# Simplifying expression containing Gamma

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?

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.

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] *)

• Even Product[Gamma[k/n + z/n], {k, 0, n - 1}] works in version 12.0. Commented Jul 13, 2020 at 16:26
• +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] & Commented Jul 13, 2020 at 16:26
• @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. Commented Jul 13, 2020 at 16:30