Ken Ward's Mathematics Pages

Series: Gaussian Integral (for Gamma)

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Because the Gaussian Integral is useful for our consideration of the gamma function, we present a simple proof here. There is not known indefinite integral of this function. However, it is possible to determine the value of a definite integral.

The graph is shown below. It may be recognized, among many applications, as the normal distribution, in statistics:

We wish to prove:

[1.01]

We assume that integrating from zero is half this function (assume it is an even function)

So, we aim to prove:

[1.02]

For convenience in referring to the integral,

[1.03]

It seems that any of the obvious ways of solving this integral do not work. For example, integration by parts, substitution, etc.

We approach this problem by dealing with the squared integral as follows:

Which we can write as:

[1.04]

We proceed with the intention of using polar coordinates.

[1.05]

In the following diagram we have r and θ as polar coordinates in the first quadrant.

We seek to integrate over r and θ, to determine the area under the curve. We note the differential area, dA=r·dθ·dr=dx·dy

We also note that we are integrating from 0 to ∞ for r and from 0 to π/2 for θ, so our integral becomes:

[1.06]

First let us consider the inner integral:

[1.07]

If we substitute u=r², du=r·dr, we have:

[1.07]

Which is a simple, standard integral, giving us:

[1.08]

Substituting this value into our integral, [1.06], we have the remaining integral to evaluate, which is very easy:

[1.09]

Taking the square root and the positive root (because we are in the first quadrant, and everything is positive here):

(We might interpret the negative root to refer to the other half of the function)

Hence:

[1.02, repeated]

And by our assumption that the curve is symmetric,

[1.01, repeated]