What's up with this integral

1. Feb 18, 2007

alba_ei

how did gauss integrate this function
to obtain the values of the normal distribution?

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2. Feb 18, 2007

arildno

By a sequence of clever tricks:

1. Let us consider the integral
$$I=\int_{-\infty}^{\infty}e^{-x^{2}}dx$$
Now, since a dummy variable's name is irrelevant, we may write:
$$I^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-x^{2}}dx)=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-y^{2}}dy)$$
2. By Fubini's theorem, we have:
$$\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy$$
3. Switching to polar coordinates, $x=r\cos\theta,y=r\sin\theta,0\leq{r}\leq\infty,0\leq\theta\leq{2\pi}$, we get:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{\infty}\int_{0}^{2\pi}re^{-r^{2}}d\theta{d}r$$
which is easily integrated to the value $\pi[/tex] 4. Hence, we have [itex]I^{2}=\pi\to{I}=\sqrt{\pi}$

Last edited: Feb 18, 2007
3. Feb 18, 2007

HallsofIvy

Staff Emeritus
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.

4. Feb 18, 2007

arildno

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

5. Feb 18, 2007

tehno

arildno,you are a mean green integration machine

Last edited: Feb 19, 2007
6. Feb 18, 2007

alba_ei

But whats happend if I want to know the integral from 0 to x. specifically from 0 to 2

7. Feb 18, 2007

arildno

then you should read HallsofIvy's post once more.

8. Feb 19, 2007

HallsofIvy

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

9. Feb 19, 2007

murshid_islam

10. Feb 20, 2007

Gib Z

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

$$Erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$$

11. Feb 20, 2007

HallsofIvy

Staff Emeritus
Right. You will need to use the given values of $\mu$ and $\sigma$ to convert to the "standard z" variable.