Closed form for the product $G(x)G(-x)$ of two Barnes functions Is there a "closed form" expression for the following product of two Barnes $G$-functions, $$G(x)G(-x),$$ where $x$ is real?
Plotting the graph I have noticed that for $-1<x<1$ we have $$\left|G(x)G(-x)-\frac{\cos(2\pi x)-1}{a}\right|<b,$$ where $a$ is some real ($a=19.476$ works quite well) and $b$ is "reasonably" small.
 A: Let me recall that the main property characterizing Barnes $G$-function is $G(x+1)=\Gamma(x)G(x)$. Now if we introduce the combinations
\begin{align}
&P(x)=G(1+x)G(1-x),\\
&R(x)=\frac{G(1+x)}{G(1-x)},
\end{align}
the corresponding functional equations will have the form
\begin{align}
&P(x+1)=\frac{\Gamma(x+1)}{\Gamma(-x)}P(x),\\
&R(x+1)=-\frac{\pi}{\sin\pi x}R(x).
\end{align}
We see that the functional equation for the symmetrized product does not simplify as compared to $G(x)$ but the equation for the ratio does.
The same can be seen at the level of integral representations:
\begin{align}
\ln G(x+1)&=\frac{x(1-x)}{2}+x\ln\sqrt{2\pi}+\int_0^{x}t\psi(t)dt,\\
\ln P(x)&=x(1-x)+\int_0^{x}t[\psi(t)+\psi(-t)]dt,\\
\ln R(x)&=x\ln 2\pi-\int_0^x\pi t\cot\pi t\,dt.
\end{align} 
In other words, the answer to your question is no. The product $P(x)=-\frac{\pi}{x\sin\pi x}G(x)G(-x)$ is a function of the same complexity as $G(x)$. However, the ratio is (a bit) simpler and can be written in terms of dilogarithms.
