Barnes G-function

In mathematics, the Barnes G-function ${\displaystyle G(z)}$ 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.^[1] It can be written in terms of the double gamma function.

Plot of the Barnes G aka double gamma function G(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D

The Barnes G function along part of the real axis

Formally, the Barnes G-function is defined in the following Weierstrass product form:^[2]

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

where ${\displaystyle \,\gamma }$ is the Euler–Mascheroni constant, exp(x) = e^x is the exponential function, and ${\displaystyle \Pi }$ denotes multiplication (capital pi notation).

The integral representation, which may be deduced from the relation to the double gamma function, is

${\displaystyle \log G(1+z)={\frac {z}{2}}\log(2\pi )+\int _{0}^{\infty }{\frac {dt}{t}}\left[{\frac {1-e^{-zt}}{4\sinh ^{2}{\frac {t}{2}}}}+{\frac {z^{2}}{2}}e^{-t}-{\frac {z}{t}}\right]}$

As an entire function, ${\displaystyle G}$ 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

${\displaystyle G(z+1)=\Gamma (z)\,G(z)}$

with normalization ${\displaystyle G(1)=1}$. Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function:

${\displaystyle \Gamma (z+1)=z\,\Gamma (z).}$

The functional equation implies that ${\displaystyle G}$ takes the following values at integer arguments:

${\displaystyle G(n)={\begin{cases}0&{\text{if }}n=0,-1,-2,\dots \\\prod _{i=0}^{n-2}i!&{\text{if }}n=1,2,\dots \end{cases}}}$

(in particular, ${\displaystyle \,G(0)=0,G(1)=1}$) and thus

${\displaystyle G(n)={\frac {(\Gamma (n))^{n-1}}{K(n)}}}$

where ${\displaystyle \,\Gamma (x)}$ denotes the gamma function and ${\displaystyle K}$ denotes the K-function. The functional equation uniquely defines the Barnes G-function if the convexity condition,

${\displaystyle (\forall x\geq 1)\,{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}\log(G(x))\geq 0}$

is added.^[3] Additionally, the Barnes G-function satisfies the duplication formula,^[4]

${\displaystyle G(x)G\left(x+{\frac {1}{2}}\right)^{2}G(x+1)=e^{\frac {1}{4}}A^{-3}2^{-2x^{2}+3x-{\frac {11}{12}}}\pi ^{x-{\frac {1}{2}}}G\left(2x\right)}$,

where ${\displaystyle A}$ is the Glaisher–Kinkelin constant.

Characterisation

Similar to the Bohr–Mollerup theorem for the gamma function, for a constant ${\displaystyle c>0}$we have for ${\displaystyle f(x)=cG(x)}$^[5]

${\displaystyle f(x+1)=\Gamma (x)f(x)}$

and for ${\displaystyle x>0}$

${\displaystyle f(x+n)\sim \Gamma (x)^{n}n^{x \choose 2}f(n)}$

as ${\displaystyle n\to \infty }$.

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):

${\displaystyle \log G(1-z)=\log G(1+z)-z\log 2\pi +\int _{0}^{z}\pi x\cot \pi x\,dx.}$

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]

${\displaystyle 2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right)+\operatorname {Cl} _{2}(2\pi z)}$

The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation ${\displaystyle \operatorname {Lc} (z)}$ for the log-cotangent integral, and using the fact that ${\displaystyle \,(d/dx)\log(\sin \pi x)=\pi \cot \pi x}$, an integration by parts gives

${\displaystyle {\begin{aligned}\operatorname {Lc} (z)&=\int _{0}^{z}\pi x\cot \pi x\,dx\\&=z\log(\sin \pi z)-\int _{0}^{z}\log(\sin \pi x)\,dx\\&=z\log(\sin \pi z)-\int _{0}^{z}{\Bigg [}\log(2\sin \pi x)-\log 2{\Bigg ]}\,dx\\&=z\log(2\sin \pi z)-\int _{0}^{z}\log(2\sin \pi x)\,dx.\end{aligned}}}$

Performing the integral substitution ${\displaystyle \,y=2\pi x\Rightarrow dx=dy/(2\pi )}$ gives

${\displaystyle z\log(2\sin \pi z)-{\frac {1}{2\pi }}\int _{0}^{2\pi z}\log \left(2\sin {\frac {y}{2}}\right)\,dy.}$

The Clausen function – of second order – has the integral representation

${\displaystyle \operatorname {Cl} _{2}(\theta )=-\int _{0}^{\theta }\log {\Bigg |}2\sin {\frac {x}{2}}{\Bigg |}\,dx.}$

However, within the interval ${\displaystyle \,0<\theta <2\pi }$, 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:

