Solve Gaussian Integral: A from \int^{-\infty}_{+\infty} \rho (x) \,dx = 1

[DukeLuke]

Homework Statement

Consider the gaussian distribution shown below

$$\rho (x) = Ae^{-\lambda (x-a)^2$$

where A, a, and $\lambda$ are positive real constants. Use $\int^{-\infty}_{+\infty} \rho (x) \,dx = 1$ 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 $u=x-a$ that left me with
$$Ae^{-\lambda} \int^{+\infty}_{-\infty} e^{u^2}\,du$$
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.

 

[Cyosis]

You have made an error.

$$e^{-\lambda u^2} \neq e^{-\lambda}e^u^2=e^{-\lambda+u^2}$$

 

[tiny-tim]

… (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!

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

(btw, there is a way to integrate ∫e^-u2du: it's √(∫e^-u2du)∫e^-v2dv), then change to polar coordinates )

 

[DukeLuke]

You have made an error.
$$e^{-\lambda u^2} \neq e^{-\lambda}e^u^2=e^{-\lambda+u^2}$$

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?

 

[D H]

(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.

 

[Cyosis]

If you still can't find it check http://en.wikipedia.org/wiki/Gaussian_integral out.

 

[DukeLuke]

$$\int_{-\infty}^{\infty} e^{-(x+b)^2/c^2}\,dx=|c| \sqrt{\pi}$$

Thanks, using the integral above from Wikipedia $c = \frac{1}{\sqrt{\lambda}}$. From there I get $A = \frac{\sqrt{\lambda}}{\sqrt{\pi}}$.

 

[Feldoh]

Looks correct, studying griffiths' quantum mechanics I see :D

 

[DukeLuke]

Yep, thought I would get a head start before the fall semester begins.