What's up with this integral

By a sequence of clever tricks:

1. Let us consider the integral
[tex]I=\int_{-\infty}^{\infty}e^{-x^{2}}dx[/tex]
Now, since a dummy variable's name is irrelevant, we may write:
[tex]I^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-x^{2}}dx)=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-y^{2}}dy)[/tex]
2. By Fubini's theorem, we have:
[tex]\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy[/tex]
3. Switching to polar coordinates, [itex]x=r\cos\theta,y=r\sin\theta,0\leq{r}\leq\infty,0\leq\theta\leq{2\pi}[/itex], we get:
[tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{\infty}\int_{0}^{2\pi}re^{-r^{2}}d\theta{d}r[/tex]
which is easily integrated to the value [itex]\pi[/tex]

4. Hence, we have [itex]I^{2}=\pi\to{I}=\sqrt{\pi}[/itex]

 

But since the question was

The answer is "numerically". In general it is impossible to find a closed form (elementary) anti-derivative but one certainly can do a numerical integration- that's how the values in a "Gaussian distribution" table are calculated.

 

Blarrgh, you read it weller than me!
I LIKE those tricks..

 

arildno,you are a mean green integration machine

 

then you should read HallsofIvy's post once more.

 

You could look it up in a table of the "Error function", Erf(x). That is the "anti-derivative" of [itex]e^{-x^2/2}[/itex] and is derived, just as I said, by numerical integration.

 

you can get the table here:
http://www.statsoft.com/textbook/sttable.html

 

If you encounter a table of values for the Error function, don't quickly take that as your value for the integral.

[tex]Erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt[/tex]

 

Right. You will need to use the given values of [itex]\mu[/itex] and [itex]\sigma[/itex] to convert to the "standard z" variable.