# Evaluate the following integral

1. Sep 25, 2005

### quasar987

Any idea on how to evaluate

$$\int_{0}^{\infty} x^n e^{-ax^2} dx$$

where n is even???

Last edited: Sep 25, 2005
2. Sep 25, 2005

### Tom Mattson

Staff Emeritus
What's that mean?

3. Sep 25, 2005

### quasar987

oh sorry, pair = french for even.

Edited.

4. Sep 25, 2005

### Tom Mattson

Staff Emeritus
I would integrate by parts.

Hint:

$$u=x^{n-1}$$

$$dv=xe^{-ax^2}dx$$

Try that and see what you come up with.

5. Sep 25, 2005

### quasar987

How could have I missed that?! :grumpy:

Thx Matt!

Btw - are you still learning diff forms?

6. Sep 25, 2005

### Tom Mattson

Staff Emeritus
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...

7. Sep 25, 2005

### quasar987

Sorry Tom.

I think I'm still missing an important point in your idea though.. in tegrating by parts like you advised I get

$$\int_{0}^{\infty} x^n e^{-ax^2} dx = \left[-\frac{x^{n-1}}{2a} e^{-ax^2} \right]_0^{\infty} + \frac{n-1}{2a} \int_0^{\infty} x^{n-2} e^{-ax^2} dx$$

So we can perform the same technique of integration on the last integral, and so on, such that after n/2 times of performing this trick, the integral we have to evaluate is

$$\int_0^{\infty} e^{-ax^2} dx$$

But what is that?! the integrator says its a constant times $\mbox{Erf}[\sqrt{5}x]$ The heck is Erf ?

8. Sep 25, 2005

### Tom Mattson

Staff Emeritus
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.

$$I=\int_0^{\infty} e^{-ax^2} dx$$

$$I=\int_0^{\infty} e^{-ay^2} dy$$

$$I^2=\int_0^{\infty}\int_0^{\infty} e^{-a(x^2+y^2)} dx$$

Now convert to polar coordinates and evaluate.

Erf(x) is the error function.

9. Sep 25, 2005

### quasar987

RIIIIGHT! I remember browsing my calculus book last year, and for some random reason, I began reading this random problem. And I found it so cute I had to do it, and of course it was proving that this integral is worth $\sqrt{\pi}/2$ by the method you outlined!

10. Sep 25, 2005

### TOKAMAK

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

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

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

$$(-1)^{\frac{n}{2}} \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}\int_{-\infty}^{\infty} dx e^{-ax^{2}}$$

So, once you've figured out what $$\int_{-\infty}^{\infty} dx e^{-ax^{2}}$$ 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.

11. Sep 25, 2005

### quasar987

I'm not familiar with the use of this method TOKAMAK (I'm not Feynman).

btw, if anyone carried out the problem til the end and want to compare, I arrive at

$$\sqrt{\frac{\pi}{2^n a^{n+1}}}\prod_{j=1,3,5,...}^{n-1} (n-j)$$