• Support PF! Buy your school textbooks, materials and every day products Here!

Integral involving Hermite polynomials

  • #1

Homework Statement


The Hermite polynomials [tex]H_n(x)[/tex] may be defined by the generating function
[tex]e^{2hx-h^2} = \sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}[/tex]

Evaluate
[tex]\int^{\infty}_{-\infty} e^{-x^2/2}H_n(x) dx[/tex]
(this should be from -infinity to infinity, but for some reason the latex won't work!)

Homework Equations


Given.

The Attempt at a Solution


I know that for odd [tex]n[/tex], this integral is 0, but I have no idea how to evaluate it. I know that the Hermite polynomials are orthogonal with respect to the given weight function, but I don't think I can use that for this integral (also, that's not in my book, I just found it on wikipedia). I'm assuming I need to use the generating function to derive my answer somehow, but I can't imagine how I would do that.
 
Last edited:

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,256
I haven't worked this problem out yet, but I'd try looking at

[tex]\int^{\infty}_{-\infty} e^{-x^2/2}e^{2hx-h^2}\,dx= \int^{\infty}_{-\infty} e^{-x^2/2}\sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}\,dx[/tex]

See if you can work out the lefthand side.
 
  • #3
I haven't worked this problem out yet, but I'd try looking at

[tex]\int^{\infty}_{-\infty} e^{-x^2/2}e^{2hx-h^2}\,dx= \int^{\infty}_{-\infty} e^{-x^2/2}\sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}\,dx[/tex]

See if you can work out the lefthand side.
Huh, I didn't even think of that. I'll try it and report back. Thank you so much!
 
  • #4
I can't get anywhere with that integral. The best alternatives I've come up with have been repeatedly integrating by parts or using a recursion relation derived from the generating function, but I can't get either of those to work out. I'm almost positive I'm supposed to use the recursion relation ([tex]H_{n+1} = 2xH_n - 2nH_{n-1}[/tex]), since I derived it in an earlier problem and the book has basically no other information on Hermite polynomials. I thought I could mash the integral into the gamma function by reducing the integral to powers of x times the exponential (and in fact I still think I can, and that this is roughly the right way to do it), but I can't make it work.
 
  • #5
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
The integral on the LHS is Gaussian if you complete the square in the exponent.
 
  • #6
Thanks. I've never heard of a gaussian integral before. So now I've got

[tex]\sqrt{2\pi}e^{h^2} = \sum_{n}^{\infty}\frac{h^n}{n!}I[/tex]
where I is the integral I'm trying to get.
Expanding the exponential on the LHS I get a series in powers of [tex]h^{2n}[/tex]. Is it kosher to simply rewrite that as [tex]\sum_{n=0}{\infty}\frac{h^n}{(n/2)!}[/tex]? If it is, equating coefficients gives [tex]I = \frac{\sqrt{2\pi}}{(n/2)!}[/tex]
Does that look right?
 
  • #7
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
The LHS is a series in even powers. Write the RHS in terms of two sums, one in even powers and the other in odd powers. You want to make sure that you're matching the correct indices in order to read off the coefficients properly.

Edit: What you write looks reasonable, but I'm too lazy to check factors of 2!
 

Related Threads on Integral involving Hermite polynomials

  • Last Post
Replies
2
Views
1K
Replies
7
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
810
  • Last Post
Replies
2
Views
774
Replies
2
Views
1K
Replies
6
Views
1K
Replies
4
Views
2K
Replies
0
Views
2K
Replies
5
Views
2K
Top