# Improper Integral

1. Jun 12, 2010

### Lancelot59

This is the problem:
$$\int_{-\infty}^{\infty} {xe^{-x^{2}}dx}$$

I noticed the function was even, so I then did this:

$$2\int_{0}^{\infty} {xe^{-x^{2}}dx}$$

I attempted to do integration by parts:

$$u=e^{-x^{2}}, du=-2e^{-x^{2}}, dv=x, v=\frac{x^{2}}{2}$$

which still left me with this at the end:
(I left out the bounds)
$$\frac{x^{2}}{2}(e^{-x^{2}} - \int_{}^{} {e^{-x^{2}}dx})$$

and of course that integral can't be done. Doing parts the other way results with the same issue, and I don't see how a substitution could work. Thanks in advance for any help.

2. Jun 12, 2010

### Dick

Why do you think x*e^(-x^2) is even?

3. Jun 12, 2010

### Staff: Mentor

The integrand is not an even function - it is odd. This should make the problem a lot easier.
That's not the best approach. An ordinary substitution is all that is required.

4. Jun 12, 2010

### Lancelot59

I see. I'll post back once I'm done with the work.

Last edited: Jun 12, 2010
5. Jun 12, 2010

### Lancelot59

Bad logic. That's why. I thought about it more and it isn't.

6. Jun 12, 2010

### Lancelot59

It worked.

$$u=e^{-x^{2}}, du=-2e^{-x^{2}}dx \rightarrow \frac{du}{-2u}=xdx$$

Then got here:

$$\lim_{t \rightarrow -\infty}\int_{t}^{0} {xe^{-x^{2}}dx} + \lim_{t \rightarrow \infty}\int_{0}^{t} {xe^{-x^{2}}dx}$$

Substituted:
$$\lim_{t \rightarrow -\infty}\int_{x=t}^{x=0} {-\frac{du}{2u}u} + \lim_{t \rightarrow \infty}\int_{x=0}^{x=t} {-\frac{du}{2u}u}$$

$$-\frac{1}{2}\lim_{t \rightarrow -\infty}\int_{x=t}^{x=0} {du} + -\frac{1}{2}\lim_{t \rightarrow \infty}\int_{x=0}^{x=t} {du}$$

Then I just got u as the integral, did the rest and found that it converged on zero. Thanks for the help.