Comparison test integral convergence

Question

$$\int_0^{\infty} \frac{e^x}{x^x} \,\mathrm dx$$

How can I tell if this integral converges or not? I was thinking of using the comparison test, but I can't think of anything to compare it to. Could you give me a hint?

Answer 1

Since $x^x=e^{x\ln{x}},$ then for $x>e$ $$\frac{e^x}{x^x}=e^{x(1-\ln x)}=\dfrac{1}{e^{x(\ln x-1)}}<\dfrac{1}{e^{x}},$$

therefore, $\displaystyle\int\limits_{e}^{\infty} \dfrac{e^x}{x^x} \,\mathrm dx$ converges. On $(0,\,e)$ integrand is bounded, because $x^x\underset{x\to{0^{+}}}{\to}{1},$ so $\displaystyle\int\limits_{0}^{\infty} \dfrac{e^x}{x^x} \,\mathrm dx$ is convergent.

Answer 2

We have $$\frac{e^x}{x^x}=e^{x-x\log x}=_\infty o(\frac{1}{x^2})$$ and $$\frac{e^x}{x^x}=e^{x-x\log x}\to_01$$ so the the given integral is convergent.

Answer 3

Since you wanted to use the comparison test:

If $x \ge e^2$, then $\ln x \ge 2$, and so $1-\ln x \le -1$ which gives $e^{x(1-\ln x)} \le e^{-x}$.

Since $t \mapsto e^t$ is a bijection of $(-\infty, 2]$ onto $(0,e^2]$, we see that $M=\sup_{x\in (0,e^2]} e^{x(1-\ln x)} = \sup_{t \in (-\infty, 2]}e^{e^t (1-t)} < \infty$, and so we have $\frac{e^x}{x^x} \le M e^{e^2} e^{-x}$ for all $x >0$.