Barnes G-function

Remarks

The product converges uniformly on compact sets. The function $G$ satisfies $G(1) = 1$ and $$G(z + 1) = \Gamma(z)\, G(z)$$ for all $z$, where $\Gamma$ is the gamma function. At a pole of $\Gamma$ the right side is read as a limit, since $G$ has a zero there.

Example: the recurrence gives $G(2) = \Gamma(1) G(1) = 1$, $G(3) = \Gamma(2) G(2) = 1$, $G(4) = \Gamma(3) G(3) = 2$ and $G(5) = \Gamma(4) G(4) = 12$. In general $G(n) = \prod_{k=0}^{n-2} k!$ for integers $n \ge 2$.

Non-example: the function $F(z) = G(z) \cdot (1 + \sin^2 \pi z)$ also satisfies $F(1) = 1$ and $F(z+1) = \Gamma(z) F(z)$, but it is not $G$. So the recurrence alone does not determine $G$; one needs an extra condition, for example that $\frac{d^3}{dx^3} \log G(x) \ge 0$ for $x > 0$.

From the product, $G$ has zeros exactly at $z = -n$ for integers $n \ge 0$, of order $n + 1$.