Prove $\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma(c)\Gamma(d) }{ \Gamma(a)\Gamma(b)}$

Question

How to prove the following identity?

$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}\tag{$a,b,c,d \in \mathbb R$}=\frac{\Gamma(c)\Gamma(d) }{ \Gamma(a)\Gamma(b)}$$

For $a+b=c+d$.

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The product can be rewritten as : $$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\prod_{n\ge0}^{ }\frac{n^{2}+\left(a+b\right)n+ab}{n^{2}+\left(c+d\right)n+cd}$$$$=\prod_{n\ge0}^{ }\frac{n^{2}+\left(a+b\right)n+ab}{n^{2}+\left(a+b\right)n+cd}=\prod_{n\ge0}^{ }\left(1+\frac{ab-cd}{n^{2}+\left(a+b\right)n+cd}\right)$$

But I think this is useless.

Or we can say:

$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}$$$$=\lim_{N \to \infty} \frac{ab}{cd}\cdot\frac{\left(a+1\right)\left(b+1\right)}{\left(c+1\right)\left(d+1\right)}\cdot...\cdot\frac{\left(a+N\right)\left(b+N\right)}{\left(c+N\right)\left(d+N\right)}$$$$=\lim_{N \to \infty}\frac{\left(a+N\right)!\left(b+N\right)!}{\left(c+N\right)!\left(d+N\right)!}\cdot\frac{\left(c-1\right)!\left(d-1\right)!}{\left(a-1\right)!\left(b-1\right)!}$$$$=\frac{ \Gamma(c)\Gamma(d)}{ \Gamma(a)\Gamma(b)}\lim_{N \to \infty}\frac{\left(a+N\right)!\left(b+N\right)!}{\left(c+N\right)!\left(d+N\right)!}$$

How to finish?

Answer 1

You can finish the proof by using the asymptotics $$ \mathop {\lim }\limits_{N \to + \infty } \frac{{(N + a)!}}{{N^{N + a + 1/2} e^{ - N} \sqrt {2\pi } }} = 1, $$ which can be derived from Stirling's formula ($a$ is an arbitrary fixed complex number).