Convergence or divergence integral question: $\int\limits^{\infty}_{-\infty} x\mathrm{e}^{-x^2}\,\mathrm{d}x$ $\displaystyle\int\limits^{\infty}_{-\infty} x\mathrm{e}^{-x^2}\,\mathrm{d}x$  Determine whether the integral is divergent or convergent.
Hi, I don't have any idea how to start. Should I start with evaluating the integral?
 A: Note that $\int xe^{-x^2}\,dx=-\frac12 e^{-x^2}+C$.  Hence, we have
$$\begin{align}
\int_{-\infty}^\infty xe^{-x^2}\,dx&=\lim_{L\to \infty}\int_0^Le^{-x^2}\,dx+\lim_{M\to -\infty}\int_M^0 e^{-x^2}\,dx\\\\
&=\lim_{L\to \infty}\left.\left(-\frac12 e^{-x^2}\right)\right|_{x=0}^{x=L}+\lim_{M\to -\infty}\left.\left(-\frac12 e^{-x^2}\right)\right|_{x=M}^{x=0}\\\\
&=\frac12-\frac12\\\\
&=0
\end{align}$$
A: $e^{-x^2}$ belongs to the Schwartz space $\mathcal{S}(\mathbb{R})$ and so does $xe^{-x^2}$. In particular, $f(x)=x e^{-x^2}$ is an integrable function over $\mathbb{R}$. Since it is an odd function, its integral is zero.
A: By definition, the integral converges if both improper integrals in the right-hand side converge:
$$\int_{-\infty}^{+\infty} xe^{-x^2}\,\mbox{d}x = \int_{-\infty}^{0} xe^{-x^2}\,\mbox{d}x+\int_{0}^{+\infty} xe^{-x^2}\,\mbox{d}x$$
For either of those (the work is similar for the other one), direct computation is straightforward:
$$\int_{0}^{+\infty} xe^{-x^2}\,\mbox{d}x = \lim_{p \to \infty}\int_{0}^{p} xe^{-x^2}\,\mbox{d}x$$
Take $u=-x^2$.
