I'm trying to prove that:
$$\prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}$$
whenever $a$ and $b$ are positive.
I know that
$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} = \frac{\int_0^{\infty}e^{-s}s^ads \int_0^{\infty}e^{-t}t^adt}{\int_0^{\infty}e^{-n}n^{a+b}dn}$$
but am confused as to where to proceed from here... Should I use the product formula for $1/\Gamma$ instead? Any direction would be appreciated. Thanks.
Another hint (bit more rigorous perhaps): The Beta Function can be written as \begin{equation} B(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1}, \end{equation} the right hand side of the above formula can be expanded \begin{eqnarray} B(x,y) &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1} \\ &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)+xy}{n(x+y+n)} \right)^{-1} \\ &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{xy+n(x+y)+n^{2}}\right) \\ &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{(n+x)(n+y)}\right) \end{eqnarray} Now; The Beta function has many other forms such as \begin{equation} B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \end{equation} From which a relationship between your left hand side and your right hand side could become readily found once you use properties of the gamma fuction.
First note that $x^{(n)}$, denotes the rising factorial and $$ x^{(n)}=x(x+1)(x+2)\cdots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)} $$ So now $$ \prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} $$ $$=\lim_{z\to\infty} \prod_{n=1}^{z}\frac{n(n+a+b)}{(n+a)(n+b)} $$ $$ = \lim_{z\to\infty} \frac{\left[\prod\limits_{n=1}^{z} n\right]\left[\prod\limits_{n=1}^{z} (n+a+b)\right]}{\left[\prod\limits_{n=1}^{z} (n+a)\right]\left[\prod\limits_{n=1}^{z} (n+b)\right]}$$ $$= \lim_{z\to\infty} \frac{z!(a+b+1)^{(z)}}{(a+1)^{(z)}(b+1)^{(z)}} $$ $$ = \lim_{z\to\infty} \frac{\Gamma(z+1)\frac{\Gamma (a+b+1+z)}{\Gamma(a+b+1)}}{\left(\frac{\Gamma(a+1+z)}{\Gamma(a+1)}\right)\left(\frac{\Gamma(b+1+z)}{\Gamma(b+1)}\right)}$$ $$=\lim_{z\to\infty} \frac{\Gamma(z+1)\Gamma (a+b+1+z)\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+z)\Gamma(b+1+z)\Gamma(a+b+1)} $$ $$ =\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\lim_{z\to\infty} \frac{\Gamma(z+1)\Gamma (a+b+1+z)}{\Gamma(a+1+z)\Gamma(b+1+z)}$$ $$=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} $$
We can use Weierstrass' formula for Gamma function.
Since $\Gamma(a+1) = a\Gamma(a)$,$\Gamma(b+1) = b\Gamma(b)$, $\Gamma(a + b + 1) = (a+b) \Gamma(a + b)$, then $$\frac{\Gamma(a + 1) \Gamma(b + 1)}{\Gamma(a + b + 1)} = \frac{ab \Gamma(a)\Gamma(b)}{(a+b) \Gamma(a + b)}$$Then according to Weierstrass' formula, we have $$\frac{1}{\Gamma(a + b)} = (a + b) e^{\gamma (a + b)} \prod_{n = 1}^{\infty}( (1 + \frac{a + b}{n}) e ^ {- \frac{a + b}{n}})$$ And we also have $$\Gamma(a) = \frac{1}{a e^{\gamma a} \prod_{n = 1}^{\infty} ((1 + \frac{a}{n}) e ^ {- \frac{a}{n}})}$$And similarly we have $$\Gamma(b) = \frac{1}{b e^{\gamma b} \prod_{n = 1}^{\infty} ((1 + \frac{b}{n})e^{- \frac{b}{n}})}$$ Then we have $$\begin{align} \frac{ab \Gamma(a) \Gamma(b)}{(a + b) \Gamma(a + b)} & = \frac{e^{\gamma (a + b)} \prod_{n = 1}^{\infty}( (1 + \frac{a + b}{n}) e ^ {- \frac{a + b}{n}})}{e^{\gamma (a+b)} \prod_{n = 1}^{\infty} ((1 + \frac{a}{n}) e ^ {- \frac{a}{n}} (1 + \frac{b}{n}) e^{- \frac{b}{n}})} \\ & = \prod_{n = 1}^{\infty} \frac{\frac{n + a + b}{n}}{\frac{n + a}{n} \cdot \frac{n + b}{n}} \\ & = \prod_{n = 1}^{\infty} \frac{n(n + a + b)}{(n + a) (n + b)} \end{align}$$
