Find $\int_{-\infty}^{\infty}xe^{-x^2/2}dx$ $$\int_{-\infty}^{\infty}xe^{-x^2/2}dx$$
I have made an attempt at this by substituting u=x^2 to get:
$$\frac12\int_{-\infty}^{\infty}e^{-u/2}du$$
This gives me:
$$\frac12[-2e^{u/2}]_{-\infty}^{{\infty}}$$
Firstly is this right and secondly, how do I plug in the limits? I realize that the answer should be zero due to symmetry but how do I go about doing it properly?  
 A: Hint The integrand is odd. $    $
As for what's wrong: When substituting, the change of limits is backward.
Here's a way to treat this rigorously without using the symmetry of the integrand. First, since the integral is improper on both ends, we decompose
$$\phantom{(\ast)} \qquad \int_{-\infty}^{\infty} x e^{-x^2 / 2} dx = \int_{-\infty}^0 x e^{-x^2 / 2} dx + \int_0^{\infty} x e^{-x^2 / 2} dx. \qquad (\ast)$$
Using the substitution $u = \frac{1}{2} x^2$, $du = x \,dx$, we find that the second integral on the r.h.s. is
$$\lim_{k \to \infty} \int_0^k x e^{-x^2 / 2} dx = \lim_{k \to \infty} \frac{1}{2} \int_0^{k^2 / 2} e^{-u} du = \lim_{k \to \infty} \left.-\frac{1}{2}e^u\right\vert_0^{k^2 / 2} = \lim_{k \to \infty} \frac{1}{2} \left(1 - \frac{1}{2} e^{-k^2 / 2}\right) = \frac{1}{2}.$$
A similar argument (instead substituting $v = -\frac{1}{2} x^2$, $dv = -x \,dx$) shows that the first integral on the r.h.s. has value $-\frac{1}{2}$. (We can justify an appeal to symmetry at this step by the way: Just like we always can when the integrand is odd and the integral of integration is symmetric about $0$, we can evaluate the first integral by substituting $y = -x$, $dy = -dx$ and the using the above result for the second integral.) Substituting in $(\ast)$ gives the 
$$\color{#bf0000}{\boxed{\int_{-\infty}^{\infty} x e^{-x^2 / 2} dx = 0}}.$$
A: To do it properly you need to make your substitution invertible, so split the range at zero. You should not be taking the square root of $\infty$ as the limit, which you should be able to see if you make the upper limit $y$ and take the limit as $y \to \infty$.  For the upper half if you take $u=\frac 12x^2, du=xdx$ you have $$\int_0^{\infty}xe^{-x^2/2}dx=\int_0^{\infty}e^{-u}du=1$$ and you can use a similar technique to show the lower half is $-1$, summing to zero as we know they must by symmetry.
A: Prove that $\int_0^\infty xe^{-x^2/2}dx$ converges (in this interval the change $u=x^2$ is right). Then use Travis' hint.
A: for odd function, the area or value of integral will be zero as shown in plot

