# Barnes G-function

In mathematics, the Barnes G-function G(z) is a function that is an extension of superfactorials to the complex numbers. It is related to the gamma function, the K-function and the Glaisher–Kinkelin constant, and was named after mathematician Ernest William Barnes. It can be written in terms of the double gamma function.
Formally, the Barnes G-function is defined in the following Weierstrass product form:
    $G(1+z)=(2\pi )^{z/2}\exp \left(-{\frac {z+z^{2}(1+\gamma )}{2}}\right)\,\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)^{k}\exp \left({\frac {z^{2}}{2k}}-z\right)\right\}$
where $\,\gamma$ is the Euler–Mascheroni constant, exp(x) = e$^x$ is the exponential function, and Π denotes multiplication (capital pi notation).
The integral representation, which may be deduced from the relation to the double gamma function, is
    $\log G(1+z)={\frac {z}{2}}\log(2\pi )+\int _{0}^{\infty }{\frac {dt}{t}}\left[{\frac {1-e^{-zt}}{4\sinh ^{2}{\frac {t}{2}}}}+{\frac {z^{2}}{2}}e^{-t}-{\frac {z}{t}}\right]$
As an entire function, G is of order two, and of infinite type. This can be deduced from the asymptotic expansion given below.

## Functional equation and integer arguments

The Barnes G-function satisfies the functional equation
    $G(z+1)=\Gamma (z)\,G(z)$
with normalisation G(1) = 1. Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function:
    $\Gamma (z+1)=z\,\Gamma (z).$
The functional equation implies that G takes the following values at integer arguments:
    $G(n)={\begin{cases}0&{\text{if }}n=0,-1,-2,\dots \\\prod _{i=0}^{n-2}i!&{\text{if }}n=1,2,\dots \end{cases}}$
(in particular, $\,G(0)=0,G(1)=1$) and thus
    $G(n)={\frac {(\Gamma (n))^{n-1}}{K(n)}}$
where $\,\Gamma (x)$ denotes the gamma function and K denotes the K-function. The functional equation uniquely defines the Barnes G-function if the convexity condition,
    $(\forall x\geq 1)\,{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}\log(G(x))\geq 0$
is added. Additionally, the Barnes G-function satisfies the duplication formula,
    $G(x)G\left(x+{\frac {1}{2}}\right)^{2}G(x+1)=e^{\frac {1}{4}}A^{-3}2^{-2x^{2}+3x-{\frac {11}{12}}}\pi ^{x-{\frac {1}{2}}}G\left(2x\right)$,
where $A$ is the Glaisher–Kinkelin constant.

## Characterisation

Similar to the Bohr–Mollerup theorem for the gamma function, for a constant $c>0$, we have for $f(x)=cG(x)$
$f(x+1)=\Gamma (x)f(x)$
and for $x>0$
$f(x+n)\sim \Gamma (x)^{n}n^{x \choose 2}f(n)$
as $n\to \infty$.

## Reflection formula

The difference equation for the G-function, in conjunction with the functional equation for the gamma function, can be used to obtain the following reflection formula for the Barnes G-function (originally proved by Hermann Kinkelin):
    $\log G(1-z)=\log G(1+z)-z\log 2\pi +\int _{0}^{z}\pi x\cot \pi x\,dx.$
The log-tangent integral on the right-hand side can be evaluated in terms of the Clausen function (of order 2), as is shown below:
    $2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right)+\operatorname {Cl} _{2}(2\pi z)$
The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation $\operatorname {Lc} (z)$ for the log-cotangent integral, and using the fact that $\,(d/dx)\log(\sin \pi x)=\pi \cot \pi x$, an integration by parts gives
    ${\begin{aligned}\operatorname {Lc} (z)&=\int _{0}^{z}\pi x\cot \pi x\,dx\\&=z\log(\sin \pi z)-\int _{0}^{z}\log(\sin \pi x)\,dx\\&=z\log(\sin \pi z)-\int _{0}^{z}{\Bigg\log(2\sin \pi x)-\log 2{\Bigg ]}\,dx\\&=z\log(2\sin \pi z)-\int _{0}^{z}\log(2\sin \pi x)\,dx.\end{aligned}}$
Performing the integral substitution $\,y=2\pi x\Rightarrow dx=dy/(2\pi )$ gives
    $z\log(2\sin \pi z)-{\frac {1}{2\pi }}\int _{0}^{2\pi z}\log \left(2\sin {\frac {y}{2}}\right)\,dy.$
The Clausen function – of second order – has the integral representation
    $\operatorname {Cl} _{2}(\theta )=-\int _{0}^{\theta }\log {\Bigg |}2\sin {\frac {x}{2}}{\Bigg |}\,dx.$
However, within the interval $\,0<\theta <2\pi$, the absolute value sign within the integrand can be omitted, since within the range the 'half-sine' function in the integral is strictly positive, and strictly non-zero. Comparing this definition with the result above for the logtangent integral, the following relation clearly holds:
    $\operatorname {Lc} (z)=z\log(2\sin \pi z)+{\frac {1}{2\pi }}\operatorname {Cl} _{2}(2\pi z).$
Thus, after a slight rearrangement of terms, the proof is complete:
    $2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right)+\operatorname {Cl} _{2}(2\pi z)\,.\,\Box$
Using the relation $\,G(1+z)=\Gamma (z)\,G(z)$ and dividing the reflection formula by a factor of $\,2\pi$ gives the equivalent form:
    $\log \left({\frac {G(1-z)}{G(z)}}\right)=z\log \left({\frac {\sin \pi z}{\pi }}\right)+\log \Gamma (z)+{\frac {1}{2\pi }}\operatorname {Cl} _{2}(2\pi z)$
Adamchik (2003) has given an equivalent form of the reflection formula, but with a different proof.
Replacing z with ⁠1/2⁠ − z in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials):
    $\log \left({\frac {G\left({\frac {1}{2}}+z\right)}{G\left({\frac {1}{2}}-z\right)}}\right)=\log \Gamma \left({\frac {1}{2}}-z\right)+B_{1}(z)\log 2\pi +{\frac {1}{2}}\log 2+\pi \int _{0}^{z}B_{1}(x)\tan \pi x\,dx$

