrt{x}(1-x)$, hence

$$ 0\leq\int_{0}^{1}\frac{dx}{x^x}=\int_{0}^{1}e^{-x\log x}\,dx \leq \int_{0}^{1} e^{\sqrt{x}(1-x)}\,dx=2\int_{0}^{1}z e^{z(1-z^2)}\,dz $$ and the last integral is clearly bounded (by $\exp\frac{2}{3\sqrt{3}}$, for instance, since the maximum of $z-z^3$ over $[0,1]$ occurs at $z=\frac{1}{\sqrt{3}}$). It might be interesting to point out that this problem is related to Sophomore's dream:

$$ \int_{0}^{1}e^{-x\log x}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{n!}\int_{0}^{1}x^n \log^n x\,dx=\sum_{n\geq 0}\frac{(-1)^n}{n!}\cdot\frac{(-1)^n n!}{(n+1)^{n+1}} $$ hence $\color{red}{\int_{0}^{1}\frac{dx}{x^x}=\sum_{n\geq 1}\frac{1}{n^n}}\leq 1+\sum_{n\geq 2}\frac{1}{2^n}=\frac{3}{2}$.

Answer 2

One can show that on $[0,3]$, $\frac{1}{x^x}<2$, so $\frac{1}{x^x}$ is bounded, hence integrable over a compact subset of $\mathbb{R}$ so your integral converges.

Answer 3

It is well known that $\lim_{x\to0+}x\log x=0$. It follows that the function $f(x):=-x\log x$ $(x>0)$ can be continuously extended to the interval $[0,3]$ by putting $f(0):=0$. One then has $$\lim_{\epsilon \to0+}\int_\epsilon^3 {1\over x^x}\>dx=\lim_{\epsilon\to0+}\int_\epsilon^3 e^{f(x)}\>dx=\int_0^3 e^{f(x)}\>dx\ .$$