${\displaystyle \operatorname {Lc} (z)=z\log(2\sin \pi z)+{\frac {1}{2\pi }}\operatorname {Cl} _{2}(2\pi z).}$

Thus, after a slight rearrangement of terms, the proof is complete:

${\displaystyle 2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right)+\operatorname {Cl} _{2}(2\pi z)}$

Using the relation ${\displaystyle \,G(1+z)=\Gamma (z)\,G(z)}$ and dividing the reflection formula by a factor of ${\displaystyle \,2\pi }$ gives the equivalent form:

${\displaystyle \log \left({\frac {G(1-z)}{G(z)}}\right)=z\log \left({\frac {\sin \pi z}{\pi }}\right)+\log \Gamma (z)+{\frac {1}{2\pi }}\operatorname {Cl} _{2}(2\pi z)}$

Adamchik (2003) has given an equivalent form of the reflection formula, but with a different proof.^[6]

Replacing ${\displaystyle z}$ with ${\displaystyle 1/2-z}$ in the previous reflection formula gives, after some simplification, the equivalent formula shown below

(involving Bernoulli polynomials):

${\displaystyle \log \left({\frac {G\left({\frac {1}{2}}+z\right)}{G\left({\frac {1}{2}}-z\right)}}\right)=\log \Gamma \left({\frac {1}{2}}-z\right)+B_{1}(z)\log 2\pi +{\frac {1}{2}}\log 2+\pi \int _{0}^{z}B_{1}(x)\tan \pi x\,dx}$

Taylor series expansion

By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained:

${\displaystyle \log G(1+z)={\frac {z}{2}}\log 2\pi -\left({\frac {z+(1+\gamma )z^{2}}{2}}\right)+\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}.}$

It is valid for ${\displaystyle \,0<z<1}$. Here, ${\displaystyle \,\zeta (x)}$ is the Riemann zeta function:

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.}$

Exponentiating both sides of the Taylor expansion gives:

${\displaystyle {\begin{aligned}G(1+z)&=\exp \left[{\frac {z}{2}}\log 2\pi -\left({\frac {z+(1+\gamma )z^{2}}{2}}\right)+\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right]\\&=(2\pi )^{z/2}\exp \left[-{\frac {z+(1+\gamma )z^{2}}{2}}\right]\exp \left[\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right].\end{aligned}}}$

Comparing this with the Weierstrass product form of the Barnes function gives the following relation:

${\displaystyle \exp \left[\sum _{k=2}^{\infty }(-1)^{k}{\frac {\zeta (k)}{k+1}}z^{k+1}\right]=\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)^{k}\exp \left({\frac {z^{2}}{2k}}-z\right)\right\}}$

Multiplication formula

Like the gamma function, the G-function also has a multiplication formula:^[7]

${\displaystyle G(nz)=K(n)n^{n^{2}z^{2}/2-nz}(2\pi )^{-{\frac {n^{2}-n}{2}}z}\prod _{i=0}^{n-1}\prod _{j=0}^{n-1}G\left(z+{\frac {i+j}{n}}\right)}$

where ${\displaystyle K(n)}$ is a constant given by:

${\displaystyle K(n)=e^{-(n^{2}-1)\zeta ^{\prime }(-1)}\cdot n^{\frac {5}{12}}\cdot (2\pi )^{(n-1)/2}\,=\,(Ae^{-{\frac {1}{12}}})^{n^{2}-1}\cdot n^{\frac {5}{12}}\cdot (2\pi )^{(n-1)/2}.}$

Here ${\displaystyle \zeta ^{\prime }}$ is the derivative of the Riemann zeta function and ${\displaystyle A}$ is the Glaisher–Kinkelin constant.

Absolute value

It holds true that ${\displaystyle G({\overline {z}})={\overline {G(z)}}}$, thus ${\displaystyle |G(z)|^{2}=G(z)G({\overline {z}})}$. From this relation and by the above presented Weierstrass product form one can show that

${\displaystyle |G(x+iy)|=|G(x)|\exp \left(y^{2}{\frac {1+\gamma }{2}}\right){\sqrt {1+{\frac {y^{2}}{x^{2}}}}}{\sqrt {\prod _{k=1}^{\infty }\left(1+{\frac {y^{2}}{(x+k)^{2}}}\right)^{k+1}\exp \left(-{\frac {y^{2}}{k}}\right)}}.}$

This relation is valid for arbitrary ${\displaystyle x\in \mathbb {R} \setminus \{0,-1,-2,\dots \}}$, and ${\displaystyle y\in \mathbb {R} }$. If ${\displaystyle x=0}$, then the below formula is valid instead:

${\displaystyle |G(iy)|=y\exp \left(y^{2}{\frac {1+\gamma }{2}}\right){\sqrt {\prod _{k=1}^{\infty }\left(1+{\frac {y^{2}}{k^{2}}}\right)^{k+1}\exp \left(-{\frac {y^{2}}{k}}\right)}}}$

for arbitrary real y.

