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It is know that $\Gamma(z)\, \Gamma{(1-z)}=\pi \csc( \pi z)$. Is there any formula for $\Gamma{(a+z)}\Gamma{(a-z)}$ where $a$ is a rational number, i. e., $a=p/k$ with $p, k$ integers and $z$ is a complex number ?

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But it does not look to me as a generalization because there is no $a$ that you can plug into $\Gamma{(a+z)}\Gamma{(a-z)}$ to get $\Gamma(z)\, \Gamma{(1-z)}$? –  user67878 Mar 27 '13 at 15:20
    
But try to explore what you have here : Gamma function as a starting point. –  user67878 Mar 27 '13 at 15:28
    
For a=0 or a=1 the expression $\Gamma{(a+z)} \Gamma{(a-z)} $ can be expressed in terms of the reflection formula above. But what about $a \neq 0$ or $a\neq 1$ ? –  ice Mar 27 '13 at 15:52

1 Answer 1

For $a\in\mathbb Z>0$ we have $\Gamma(a+z)\Gamma(a-z)=-\dfrac{\pi}{z\sin(\pi z)}~\cdot~\displaystyle\prod_{k=0}^{a-1}(k^2-z^2)$.

For $a\in\mathbb Z<0$ we have $\Gamma(a+z)\Gamma(a-z)=-\dfrac{\pi}{z\sin(\pi z)}\bigg/\,\displaystyle\prod_{k=a}^{-1}(k^2-z^2).$

This is obvious from the fundamental property of the $\Gamma$ function, $\Gamma(z+1)=z\,\Gamma(z)$, combined with the afore-mentioned reflection formula. But for $a\in\mathbb Q\setminus\mathbb Z$ no such general formula is known, except for $a=k+\dfrac12$ , in which case we use the fact that $\Gamma\bigg(\dfrac12+z\bigg)\Gamma\bigg(\dfrac12-z\bigg)=\dfrac\pi{\cos\pi z}$ as a starting point, to arrive at two other formulas, very similar to the ones presented above.

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