What's up with this integral

[alba_ei]

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

 

[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 [itex]\pi[/tex]

4. Hence, we have $I^{2}=\pi\to{I}=\sqrt{\pi}$

 

[HallsofIvy]

But since the question was

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

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.

 

[arildno]

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

 

[tehno]

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 [itex]\pi[/tex]

4. Hence, we have $I^{2}=\pi\to{I}=\sqrt{\pi}$

arildno,you are a mean green integration machine

 

[alba_ei]

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

 

[arildno]

then you should read HallsofIvy's post once more.

 

[HallsofIvy]

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.

 

[murshid_islam]

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

 

[Gib Z]

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

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

 

[HallsofIvy]

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