# Showing that the Barnes G-function satisfies the functional equation $G(z+1) = \Gamma(z) G(z)$

Let $G(z)$ be the Barnes G-function.

I want to use the infinite product representation

$$G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z(z+1)}{2}- \frac{\gamma z^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{k}\text{exp}\left(\frac{z^2}{2k}-z\right)$$

to show that the Barnes G-function satisfies the functional equation

$$G(z+1) = \Gamma(z) G(z).$$

Specifically, I want to show that $$\frac{G(z+1)}{G(z)} = \frac{e^{-\gamma z}}{z} \prod_{k=1}^{\infty} \frac{e^{\frac{z}{k}}}{1+\frac{z}{k}}$$ where the right side of the equation is the Weierstrass infinite product representation of the gamma function.

Obviously,

\begin{align} \frac{G(z+1)}{G(z)} &= \frac{(2\pi)^{z/2}\text{exp}\left(-\frac{z(z+1)}{2}- \frac{\gamma z^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{k}\text{exp}\left(\frac{z^2}{2k}-z\right)}{(2\pi)^{(z-1)/2}\text{exp}\left(-\frac{(z-1)z}{2}- \frac{\gamma (z-1)^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z-1}{k}\right)^{k}\text{exp}\left(\frac{(z-1)^2}{2k}-(z-1)\right)} \\ &= \sqrt{2 \pi} \exp \left(-z - \gamma z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( \frac{2z-1-2k}{2k} \right). \end{align}

But it's not clear to me what to do next.

• Functional equation only does not fix the Barnes function. For example, $G(z)\sin2\pi z$ will satisfy the same functional equation. – Start wearing purple Oct 1 '13 at 12:32
• I actually knew that. The question was badly worded. I just want to show that it satisfies the functional equation. But that alone seems to be very difficult to show. After doing a bunch of searching, it seems to be shown nowhere except in Barnes original papers. – Random Variable Oct 1 '13 at 15:53
• Concerning your last question: it depends on how do you define $G(z)$ itself. I never liked this convexity thing. What is really important about $G(z)$ is not the convexity, but the set of poles/zeros which is, in a sense, minimal. Another way to introduce $G(z)$ is the integral representation. Yet another thing is that $G(z)$ is a limiting case of a much nicer and rigid function $\Gamma_q(z)$ in the case where the two periods of the latter coincide. – Start wearing purple Oct 4 '13 at 19:00
• Are you referring to the above integral representation or some other integral representation? I derived that by using the infinite product representation. So in a sense I feel like I'm going in circles. – Random Variable Oct 4 '13 at 19:38
• No, for me the defining integral representation of the Barnes $G$-function would be $$G(1+z)= \left(2\pi\right)^{\frac{z}{2}} \exp \int_0^{\infty} \frac{ds}{s} \Biggl[ \frac{1-e^{-zs}}{ 4\sinh^2 \frac{s}{2}} - \frac{z}{s}+ \frac{z^2}{2} \,e^{-s} \Biggr]$$ – Start wearing purple Oct 5 '13 at 0:00

I will start from your last expression. The idea is, roughly, to write $(k+z-1)^k$ in the denominator as $(k+z-1)^{k-1}\cdot (k+z-1)$, then to replace $k-1$ by $k'$ and observe that the product telescopically simplifies. Also, to avoid mistakes, it is better to start with a finite product.

We have \begin{align} \prod_{k=1}^N\left(\frac{k+z}{k-1+z}\right)^k \exp\frac{2z-1-2k}{2k}=\frac{\prod_{k=1}^N(k+z)^k\exp\frac{2z-1-2k}{2k}}{\prod_{k=1}^N(k-1+z)^{k-1}(k-1+z)}=\\ =\frac{\prod_{k=1}^{N}(k+z)^k\exp\frac{2z-1-2k}{2k}}{\prod_{k=0}^{N-1}(k+z)^{k}\prod_{k=0}^{N-1}(k+z)}=\\ =\frac{(N+z)^N\prod_{k=1}^{N}\exp\frac{2z-1-2k}{2k}}{z\prod_{k=1}^{N-1}(k+z)}=\\ =\frac{N^N\left(1+\frac{z}{N}\right)^N\prod_{k=1}^{N}\exp\frac{2z-1-2k}{2k}}{\frac{z}{N+z}\prod_{k=1}^{N}k(1+\frac{z}{k})}=\\ =\frac{(N/e)^N}{N!}\times\left(1+\frac{z}{N}\right)^N\times\frac{N+z}{z}\times\prod_{k=1}^{N}\frac{e^{z/k}}{1+z/k}\times \exp\left\{-\frac12\sum_{k=1}^N\frac1k\right\}.\tag{1} \end{align} Now consider the asymptotics of different factors in the last expression as $N\rightarrow\infty$:

• From Stirling's formula we get $$\frac{(N/e)^N}{N!}=\frac{1}{\sqrt{2\pi N}}+O\left(N^{-3/2}\right).$$

• $\left(1+\frac{z}{N}\right)^N$ tends to $e^z$.

• Using product representation of the gamma function, in the limit we can replace $\prod_{k=1}^{N}\frac{e^{z/k}}{1+z/k}$ by $ze^{\gamma z}\Gamma(z)$.

• Harmonic series is known to behave as $$\sum_{k=1}^N\frac1k=\ln N+\gamma+O\left(N^{-1}\right)$$

Therefore, the limit of (1) gives $$\frac{1}{\sqrt{2\pi N}}\times e^z\times N\,e^{\gamma z}\Gamma(z)\times \frac{e^{-\gamma/2}}{\sqrt{N}}\sim\frac{\exp\left\{z(\gamma+1)-\frac{\gamma}{2}\right\}}{\sqrt{2\pi}}\Gamma(z).$$ This is equivalent to the functional relation you want to prove.