Evaluate the following integral

What's that mean?

 

I would integrate by parts.

Hint:

[tex]
u=x^{n-1}
[/tex]

[tex]
dv=xe^{-ax^2}dx
[/tex]

Try that and see what you come up with.

 

Can happen to anyone.

The name's Tom, not Matt. How would you like it if I called you "nebula"?

Yeah, my next set of notes is in preparation. I stopped posting to the thread because I was asked to teach a couple of summer courses. They're condensed into 6 weeks, so I didn't have time for anything else.

Keep watching...

 

That is perhaps the most famous integral ever taught to students of multivariable calculus, at least in America. It is a standard way to introduce polar coordinates in double integration.

[tex]I=\int_0^{\infty} e^{-ax^2} dx [/tex]

[tex]I=\int_0^{\infty} e^{-ay^2} dy [/tex]

[tex]I^2=\int_0^{\infty}\int_0^{\infty} e^{-a(x^2+y^2)} dx [/tex]

Now convert to polar coordinates and evaluate.

Erf(x) is the error function.

 

If you want to skip the integration by parts, you can choose to use the differentiation under the integral sign:

[tex] \int_{-\infty}^{\infty} x^{n} dx e^{-ax^{2}} = \int_{-\infty}^{\infty} dx \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}(-1)^{\frac{n}{2}}e^{-ax^{2}} [/tex]

So now you can take the derivative with respect to a out in front:

[tex] (-1)^{\frac{n}{2}} \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}\int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex]

So, once you've figured out what [tex] \int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex] is, you could also do it in this way.
Notice of course, that this is only useful if n is even, which I think you stipulated, and if the integral is a definite integral.