How can I integrate e^(-(x^2)/2)?

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So there is no way to get an exact answer to this integral, but it can be approximatedIn summary, the conversation discusses the difficulty of integrating the function y=e^(-(x^2)/2) and the use of numerical approximations such as the Simpson's Rule. It is mentioned that there is no elementary antiderivative of the function, but a double integral method can be used for evaluating the definite integral of the function from -infinity to infinity. Several approximation methods are also mentioned, such as using polar coordinates and the erf function. In the end, it is concluded that while an exact answer may not be attainable, the integral can be approximated.
  • #1
prasannapakkiam
I fell very silly posting this, but I am making Normal Distribution tables. I tried to Integrate what seems to be quite a simple function:
y=e^(-(x^2)/2)
Well... I can't I have tried Integration by Parts, Algebraic Substitution etc. But I cannot do it! Can somebody help me?
 
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  • #3
Ok. But I don't intitutively see why... Can anyone answer this?

So all tables are created as an approximation - e.g. the Simpson's Rule?
 
  • #4
Yes, they are numerical approximations. I can assure you they weren't calculated via Simpson's Rule.

There is no elementary antiderivative of the pdf. Elementary being sums/products/compositions of 'nice' functions.
 
  • #5
It can be very hard to prove that a certain function has non-elementary antiderivatives. We know in this case that no known function has its derivative as that integral, so we defined one.
 
  • #6
prasannapakkiam said:
I fell very silly posting this, but I am making Normal Distribution tables. I tried to Integrate what seems to be quite a simple function:
y=e^(-(x^2)/2)
Well... I can't I have tried Integration by Parts, Algebraic Substitution etc. But I cannot do it! Can somebody help me?
One could do it by numerical integration, but try a double integral, as in

[tex]\int_{0}^{x} e^{-x^2} dx\,\int_{0}^{y} e^{-y^2} dy[/tex], where I is each integral.

combine the two and use the transformation from Cartesian (x, y) to polar coordinates (r, [itex]\theta[/itex]).

Limits (0,x) and (0, y) become (0, r) and (0, 2[itex]\pi[/itex])
 
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  • #7
I am not too sure about Astronuc's suggestion. But anyway, can someone give me a reasonable function for an approximation. So far I have only come up with tanh(x), with a precding constant...
 
  • #8
Astronuc said:
... but try a double integral, as in...

The double integral method is good for evaluating the definite integral of exp(-x^2/2) from -infinty to infinity, but it is of no use for integrating over finite limits.

The double integral method works by transforming the square of the integral into a double integral over a region of the x,y plane. Since the element of area dx dy becomes [tex]r \, dr \, d\theta[/tex] then it follows that the inner of the two dimensional ([tex]dr \, d\theta[/tex]) integral becomes the trivial [tex]\int r e^{-r^2/2} dr[/tex].

The catch is that the limits of the integration correspond to a square region of the x,y plane, so r is not a constant! Bascially the difficulty just gets tranferred to the outer [tex]d\theta[/tex] integral, so in general this is no solution.

For the specific case of the integral from -infinity to +infinity however the double integral is over the entire x,y plane and therefore the difficulty with the rectangular limits vanishes. This is the standard method of proving that [tex]\int_{-\infty}^{+\infty} e^{-x^2/2} dx = \sqrt{2 \pi}[/tex]
But anyway, can someone give me a reasonable function for an approximation.
Goolging for 'erf approximations' gives several very good approximations in the first few hits. One nice simple one is :[tex]\frac{1}{\sqrt{2 \pi}} \int_x^{\infty} e^{-x^2/2} \, \simeq \frac{ e^{-x^2/2} } {1.64 x + \sqrt{0.76 x^2 + 4}} [/tex]
 
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  • #9
I see...
 

FAQ: How can I integrate e^(-(x^2)/2)?

What is the "Integral of the S.N.D.F."?

The Integral of the S.N.D.F. stands for the "Integral of the Standard Normal Distribution Function." It is a mathematical concept used in statistics to calculate the area under the curve of a normal distribution.

How is the Integral of the S.N.D.F. calculated?

The Integral of the S.N.D.F. is calculated using integration, a mathematical process that finds the area under a curve. In this case, the curve is the bell-shaped curve of the normal distribution. The formula for the Integral of the S.N.D.F. is: ∫ e^(-x^2/2) dx, where e is the mathematical constant and x is the variable.

Why is the Integral of the S.N.D.F. important in statistics?

The Integral of the S.N.D.F. is important in statistics because it allows us to calculate the probability of a random variable falling within a certain range. This is useful in analyzing data and making predictions about future outcomes in various fields such as finance, economics, and science.

What is the relationship between the Integral of the S.N.D.F. and the normal distribution?

The Integral of the S.N.D.F. is directly related to the normal distribution. In fact, it is the cumulative distribution function of the normal distribution. This means that the Integral of the S.N.D.F. gives us the probability of a random variable falling within a certain range on the normal distribution curve.

Can the Integral of the S.N.D.F. be used for non-normal distributions?

No, the Integral of the S.N.D.F. is only applicable to the normal distribution. Other types of distributions have their own specific integral functions that are used to find the area under their respective curves. However, in some cases, the normal distribution can be used as an approximation for non-normal distributions, allowing the Integral of the S.N.D.F. to be used indirectly.

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