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- Thread starter alba_ei
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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]

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]

Last edited:

- #3

HallsofIvy

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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.how did gauss integrate this function

to obtain the values of the normal distribution?

- #4

arildno

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Blarrgh, you read it weller than me!

I LIKE those tricks..

I LIKE those tricks..

- #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]

arildno,you are a mean green integration machine

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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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- #9

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you can get the table here:

http://www.statsoft.com/textbook/sttable.html

http://www.statsoft.com/textbook/sttable.html

- #10

Gib Z

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

- #11

HallsofIvy

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