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According to this site this expression involving Gamma is valid:

$$\prod_{k=0}^{n-1}\Gamma\left(\dfrac{k+z}{n}\right) = n^{\frac{1}{2}-z}(2\pi)^{\frac{n-1}{2}}\Gamma(z)$$

However it seems that Mathematica cannot simplify it (at least not version 10.0). Is thre a way to enforce it?

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2 Answers 2

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With version 12.1.0, for any positive integer value of $\,n\,$ you can use FunctionExpand to verify with

ex[n_] := Product[Gamma[(k - 1 + z)/n], {k, n}] ==
   Gamma[z](2Pi)^(n/2 - 1/2)n^(1/2 - z) // FunctionExpand

For example, ex[3] returns True. However, using ex[n] does not evaluate to a truth value.

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Workaround:

$$\Gamma \left(\frac{k}{n}+\frac{z}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+\frac{z}{n}+1\right)}{\frac{k}{n}+\frac{z}{n}}$$

$Version
(* 12.1.1 for Microsoft Windows (64-bit) (June 9, 2020) *)

Product[Gamma[k/n + z/n + 1]/(k/n + z/n), {k, 0, n - 1}] // FunctionExpand

(* n^(1/2 - z) (2 \[Pi])^(-(1/2) + n/2) Gamma[z] *)
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    $\begingroup$ Even Product[Gamma[k/n + z/n], {k, 0, n - 1}] works in version 12.0. $\endgroup$
    – user64494
    Commented Jul 13, 2020 at 16:26
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    $\begingroup$ +1 Working from the original form: Product[Gamma[(k + z)/n], {k, 0, n - 1}] /. Gamma[x_] :> Gamma[x + 1]/x // FunctionExpand // Simplify[#, n > 0] & $\endgroup$
    – Bob Hanlon
    Commented Jul 13, 2020 at 16:26
  • $\begingroup$ @user64494.If: Product[Gamma[(k + z)/n], {k, 0, n - 1}] dosen't work, but: Product[Gamma[(k + z)/n] // ExpandAll, {k, 0, n - 1}] works fine. $\endgroup$ Commented Jul 13, 2020 at 16:30

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