Integrating the Normal Distribution Curve

[Brad_Ad23]

Given

$$P(x)= \frac{1}{\sigma \sqrt{2\pi}} e ^ \frac { -(x - \mu )^2}{2 \sigma ^2 }$$

This is of course the normal distribution curve. When $$\mu = 0$$ and $$\sigma = 1$$ I can integrate this from minus infinity to positive infinity no problem using polar coordinates and a bit of multivariable calculus. The question I have, is, is it at all possible to do this if one leaves $$\mu$$ and $$\sigma$$ in as generic parameters? I would think so, but I'm not sure. No need to give a worked through example, just, is it possible at all to fit it to some form? Or is it just that with those parameters set to 0 and 1 that this is an integrable function?

 

[cookiemonster]

Looks like Mathworld has a little blob on this.

http://mathworld.wolfram.com/NormalDistribution.html

cookiemonster

 

[HallsofIvy]

Certainly. Just do a little creative substitution:
Let $$u= \frac{x-\mu}{\sqrt(2)\sigma^2}$$.

Then $$du= \frac{1}{\sqrt(2)\sigma^2}dx$$

$$P(x)= \frac{1}{\sigma \sqrt{2\pi}} e ^ \frac { -(x - \mu )^2}{2 \sigma ^2 }$$

becomes $$P(u)= \frac{1}{\sqrt{\pi}} e ^{-u^2}$$

which you can integrate from negative to positive infinity as usual.
(That's why you probably have only seen that case.)

 

[Brad_Ad23]

Yep.

 

[allergic]

can it be integrated from a number to anohter... say -2 and 2?

 

[suyver]

Numerically, yes.
Analytically, no (unless you count functions like Erf(x), which is cheating in my view).

 

[Brad_Ad23]

Well wait a minute. Wouldn't you just change the bounds on the integral with respect to radius from 0 to infinity to 0 to 2? The theta integral would remain the same, and thus you'd get the analytical answer?

 

[suyver]

Maybe I misunderstood allergic's question. I thought that he wanted to know

$$\int_0^2 \frac{1}{\sqrt{\pi}} e ^{-u^2} \,{\rm d}u$$

I don't think there is an analytical expression for that. Numerically it's easy (0.497661).