Evaluate the product:

$$\Gamma \left ( \frac{1}{n} \right )\Gamma \left ( \frac{2}{n} \right )\cdots \Gamma \left ( \frac{n-1}{n} \right )$$

A path to a solution: $$\begin{aligned}\Gamma\left (\frac{1}{m}\right )\Gamma\left (\frac{2}{m}\right )\cdots\Gamma\left (\frac{m-2}{m}\right )\Gamma\left (\frac{m-1}{m}\right ) &=\\ \Gamma\left (\frac{1}{m}\right )\Gamma\left (\frac{2}{m}\right )\cdots\Gamma\left ( 1-\frac{2}{m}\right )\Gamma\left ( 1-\frac{1}{m}\right ) &=\\ \Gamma\left (\frac{m}{2}\right )\cdot\prod_{i=1}^{m-1}\Gamma\left (\frac{i}{m}\right )\Gamma\left ( 1-\frac{i}{m}\right ) &=\\ \Gamma\left (\frac{m}{2}\right )\prod_{i=1}^{m-1}\frac{\pi}{\sin\frac{\pi i}{m}}&\end{aligned}$$

However, since $\displaystyle \Gamma \left(\frac{m}{2} \right) $ is quite known , that product is a bonner... I cannot do anything with that.

By the way, Gauss multiplication formula does not seem to work since we have to plug in $z=0$. Doing so, we get $\Gamma(0)$ which clearly is not defined but $\displaystyle \lim_{x \rightarrow 0+}\Gamma(x)=+\infty$.

Any other ideas? Can we evaluate that product?


1 Answer 1


Gauss multiplication formula works, indeed. If we take $z=\frac{1}{n}$ in: $$\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right) = (2\pi)^{\frac{n-1}{2}} \,n^{\frac{1}{2}-nz}\,\Gamma(nz)\tag{1}$$ we get: $$ \prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right) = (2\pi)^{\frac{n-1}{2}} \,n^{-\frac{1}{2}}=\frac{(\sqrt{2\pi})^n}{\sqrt{2\pi n}}.\tag{2} $$


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