Beta Function Extension?

1. Mar 7, 2013

FeDeX_LaTeX

I was looking at some integration problems the other day and I came across this identity:

$$\int_{0}^{\frac{\pi}{2}} \sin^{p}x \cos^{q}x dx = \frac{1}{2} \mbox{B} \left( \frac{p+1}{2},\frac{q+1}{2}\right)$$

where B(x,y) is the Beta function, for Re(x) and Re(y) > 0. From the way in which the above formula is derived, it turns out that this is valid for p > -1 and q > -1.

However, I'm interested in seeing if this identity can be extended to allow any real power of p or q. Does anyone know of a similar identity that enables me to solve this for any real p or q, or over any interval?

Last edited: Mar 7, 2013
2. Mar 7, 2013

fzero

For ${\text Re}(x),{\text Re}(y)>0$, we can use the integral representations to derive

$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{ \Gamma(x+y)} .$$

Since the gamma function has an extension to complex numbers apart from zero or the negative integers, we can use this formula to extend the beta function to a similar domain.

3. Mar 7, 2013

FeDeX_LaTeX

Thanks for the reply. That's the formula I've been using generally to find values of the beta function; so would it be correct to say that the identity given in the OP is valid for all real p and q, unless p or q are negative integers? So one could use this for negative reals (apart from the integers)?

4. Mar 7, 2013

Mute

No, the logic doesn't run that way. The integral in your original post is true for p, q > -1. It doesn't matter that you can extend the Beta function to non-(negative integer) values, proving that the integral holds for those values is something you would have to separately show.

For instance, the Gamma function itself is defined via an integral, but only for non-negative arguments. We can analytically continue the Gamma function to complex values, but the integral form is still only equal to the Gamma function along the non-negative real line.

Similarly, I expect that your integral simply won't converge for p, q < -1. If you make the change of variables $t = \sin x$, I think you'll find that for p or q < -1 the integrand diverges at at least one of the limits (and can't be integrated to give a finite area).

5. Mar 7, 2013

fzero

Mute is correct, outside of the positive argument domain, the integral representations of either the Beta or Gamma function are not convergent. The extension here is called analytic continuation. We use other properties of the function that make sense on a larger domain to extend (or continue) the definition of the function to the larger domain. For instance (see http://en.wikipedia.org/wiki/Gamma_function#The_gamma_function_in_the_complex_plane), for the Gamma function we can use

$$\Gamma(x) = \frac{\Gamma(x+n)}{x (x+1)\cdots (x+n-1)}$$

to define the value at a nonpositive value of $x$ in terms of the Gamma function at a positive value of $x$. The expression that we use here is well-defined, while the integral formula is not.