TITLE: Is there a way to prove convergence of an improper integral without calculating the integral itself?
QUESTION [0 upvotes]: I'm working on a problem right now, that requires to prove convergence of an improper integral but without calculating the integral itself. I was thinking of applying some sort of comparison test but I'm not very sure how to start with it. Am I even allowed to do that like it's done with series, before calculating the integral itself?
The problem is the following:
$$- \int_{-\infty}^{\infty}u(x)v'(x)e^{-x^2}dx - 2 \int_{-\infty}^{\infty}xe^{-x^2}u(x)v(x)dx$$
where $u$ and $v$ are polynomials and $v'$ is the derivative with respect to $x$. I mean it is kind of obvious that it converges because of the $e^{-x^2}$ term but I can't think of any proper method to prove it in a nice way.

REPLY [2 votes]: If $p(x)$ is any polynomial then
$$p(x)e^{-x^2}=\bigl(p(x)e^{-x^2/2}\bigr)\cdot e^{-x^2/2}=:g(x)\cdot e^{-x^2/2}\ .$$
Here $\lim_{|x|\to\infty}g(x)=0$, and $\int_{-\infty}^\infty e^{-x^2/2}\>dx$ is, as we all know, convergent.
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