# Homework Help: Indefinite integral (Hermite polynomials)

1. Mar 30, 2010

### PhysicsMark

1. The problem statement, all variables and given/known data
I need to evaluate the following integral:

$$\int_{-\infty}^{\infty}x^mx^ne^{-x^2}dx$$

I need the result to construct the first 5 Hermite polynomials.
2. Relevant equations

3. The attempt at a solution
First I tried arbitrary values for "m" and "n". I was not able to evaluate the integral when (m+n) was even. I found that when (m+n) is odd, the integral is equal to 0. This makes sense since e^(-x^2) is an even function.

Then I looked it up. Using wolfram integral calculator left me with something known as the incomplete gamma function (Not very sure what that is) in the solution. Maple gave me a very long answer that involved something labeled as Whittaker M.

Finally I went to my professor to ask him how to evaluate it. He said that it is an integral that can be looked up and I should have a square root pi in my answer. He also said that the incomplete gamma function should not be there.

I am not too sure of how to evaluate the integral myself. Maybe more importantly, I am not sure where to find the result of the integral if it can indeed be "looked up". Thanks for reading.

2. Mar 30, 2010

### vela

Staff Emeritus
When the integrand is even, you can rewrite it with limits from 0 to infinity. Then use the substitution u=x^2, and it'll be in the form of a gamma function.

3. Mar 30, 2010

### PhysicsMark

Thanks vela!

I am not sure how the gamma function works. I have just looked over articles pertaining to it online, but I am not confident enough in my understanding of it to be able to identify it in this case.

Here is what I have following vela's advice:

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

Letting u=x^2 I get:

$$\int_{0}^{\infty}\frac{u^{n-1}}{2}e^{-u}du$$

Integrating by parts yields:

$$\int_{0}^{\infty}\frac{u^{n-1}}{2}e^{-u}du= -\frac{u^{n-1}e^{-u}}{2}+\int_{0}^{\infty}\frac{(n-1)u^{n-2}}{2}e^{-u}du$$

Integrating by parts yields more repeating integrals (as far as I can tell from here). Each time I integrate by parts I can pull out a factor. For the first one it was:

$$\frac{n-1}{2}$$

The next factor (I guess I'm talking about the coefficient of the "uv" of the integration by parts...if you use uv-int(vdu)) I get is:

$$\frac{(n-1)(n-2)}{2}$$

This keeps repeating. I see that according to wikipedia, the gamma function can be defined as:

$$\Gamma(n)=(n-1)!$$

As of this point, I do not see how this will get me where I want to go. I am looking for the result of this integral in order to establish rules to determine the orthogonality relationships between (x^m,x^n).

For instance when finding the Legendre polynomials, I found that when (m+n) was odd, (x^m,x^n) = 0. When (m+n) was even, (x^m,x^n) = (2/(m+n+1)). I determined this by performing the integral in the first post. The difference being the weight function.

Thanks again for taking the time to read this.

4. Mar 30, 2010

### vela

Staff Emeritus
You made a mistake when you did the substitution. Because du=2x dx, you should get

$$\frac{1}{2}\int_0^\infty u^{n-1/2}e^{-u} du = \frac{1}{2}\int_0^\infty u^{(n+1/2)-1} e^{-u} du = \frac{1}{2}\Gamma(n+\frac{1}{2})$$

You can look up what that evaluates to, or deduce it using the facts that $\Gamma(n+1)=n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$.

I didn't realize you weren't familiar with the gamma function. I would have suggested a different way to go about this integral. For instance, you could use the trick where you say

$$\int x^{2n} e^{-x^2} dx = \left[\int (-1)^n \frac{d^n}{d\alpha^n} e^{-\alpha x^2} dx\right]_{\alpha=0}$$

and then interchange the order of integration and differentiation. But then again, that's kind of tedious, so maybe learning about the gamma function isn't so bad. :)

5. Mar 30, 2010

### PhysicsMark

Thanks for catching my mistake! So after scouring the internet (I wasn't able to figure out the quantity by searching or using the 2 bits of information) and putting together all I found, I think:

$$\Gamma(n+\frac{1}{2})=\frac{2n!}{n!2^{2n}}\sqrt\pi$$

I realize the above quantity needs to be multiplied by 1/2 to get the quantity I am looking for. Is this correct? If so, can it be simplified. I tried but so far have not been able to re-write it. In fact, I was only able to get the above equation by putting 3 or so different equations together. I don't fully understand why the (pi)^.5 can "come out" of the gamma function.

Also, I didn't understand your other method of finding this relation. How could you do this without knowledge of the gamma function?

6. Mar 30, 2010

### vela

Staff Emeritus
I think it's right. Wikipedia's page on the gamma function actually has the result you were looking for, although it's written slightly differently.

When n=0, the integral reduces to a gaussian. That's why you get the square root of pi in the answer.
It's just a common trick for doing these sorts of integrals. There's a mistake in what I wrote. You want to set alpha=1 at the end, not to zero.

7. Mar 31, 2010

### PhysicsMark

Now it seems that I either have the wrong solution, or I do not know what I am doing. When I use the relation :

$$\int_0^\infty{x^m}x^n{e^{-x^2}dx=\frac{1}{2}((m+n)-1/2)!$$ if (m+n) = even

I'm assuming the gamma function below is equal to the right hand side above

$$\frac{1}{2}\Gamma(n+\frac{1}{2})=\frac{1}{2}\frac{2n!}{n!2^{2n}}\sqrt\pi$$

I thought this value would allow me to generate the first few Hermite polynomials but I am not sure now how to do that.

Using the Schmidt orthogonalization process, as outlined in my tutorial, I take:

$$H_0(x)=\frac{x^0}{\sqrt{(x^0,x^0)}}$$

Using the above equations for n=0 gives me:

$$(x^0,x^0)=\frac{\sqrt{\pi}}{2}$$

Shouldn't the first polynomial = 1?

To continue, the next step is to take

$$H_1(x)=x^1+\alpha{H_0}$$

and solve for alpha to find H_1. Am I doing something incorrect here?

Last edited: Mar 31, 2010
8. Mar 31, 2010

### vela

Staff Emeritus
The difference is normalization. The Gram-Schmidt method needs the functions to be normalized such that (Hn,Hn)=1, but the Hermite polynomials are typically normalized to make them more convenient to use. The http://en.wikipedia.org/wiki/Hermite_polynomials" [Broken] discusses this a bit. What you will find using your method are polynomials that are proportional to the Hermite polynomials.

Last edited by a moderator: May 4, 2017