I need to simplify Division of two gamma functions whose inputs differ by a positive real number as follow:
$\dfrac{\Gamma(x+y)}{\Gamma(y)},\qquad x,y\in (0,+\infty)$
If $x$ and $y$ were positive integers and greater than one, I would use the equality $\Gamma(y)=(y-1)\Gamma(y-1)$ and simplify the functin. But they can be every positive real number. Also, I tried to simplify the expression by using of the q-gamma function, or basic gamma function as follow:
$\mathop {\lim }\limits_{q \to {1^ + }}\dfrac{\Gamma(x+y)}{\Gamma(y)}= \mathop {\lim }\limits_{q \to {1^ + }} \frac{{{\Gamma _q}(x + y)}}{{{\Gamma _q}(y)}}$
$= \mathop {\lim }\limits_{q \to {1^ + }} \frac{{{{(1 - q)}^{1 - (x + y)}}}}{{{{(1 - q)}^{1 - y}}}} \times \frac{{\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + 1}}}}{{1 - {q^{i + x + y}}}}} }}{{\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + 1}}}}{{1 - {q^{i + y}}}}} }}$
$= \mathop {\lim }\limits_{q \to {1^ + }} {(1 - q)^{ - x}}\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + y}}}}{{1 - {q^{i + x + y}}}}} .$
But I could not be successful. Please, somebody help me!!!
You could call it $\text{pochhammer}(y,x)$.