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

Question

I'm trying to proof the equality $\operatorname B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ when $x,y>0,$ without using calculus in many variables. I've investigated about the topic but all references make use of Fubini Theorem. Anyone knows a proof or reference using calculus in single variable?

I'd appreciate any kind of help.

Answer 1

You can try the following

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$

Substitution now:

$$u:=\frac t{1-t}=\frac1{1-t}-1\implies du=\frac{dt}{(1-t)^2}$$

so we get

$$B(x,y)=\int_0^\infty \frac{u^{x-1}}{(u+1)^{x-1}}\frac1{(u+1)^{y-1}} (u+1)^2du=\int_0^\infty\frac{u^{x-1}}{(u+1)^{x+y}}du$$

Now, we have

$$\Gamma(x):=\int_0^\infty t^{x-1}e^{-t}dt\stackrel{nu=t\implies ndu=dt}=n^x\int_0^\infty u^{x-1}e^{-un}du$$

Now put $\;n=s+1\;,\;\;x=w+z\;$ , so that we have:

$$\frac1{(s+1)^{w+z}}=\frac1{\Gamma(w+z)}\int_0^\infty u^{w+z-1}e^{-(s+1)u}du$$

Well, now substitute all this mess in the new expression we got for the Beta function above...