My interest in “beta” functions comes from their utility in Bayesian statistics. A nice 78 minute introduction to Bayesian statistics and how the beta distribution is used can be found here; you need to understand basic mathematical statistics concepts such as “joint density”, “marginal density”, “Bayes’ Rule” and “likelihood function” to follow the youtube lecture. To follow this post, one should know the standard “3 semesters” of calculus and know what the gamma function is (the extension of the factorial function to the real numbers); previous exposure to the standard “polar coordinates” proof that would be very helpful.

So, what it the beta function? it is where . Note that for integers The gamma function is the unique “logarithmically convex” extension of the factorial function to the real line, where “logarithmically convex” means that the logarithm of the function is convex; that is, the second derivative of the log of the function is positive. Roughly speaking, this means that the function exhibits growth behavior similar to (or “greater”) than

Now it turns out that the beta density function is defined as follows: as one can see that the integral is either proper or a convergent improper integral for such values.

I’ll do this in two steps. Step one will convert the beta integral into an integral involving powers of sine and cosine. Step two will be to write as a product of two integrals, do a change of variables and convert to an improper integral on the first quadrant. Then I’ll convert to polar coordinates to show that this integral is equal to

**Step one: converting the beta integral to a sine/cosine integral.** Limit and then do the substitution . Then the beta integral becomes:

**Step two: transforming the product of two gamma functions into a double integral and evaluating using polar coordinates.**

Write

Now do the conversion to obtain:

(there is a tiny amount of algebra involved)

From which we now obtain

Now we switch to polar coordinates, remembering the that comes from evaluating the Jacobian of

This splits into two integrals:

The first of these integrals is just so now we have:

The second integral: we just use to obtain:

(yes, I cancelled the 2 with the 1/2)

And so the result follows.

That seems complicated for a simple little integral, doesn’t it?