# Normalizatoin Constant and a Gaussian Integral.

• Oct 21st 2011, 01:59 PM
atomicpedals
Normalizatoin Constant and a Gaussian Integral.
Hello! I'm attempting to compute the normalization constant N for the following function $\displaystyle f(x)=N e^{-(x-x_0)/2k^{2})}$. This is generally done by evaluating $\displaystyle \int |f(x)|^{2}dx=1$, where the limits of integration are -inf to +inf.
I can get this right up until the point where I need to make a u substitution.
$\displaystyle 1=\int |N e^{-(x-x_0)/2k^{2})}|^{2}dx$
$\displaystyle 1=\int N^2 e^{-(x-x_0)^2/k^2} dx$
$\displaystyle 1=N^2 \int e^{-(x-x_0)^2/k^2} dx$
making the substitution $\displaystyle u=x-x_0, du=dx$
$\displaystyle 1=N^2 \int e^{-u^2/k^2}du$
This integral is only slightly more clear to me. It looks like a Gaussian integral, but, how do I go about solving it? Any help is much appreciated.
• Oct 21st 2011, 06:06 PM
Ackbeet
Re: Normalizatoin Constant and a Gaussian Integral.
I would actually make the subtitution

$\displaystyle u=\frac{x-x_{0}}{k}.$

That should make the integral a more standard integral.
• Oct 22nd 2011, 06:24 AM
atomicpedals
Re: Normalizatoin Constant and a Gaussian Integral.
So, that would yield: $\displaystyle 1=N^{2} \int e^{-u^{2}}du=N^{2}\sqrt{\pi}$ ?
• Oct 22nd 2011, 07:52 AM
Ackbeet
Re: Normalizatoin Constant and a Gaussian Integral.
Quote:

Originally Posted by atomicpedals
So, that would yield: $\displaystyle 1=N^{2} \int e^{-u^{2}}du=N^{2}\sqrt{\pi}$ ?

I don't think you've quite got it yet. The substitution I suggested requires a change in the differential as well. The final answer had better have a 'k' in it somewhere.
• Oct 23rd 2011, 06:22 AM
atomicpedals
Re: Normalizatoin Constant and a Gaussian Integral.
Woops! $\displaystyle 1=N^{2}\frac{\sqrt{\pi}}{k}$ which then yeilds a normalization constant $\displaystyle N=\pm\frac{\sqrt{k}}{\pi^{1/4}}$ .
• Oct 24th 2011, 02:18 AM
Ackbeet
Re: Normalizatoin Constant and a Gaussian Integral.
Better, but still not quite there yet. Be careful as to numerator and denominator. Can you show your working, please?
• Oct 24th 2011, 02:32 AM
atomicpedals
Re: Normalizatoin Constant and a Gaussian Integral.
So my line of thinking was as follows:

$\displaystyle u=\frac{(x-x_{0})}{k}$ which I think yields $\displaystyle du=\frac{1}{k}dx$ and so $\displaystyle 1=N^{2}\int \frac{1}{k}e^{-u^{2}}du=1=N^{2}\frac{1}{k}\int e^{-u^{2}}du=N^{2}\frac{1}{k}\sqrt{\pi}$
• Oct 24th 2011, 02:33 AM
atomicpedals
Re: Normalizatoin Constant and a Gaussian Integral.
Wait... I should have $\displaystyle kdu=dx$ shouldn't I...
• Oct 24th 2011, 02:35 AM
Ackbeet
Re: Normalizatoin Constant and a Gaussian Integral.
Quote:

Originally Posted by atomicpedals
So my line of thinking was as follows:

$\displaystyle u=\frac{(x-x_{0})}{k}$ which I think yields $\displaystyle du=\frac{1}{k}dx$

Fine so far.

Quote:

and so $\displaystyle 1=N^{2}\int \frac{1}{k}e^{-u^{2}}du$
The problem here is that what you start with is a dx, not a du. Solve du = dx/k for dx. That's what you must plug in.

Quote:

$\displaystyle =1=N^{2}\frac{1}{k}\int e^{-u^{2}}du=N^{2}\frac{1}{k}\sqrt{\pi}$