How do I prove that β(m,n) =∫_0^1 {x^(m-1) +x^(n-1)} /(1+x) ^(m+n) dx?

Answer 1

To derive the beta function, we NEED to understand what it's asking for. What does B(x,y) mean? Well, it is closely related to the binomial coefficient, n choose k. More specifically, B(n,n+1)=1/[n*(2n choose n)] -- which is quite a useful relation. So we now that B(m,n) has something to do with factorials. To derive it, we start with m!n!, just for giggles. From the gamma function, we can write m!n! = integral from 0 to infinity of (e^-u)(u^m)du whole thing times integral from 0 to infinity of (e^-v)(v^n)dv. We changed from u to v because n is not necessarily equal to m, so we need a whole new integral expression. (Take a few moments to let that last sentence sink in and then move on.)

Now, products of integrals are not very fun to do, so let's see if we can re-write this as an iterated integral. In this we have the double integral from 0 to infinity of (e^-u)(u^m)*(e^-v)(v^n)dudv. Few things about this: we have an e^something times e^something else. How nice would it be if we could just express it as e^one thing? Quite nice! So let's do that:

=double integral from 0 to infinity of [e^(-u-v)](u^m)*(v^n)dudv.

At this point, you see how we have an e^something plus something? We know from polar coordinates that e and squared trigonometric bois that add each other go along REAL well! So let u=x^2 and v=y^2 and then we'll translate it into polar coordinates. But first, let's simplify this integral as much as we can with our new substitutions:

=double integral from -infinity to infinity of [e^-(x^2 + y^2)] |x^(2m+1)| * |y^(2n+1)| dxdy. (You'll hate me, but the simplification is trivial and left as an exercise lol. I've always wanted to do that. If you're confused, hmu.) Btw, the x^thingy and y^thingy terms are in absolute value because we defined u and v in terms of x^2 and y^2, so we end up with x = sqrt(x^2) and y = sqrt(y^2), which is a common definition of absolute value(|n| = sqrt(n^2)).

Now we can take this into polar coordinates!

=integral from 0 to 2pi of [integral from 0 to infinity of [e^-r^2][ |rcos(theta)|^(2m+1) * |rsin(theta)|^(2n+1) * r*dr*d(theta)]. (Again, the proof is trivial and left as an exercise *wink wink*).

Now we can separate the r's and the theta's like such:

[integral from 0 to infinity of e^(-r^2)*r*r^(2n+1)*r^(2m+1)]*[integral from 0 to 2pi of |cos(theta)^(2m+1)*sin(theta)^(2n+1)| d(theta)]

Little trick to simplify this bad boi: think of |sin(x)| in terms of the unit circle. |sin(pi/4)| happens at pi/4, and 3pi/4, and since |sin(x)| can't be negative, it repeats itself in the third and fourth quadrants for 5pi/4 and 7pi/4. So, in total, we have |sin(pi/4)| = 4*sin(pi/4),, where the pi/4 lies between 0 and pi/2. Let's apply that to get rid of our absolute values:

= [integral from 0 to infinity of e^(-r^2)*r*r^(2n+1)*r^(2m+1)]*[4 * integral from 0 to pi/2 of cos(theta)^(2m+1)*sin(theta)^(2n+1) d(theta)]

Now, from the gamma function, we now what the dr integral is in terms of factorials. (trivial and left as an exercise, please don't hate me):

= [2*(m+n+1)!] * [integral from 0 to pi//2 of cos(theta)^(2m+1)*sin(theta)^(2n+1) d(theta)], which is a definition of B(n+1, m+1).

Now, to answer your question, we can let m = m+1 and n = n+1 to find B(n,m) = [2*(m+n-1)!] * [integral from 0 to pi//2 of cos(theta)^(2(m-1)+1)*sin(theta)^(2(n-1)+1) d(theta)].

Here's your exercise: back-substitute x and y into this equation(x = rcos(theta), y = rsin(theta)), finesse this boi by simplification, and you'll see how the question is true.