Integrating the Normal Distribution: Uncovering Gauss' Method

[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 \pi[/tex]<br /> <br /> 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 \pi[/tex]<br /> <br /> 4. Hence, we have I^{2}=\pi\to{I}=\sqrt{\pi}

<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":cool:" title="Cool :cool:" data-smilie="6"data-shortname=":cool:" /> <br /> arildno,you are a mean green integration machine <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/approve.gif" class="smilie" loading="lazy" alt=":approve:" title="Approve :approve:" data-shortname=":approve:" />

 

[alba_ei]

But what's happened 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.