Solving Improper Integral: \int_{-\infty}^{\infty} {xe^{-x^{2}}dx}

[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.

 

[Dick]

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

 

[Mark44]

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}

The integrand is not an even function - it is odd. This should make the problem a lot easier.

I attempted to do integration by parts:

That's not the best approach. An ordinary substitution is all that is required.

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.

 

[Lancelot59]

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

 

[Lancelot59]

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

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

 

[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.