A naive change of integration limit

Question

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?

Answer 1

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.

Answer 2

The function is odd so you have that

$\int_{-\infty}^{\infty}x^3e^-\frac{x^2}{2}dx=\lim_{k\to \infty}\int_{-k}^{k}x^3e^-\frac{x^2}{2}dx=$

$=\lim_{k\to \infty} 0=0 $

Answer 3

As others have already correctly pointed out, the integrand is odd, so if the integral exists, which it does, then it is nothing. But to answer your second question, the limits on an integral indicate the region of values being integrated over and also the direction of integration. Integrating from infinity to zero is exactly the opposite of integrating from zero to infinity. $$ \int_\infty^0 f(x) \, \mathrm{d}x = -\int_0^\infty f(x)\, \mathrm{d}x, $$ provided this integral exists.