Q:

A naive change of integration limit

I am trying to understand something pretty dumb. 
First question:
Assuming I have the following integral: 
$$
\int_{-\infty}^{\infty}x^3e^{-\frac{-x^2}{2}}dx
$$
If it was an indefinite integral I would solve this one with a simple change of variable: $x^2=t$ and the solution is pretty much simple. 
The problem is though that this is an improper integral, and since my final  answer is wrong I guess I am not changing the limits of the integral properly. 
My dumb, naive approach would be:
when $x\rightarrow \infty$ then $(t=x^2)\rightarrow \infty$, and when $x\rightarrow -\infty$ then $(t=x^2)\rightarrow = \infty$
.  Is that correct to say that the integral is actually:
$$ 
\int_{\infty}^{\infty}te^{-\frac{t}{2}}dt=0? 
$$
Right? 
Second question:
Assuming again I have the following integral: 
$$
\int_{-\infty}^{\infty}x^3e^{-\frac{-x^2}{2}}dx
$$
Now I tried to separate this integral:
$$
\int_{-\infty}^{\infty}x^3e^{-\frac{-x^2}{2}}dx = \int_{-\infty}^{0}x^3e^{-\frac{-x^2}{2}}dx + \int_{0}^{\infty}x^3e^{-\frac{-x^2}{2}}dx
$$
Now the change of variables gives: 
$$
 \int_{\infty}^{0}te^{-\frac{-t}{2}}dt + \int_{0}^{\infty}te^{-\frac{-t}{2}}dt
$$
My question is: what does $ \int_{\infty}^{0}te^{-\frac{-t}{2}}dt$ mean? 

A:

First of all, note that you are integrating an odd function.  So, all you need to do is that the integral converges and then it will be automatically equal to $0$.
But if you do $t=x^2$, well… This means that $x=\sqrt t$, right?!  So, use it to compute$$\int_0^\infty x^3e^{-\frac{x^2}2}\,\mathrm dx$$insted.  Then$$\int_{-\infty}^0 x^3e^{-\frac{x^2}2}\,\mathrm dx$$will be the symmetric of that.