Simplify ${\Gamma(\beta)\Gamma(u) \over \Gamma(\beta+u)}$

Question

Is it possible that the expression below be simplified to a simpler form:

$${\Gamma(\beta)\Gamma(u) \over \Gamma(\beta+u)}$$

whereby $\beta$ is a variable

Answer 1

This is the Beta function - not because you have an argument denoted with beta i.e. $β$ - but because it is called so. One form of the Beta function is the one you posted $$B(β,u):=\dfrac{\Gamma(β)\,\Gamma(u)}{\Gamma(β+u)} $$

whereas in the link that I gave you can see its general definition and its properties. If the arguments $β,u$ are for example integers then, yes it can be furthered simplified.

Answer 2

Hint: For $\;\text{Re}(a)\,,\,\,\text{Re}(b)>0$ :

$$\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(a,b):=\int_0^1x^{a-1}(1-x)^{b-1}dx=2\int_0^{\pi/2}\sin^{2a-1}(t)\;\cos^{2b-1}(t)\;dt\;$$

Answer 3

Well, that's basically the beta function.
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $\beta(m,n)$=$\dfrac{\Gamma(n).\Gamma(m)}{\Gamma(m+n)}$=$\beta(n,m)$

$\beta(\beta_1,u)=\dfrac{\Gamma(\beta_1).\Gamma(u)}{\Gamma(\beta_1+u)}$

If $\beta_1,u$ are integers then,

The above expression is simply, $\dfrac{(\beta_1-1)!*(u-1)!}{(\beta_1+u-1)!}$ (as far as I remember!)