Integral of the S.N.D.F.

1. May 20, 2007

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?

2. May 20, 2007

Office_Shredder

Staff Emeritus
3. May 20, 2007

prasannapakkiam

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. May 23, 2007

ZioX

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. May 23, 2007

Gib Z

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. May 23, 2007

Staff: Mentor

One could do it by numerical integration, but try a double integral, as in

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

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

Limits (0,x) and (0, y) become (0, r) and (0, 2$\pi$)

Last edited: May 23, 2007
7. May 23, 2007

prasannapakkiam

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. May 24, 2007

uart

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 $$r \, dr \, d\theta$$ then it follows that the inner of the two dimensional ($$dr \, d\theta$$) integral becomes the trivial $$\int r e^{-r^2/2} dr$$.

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 $$d\theta$$ 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 $$\int_{-\infty}^{+\infty} e^{-x^2/2} dx = \sqrt{2 \pi}$$

Goolging for 'erf approximations' gives several very good approximations in the first few hits. One nice simple one is :

$$\frac{1}{\sqrt{2 \pi}} \int_x^{\infty} e^{-x^2/2} \, \simeq \frac{ e^{-x^2/2} } {1.64 x + \sqrt{0.76 x^2 + 4}}$$

Last edited: May 24, 2007
9. May 25, 2007

I see......