What's up with this integral

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  • #1
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how did gauss integrate this function
to obtain the values of the normal distribution?:confused:
 

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  • #2
arildno
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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]
 
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  • #3
HallsofIvy
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But since the question was
how did gauss integrate this function
to obtain the values of the normal distribution?:confused:
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
arildno
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Blarrgh, you read it weller than me!
I LIKE those tricks..:frown: :cry:
 
  • #5
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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]
:cool:
arildno,you are a mean green integration machine :approve:
 
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  • #6
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But whats happend if I want to know the integral from 0 to x. specifically from 0 to 2
 
  • #7
arildno
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then you should read HallsofIvy's post once more.
 
  • #8
HallsofIvy
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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.
 
  • #10
Gib Z
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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]
 
  • #11
HallsofIvy
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Right. You will need to use the given values of [itex]\mu[/itex] and [itex]\sigma[/itex] to convert to the "standard z" variable.
 
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