Positive to negative infinity integration $$\int_{-\infty}^{\infty} xe^{-x^2} dx$$
I guess let $u = -x^2$ , hence $\frac{du}{dx}=-2x$ 
$$dx = \frac{-1}{2x}du$$
$$\int_{-\infty}^{\infty} \frac{e^{u}}{2} du$$
this is where i get lost
 A: Also you could use;
\begin{equation}
\frac{d (e^{-x^{2}})}{dx} = -2xe^{-x^{2}}
\end{equation}
A: When you have a “doubly improper” integral, you must split it into two integrals (or do a double limit). Thus
$$
\int_{-\infty}^\infty xe^{-x^2}\,dx=
\int_{-\infty}^a xe^{-x^2}\,dx+
\int_a^\infty xe^{-x^2}\,dx
$$
where you can choose $a$ arbitrarily. In this case, a good choice is $a=0$, because in the first integral you can make the substitution $x=-t$ that produces
$$
\int_{\infty}^0 te^{-t^2}\,dt+
\int_0^\infty xe^{-x^2}\,dx
$$
Now, the second one is a converging integral and the first one is its opposite, so the result is zero.
Beware that simply reasoning by “the function is odd” can lead to wrong results: the function $f(x)=xe^{x^2}$ is odd, but
$$
\int_{-\infty}^{\infty}xe^{x^2}\,dx
$$
doesn't exist in the sense of improper integrals; there is a “principal value integral”, but it's a different matter.
A: Your function is odd. So when you integrate all over the real axis ($(-\infty,\infty)$) all the areas cancel each other out. Even if you don't notice that, after the substitution your limits of integration change to $\displaystyle\int_{-\infty}^{-\infty}$ and $\int_a^a f(x) \mathrm{d}x$ is always $0$.
