TITLE: Can we use symmetry rules in improper integrals?
QUESTION [30 upvotes]: I wish to evaluate the integral $$I=\int^{\infty}_{-\infty}xe^{-x^2}dx$$
Can I simply note that that $f(x)=xe^{-x^2}$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $\mathbb{R}$," it tempts me to accept the symmetry argument. 
The actual answer via limits is $0$.

REPLY [2 votes]: Since
$$
\int_{-\infty}^\infty\left|\,xe^{-x^2}\right|\,\mathrm{d}x=1
$$
we have that $xe^{-x^2}\in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $\int_{-\infty}^\infty\frac{x\,\mathrm{d}x}{x^2+1}$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.