## Taylor series expansion

By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained:
    $\log G(1+z)={\frac {z}{2}}\log 2\pi -\left({\frac {z+(1+\gamma )z^{2}}{2}}\right)+\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}.$
It is valid for $\,0<z<1$. Here, $\,\zeta (x)$ is the Riemann zeta function:
    $\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.$
Exponentiating both sides of the Taylor expansion gives:
    ${\begin{aligned}G(1+z)&=\exp \left[{\frac {z}{2}}\log 2\pi -\left({\frac {z+(1+\gamma )z^{2}}{2}}\right)+\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right]\\&=(2\pi )^{z/2}\exp \left[-{\frac {z+(1+\gamma )z^{2}}{2}}\right]\exp \left[\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right].\end{aligned}}$
Comparing this with the Weierstrass product form of the Barnes function gives the following relation:
    $\exp \left[\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right]=\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)^{k}\exp \left({\frac {z^{2}}{2k}}-z\right)\right\}$

## Multiplication formula

Like the gamma function, the G-function also has a multiplication formula:
    $G(nz)=K(n)n^{n^{2}z^{2}/2-nz}(2\pi )^{-{\frac {n^{2}-n}{2}}z}\prod _{i=0}^{n-1}\prod _{j=0}^{n-1}G\left(z+{\frac {i+j}{n}}\right)$
where $K(n)$ is a constant given by:
    $K(n)=e^{-(n^{2}-1)\zeta ^{\prime }(-1)}\cdot n^{\frac {5}{12}}\cdot (2\pi )^{(n-1)/2}\,=\,(Ae^{-{\frac {1}{12}}})^{n^{2}-1}\cdot n^{\frac {5}{12}}\cdot (2\pi )^{(n-1)/2}.$
Here $\zeta ^{\prime }$ is the derivative of the Riemann zeta function and $A$ is the Glaisher–Kinkelin constant.

## Absolute value

It holds true that $G({\overline {z}})={\overline {G(z)}}$, thus $|G(z)|^{2}=G(z)G({\overline {z}})$. From this relation and by the above presented Weierstrass product form one can show that
    $|G(x+iy)|=|G(x)|\exp \left(y^{2}{\frac {1+\gamma }{2}}\right){\sqrt {1+{\frac {y^{2}}{x^{2}}}}}{\sqrt {\prod _{k=1}^{\infty }\left(1+{\frac {y^{2}}{(x+k)^{2}}}\right)^{k+1}\exp \left(-{\frac {y^{2}}{k}}\right)}}.$
This relation is valid for arbitrary $x\in \mathbb {R} \setminus \{0,-1,-2,\dots \}$, and $y\in \mathbb {R}$. If $x=0$, then the below formula is valid instead:
    $|G(iy)|=y\exp \left(y^{2}{\frac {1+\gamma }{2}}\right){\sqrt {\prod _{k=1}^{\infty }\left(1+{\frac {y^{2}}{k^{2}}}\right)^{k+1}\exp \left(-{\frac {y^{2}}{k}}\right)}}$
for arbitrary real y.

## Asymptotic expansion

The logarithm of G(z + 1) has the following asymptotic expansion, as established by Barnes:
    ${\begin{aligned}\log G(z+1)=&{\frac {z^{2}}{2}}\log z-{\frac {3z^{2}}{4}}+{\frac {z}{2}}\log 2\pi -{\frac {1}{12}}\log z\\&+\left({\frac {1}{12}}-\log A\right)+\sum _{k=1}^{N}{\frac {B_{2k+2}}{4k\left(k+1\right)z^{2k}}}~+~O\left({\frac {1}{z^{2N+2}}}\right).\end{aligned}}$
Here the $B_{k}$ are the Bernoulli numbers and $A$ is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes the Bernoulli number $B_{2k}$ would have been written as $(-1)^{k+1}B_{k}$, but this convention is no longer current.) This expansion is valid for $z$ in any sector not containing the negative real axis with $|z|$ large.

## Relation to the log-gamma integral

The parametric log-gamma can be evaluated in terms of the Barnes G-function:
    $\int _{0}^{z}\log \Gamma (x)\,dx={\frac {z(1-z)}{2}}+{\frac {z}{2}}\log 2\pi +z\log \Gamma (z)-\log G(1+z)$