Asymptotic expansion

The logarithm of G(z + 1) has the following asymptotic expansion, as established by Barnes:

${\displaystyle {\begin{aligned}\log G(z+1)={}&{\frac {z^{2}}{2}}\log z-{\frac {3z^{2}}{4}}+{\frac {z}{2}}\log 2\pi -{\frac {1}{12}}\log z\\&{}+\left({\frac {1}{12}}-\log A\right)+\sum _{k=1}^{N}{\frac {B_{2k+2}}{4k\left(k+1\right)z^{2k}}}~+~O\left({\frac {1}{z^{2N+2}}}\right).\end{aligned}}}$

Here the ${\displaystyle B_{k}}$ are the Bernoulli numbers and ${\displaystyle A}$ is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes ^[8] the Bernoulli number ${\displaystyle B_{2k}}$ would have been written as ${\displaystyle (-1)^{k+1}B_{k}}$, but this convention is no longer current.) This expansion is valid for ${\displaystyle z}$ in any sector not containing the negative real axis with ${\displaystyle |z|}$ large.

Relation to the log-gamma integral

The parametric log-gamma can be evaluated in terms of the Barnes G-function:^[9]

${\displaystyle \int _{0}^{z}\log \Gamma (x)\,dx={\frac {z(1-z)}{2}}+{\frac {z}{2}}\log 2\pi +(z-1)\log \Gamma (z)-\log G(z)}$

A proof of the formula

The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function: ${\displaystyle z\log \Gamma (z)-\log G(1+z)}$ where ${\displaystyle {\frac {1}{\Gamma (z)}}=ze^{\gamma z}\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)e^{-z/k}\right\}}$ and ${\displaystyle \,\gamma }$ is the Euler–Mascheroni constant. Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives: ${\displaystyle {\begin{aligned}&z\log \Gamma (z)-\log G(1+z)=-z\log \left({\frac {1}{\Gamma (z)}}\right)-\log G(1+z)\\[5pt]={}&{-z}\left[\log z+\gamma z+\sum _{k=1}^{\infty }{\Bigg \{}\log \left(1+{\frac {z}{k}}\right)-{\frac {z}{k}}{\Bigg \}}\right]\\[5pt]&{}-\left[{\frac {z}{2}}\log 2\pi -{\frac {z}{2}}-{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}+\sum _{k=1}^{\infty }{\Bigg \{}k\log \left(1+{\frac {z}{k}}\right)+{\frac {z^{2}}{2k}}-z{\Bigg \}}\right]\end{aligned}}}$ A little simplification and re-ordering of terms gives the series expansion: ${\displaystyle {\begin{aligned}&\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}\\[5pt]={}&{-z}\log z-{\frac {z}{2}}\log 2\pi +{\frac {z}{2}}+{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}-z\log \Gamma (z)+\log G(1+z)\end{aligned}}}$ Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval ${\displaystyle \,[0,\,z]}$ to obtain: ${\displaystyle {\begin{aligned}&\int _{0}^{z}\log \Gamma (x)\,dx=-\int _{0}^{z}\log \left({\frac {1}{\Gamma (x)}}\right)\,dx\\[5pt]={}&{-(z\log z-z)}-{\frac {z^{2}\gamma }{2}}-\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}\end{aligned}}}$ Equating the two evaluations completes the proof: ${\displaystyle \int _{0}^{z}\log \Gamma (x)\,dx={\frac {z(1-z)}{2}}+{\frac {z}{2}}\log 2\pi +z\log \Gamma (z)-\log G(1+z)}$ And since ${\displaystyle \,G(1+z)=\Gamma (z)\,G(z)}$ then, ${\displaystyle \int _{0}^{z}\log \Gamma (x)\,dx={\frac {z(1-z)}{2}}+{\frac {z}{2}}\log 2\pi -(1-z)\log \Gamma (z)-\log G(z)\,.}$

Taking the logarithm of both sides introduces the analog of the Digamma function ${\displaystyle \psi (x)}$,

${\displaystyle \varphi (x)\equiv {\frac {d}{dx}}\log G(x),}$

where ^{[2][1] [10]}

${\displaystyle \varphi (x)=(x-1)[\psi (x)-1]+\varphi (1),\quad \varphi (1)={\frac {\ln(2\pi )-1}{2}}}$

with Taylor series

${\displaystyle \varphi (x)=\varphi (1)-(\gamma +1)(x-1)+\sum _{k\geq 2}(-1)^{k}\zeta (k)(x-1)^{k}.}$