Originally Posted by

**Mentia** I suppose there are also problems at zero, but take the limit of the function at zero and you get zero so all is well.

Originally Posted by

**HenryB** Comparison to $\displaystyle 1/x^2$ will only show integrability over $\displaystyle [1,\infty)$. I think the real difficulty is showing integrability over $\displaystyle (0,1)$. Indeed I'm not even sure if it integrable over this interval.

Mentia is right, there should be no problem at 0: $\displaystyle \int_0^1 \int_x^{x+x^2} e^{-xy} dy dx$ amounts to integrating a continuous function on a compact subset of $\displaystyle \mathbb{R}^2$, so it is finite, without computation. It only remains to show that $\displaystyle \int_1^\infty \int_x^{x+x^2} e^{-xy} dy dx$ is finite, and a comparison with $\displaystyle \frac{1}{x^2}$ works fine. In order to simplify the computation, you can even say: for $\displaystyle x\geq 1$,

$\displaystyle 0\leq \int_x^{x+x^2} e^{-xy} dy\leq \int_x^\infty e^{-xy} dy = \frac{e^{-x^2}}{x}\leq \frac{C}{x^2}$,

where $\displaystyle C$ is some constant. So that the integral on $\displaystyle x\in[1,+\infty)$ converges.

Just for the sake of not letting a question answered: the function $\displaystyle \frac{e^{-x^2}}{x}$ is clearly not integrable on $\displaystyle (0,1]$: the numerator converges to 1 as $\displaystyle x\to0^+$, hence the function is greater than $\displaystyle \frac{1/2}{x}$ for small enough $\displaystyle x$ (or we can say we have the asymptotic equivalence $\displaystyle \frac{e^{-x^2}}{x}\sim_{x\to 0^+} \frac{1}{x}$), and $\displaystyle \frac{1}{x}$ is not integrable on $\displaystyle (0,1]$.

So how come I proved the function is integrable ? Just because there is another term than $\displaystyle \frac{e^{-x^2}}{x}$, and their *difference* tends to 0 when $\displaystyle x\to 0$: no divergence.