Explicit form Gamma function

Question

For some time now I have been stuck evaluating products of the form $$\frac{\Gamma(l+1+ix)}{\Gamma(l+1-ix)}\frac{\Gamma(1-ix)}{\Gamma(1+ix)}, $$ where $l\in\mathbb{N}_0 $. I really tried all possible combinations of formulas in the wikipedia page for the Gamma function to try to simplify this. I expect there is some $l$ dependence on the result. Does anyone have an idea/hint on how to proceed? Thank you!

Answer 1

Repeatedly use the recursive formula for the gamma function $\Gamma(x+1)=x \Gamma(x)$ for both the denominator and the numerator terms. You'll get a cancellation of terms with $\Gamma(\cdot)$.

$\frac{\Gamma(l+1+ix)\Gamma(1-ix)}{\Gamma(l+1-ix)\Gamma(1+ix)}$= $\frac{\Pi_{k=1}^l(l-k+1+ix)\Gamma(1+ix) \Gamma(1-ix)}{\Pi_{k=1}^l(l-k+1-ix)\Gamma(1+ix) \Gamma(1+ix)}$=$\frac{\Pi_{k=1}^l(l-k+1+ix)}{\Pi_{k=1}^l(l-k+1-ix)}$