How do I integrate Gaussian form equations without prior knowledge?

[sanitykey]

[SOLVED] Gaussian Form Integration

Homework Statement

http://img153.imageshack.us/img153/9489/001km6.jpg

I'm trying to do the last part.

Homework Equations

Stated in the question.

The Attempt at a Solution

For (iii) i know the probability density is:

$$A^2e^{- \frac{x^2}{\sigma ^2}}$$

Edit: Deleted my first attempt which I'm pretty sure was wrong and I tried using integration by parts which looked quite promising:

The integral I'm trying to do is:

$$\int_{0}^{\sigma} A^2e^{- \frac{x^2}{\sigma ^2}} dx$$

(Take the $$A^2$$ out for now)

Which i can split into:

$$A^2\int_{0}^{\sigma} e^{- \frac{x^2}{2\sigma ^2}} e^{- \frac{x^2}{2\sigma ^2}} dx$$

Let u = $$e^{- \frac{x^2}{2\sigma ^2}}$$ and dv = $$e^{- \frac{x^2}{2\sigma ^2}}$$

then du = $$-\frac{x}{\sigma ^2}e^{- \frac{x^2}{2\sigma ^2}} dx$$ and v = $$(0.68) \sqrt{2\pi} \sigma$$

Using $$\int udv = uv - \int vdu$$ The integral becomes:

$$\left[e^{- \frac{x^2}{2\sigma ^2}}(0.68) \sqrt{2\pi}\sigma \right]_{0}^{\sigma}$$ - $$(0.68) \sqrt{2\pi} \sigma \int_{0}^{\sigma} -\frac{x}{\sigma ^2}e^{- \frac{x^2}{2\sigma ^2}} dx$$

= $$\left[e^{- \frac{x^2}{2\sigma ^2}}(0.68) \sqrt{2\pi} \sigma\right]_{0}^{\sigma}$$ - $$\left[e^{- \frac{x^2}{2\sigma ^2}}(0.68) \sqrt{2\pi} \sigma\right]_{0}^{\sigma}$$

Which i think = 0 and then multiplied by $$A^2$$ still = 0

Is this right?

Thanks
Richy

Edit 2: Woops i put xs instead of es that was probably really confusing to anyone who read this before my edit, sorry!

Edit 3: Actually when i think about the graph for this which would generally be a gaussian shape and it wants me to integrate or find the area between 0 and $$\sigma$$ i don't see how the result can be 0, i think i might have went wrong somewhere but i can't see where.

Edit 4: Ok i just found these answers in a different paper apparently:

$$\int_{0}^{\sigma} e^{- \frac{x^2}{\sigma ^2}} dx = (0.421)\sigma \sqrt{\pi}$$

and

$$\int_{-\infty}^{\infty} e^{- \frac{x^2}{\sigma ^2}} dx = \sigma \sqrt{\pi}$$

From this i can see that the probability of finding the particle is 0.421 but why didn't my integration work? How would i work this out without this knowledge?

 

[Dick]

Great. You found it. You can't figure that out by integrating by parts like that. You do it by changing variables so the integrand becomes exp(-t^2), so in this case t=x/sigma. Once you have done that, you STILL can't do the integral but the function, integral from 0 to x of (2/sqrt(pi))exp(-t^2) is a standard function called erf(x). You can find tables of it or you may find it on your calculator. In your case the 0.421 is erf(1)/2. BTW, I think the two integrals given in your hint are both off by a factor of two.

 

[sanitykey]

Thanks for the help Dick,

I've never come across the erf() function before but i understand what it is now, apparently my calculator isn't that good :P (unless perhaps it has a different name i couldn't see any "erf" button) mathcad can probably do it though or i could find some tables like you say.

Which integrals are you talking about by the way? I'm sure the ones i copied down are correct.

Also both these questions are exactly the same (same marks, same wording) the only difference is one has $$2\sigma ^2$$ whereas the other has $$\sigma ^2$$ yet the methods seem to be completely different? I mean with this the ratio would give me an answer of 0.68/2 = 0.34 but this is for the $$2\sigma ^2$$ function and therefore not at all helpful?

I guess what I'm trying to ask is with the question i originally posted can i use those integrals given to me in any useful way? Or is the erf() function the only way to go about this?

 

[Dick]

The integrals you were given in the problem were not the integrals you needed to do the problem (as you know). And you would compute them the same way as you would the other one. For the first one, I get erf(1/sqrt(2))/2 times the sqrt(2*pi)*s. That numerical erf part comes out to 0.34 - not 0.68. So they aren't only the wrong integrals, they aren't even correct. It's a pretty sloppy problem.

 

[sanitykey]

Thanks again for the help, that's cleared things up for me.