What's up with this integral

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In summary: Gauss used a standard z-table to calculate these values.In summary, Gauss integrated this function to obtain the values of the normal distribution.
  • #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
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
But since the question was
alba_ei said:
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
Blarrgh, you read it weller than me!
I LIKE those tricks..:frown: :cry:
 
  • #5
arildno said:
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
But what's happened if I want to know the integral from 0 to x. specifically from 0 to 2
 
  • #7
then you should read HallsofIvy's post once more.
 
  • #8
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
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
Right. You will need to use the given values of [itex]\mu[/itex] and [itex]\sigma[/itex] to convert to the "standard z" variable.
 

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to solve problems involving rates of change, such as finding the distance traveled by an object with a given velocity.

Why is integration important?

Integration is an important tool in mathematics and science because it allows us to find the exact value of a quantity, rather than just approximating it. It is also used to solve a wide range of real-world problems, such as calculating volumes and areas.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, meaning that it calculates the area under a curve between two given points. An indefinite integral does not have limits of integration and is used to find a general solution to a specific problem.

How is integration related to differentiation?

Integration and differentiation are inverse operations. This means that integration is the reverse process of differentiation. Integration allows us to find the original function when given the derivative, while differentiation allows us to find the rate of change of a function.

What are some common techniques for solving integrals?

Some common techniques for solving integrals include substitution, integration by parts, trigonometric substitution, and partial fractions. These techniques are used to simplify the integrand and make it easier to evaluate the integral.

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