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

Gaussian Integral

  • Thread starter DukeLuke
  • Start date
  • #1
94
0

Homework Statement


Consider the gaussian distribution shown below

[tex] \rho (x) = Ae^{-\lambda (x-a)^2 [/tex]

where A, a, and [itex] \lambda [/itex] are positive real constants. Use [itex] \int^{-\infty}_{+\infty} \rho (x) \,dx = 1 [/itex] to determine A. (Look up any integrals you need)

Homework Equations


Given in question above

The Attempt at a Solution


My plan was to integrate the probability density set it equal to one and then solve for A. The problem is I'm getting stuck on the integration. I started by pulling the constants out of the integral and doing the substitution [itex] u=x-a [/itex] that left me with
[tex] Ae^{-\lambda} \int^{+\infty}_{-\infty} e^{u^2}\,du [/tex]
It's been a while since calc II and I can't figure out how to do this one (even though it looks so simple). I also tried looking it up in a integral table but couldn't find it. Any help would be appreciated.
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
You have made an error.

[tex]
e^{-\lambda u^2} \neq e^{-\lambda}e^u^2=e^{-\lambda+u^2}
[/tex]
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
251
… (Look up any integrals you need) …

I also tried looking it up in a integral table but couldn't find it. Any help would be appreciated.
Hi DukeLuke! :wink:

You need the erf(x) function … see http://en.wikipedia.org/wiki/Error_function :smile:

(btw, there is a way to integrate ∫e-u2du: it's √(∫e-u2du)∫e-v2dv), then change to polar coordinates :wink:)
 
  • #4
94
0
You have made an error.
[tex]
e^{-\lambda u^2} \neq e^{-\lambda}e^u^2=e^{-\lambda+u^2}
[/tex]
Thanks, man am I getting rusty over the summer

You need the erf(x) function
I looked at it but I'm lost on how to use it solve this problem. Could you help me get started?
 
  • #5
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
(Look up any integrals you need)
Have you tried this bit of advise? You even know the relevant keywords (hint: use the title of this thread). Google is your friend.
 
  • #7
94
0
[tex] \int_{-\infty}^{\infty} e^{-(x+b)^2/c^2}\,dx=|c| \sqrt{\pi} [/tex]

Thanks, using the integral above from Wikipedia [itex] c = \frac{1}{\sqrt{\lambda}} [/itex]. From there I get [itex] A = \frac{\sqrt{\lambda}}{\sqrt{\pi}} [/itex].
 
  • #8
1,341
3
Looks correct, studying griffiths' quantum mechanics I see :D
 
  • #9
94
0
Yep, thought I would get a head start before the fall semester begins.
 

Related Threads on Gaussian Integral

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
9
Views
9K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
615
  • Last Post
Replies
4
Views
1K
Replies
12
Views
6K
Top