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Barnes G-function

Extension of superfactorials to the complex numbers

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Source updated June 8, 2026

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 $\Pi$ 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.

01Functional equation and integer arguments

The Barnes G-function satisfies the functional equation

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

with normalization $G(1)=1$ . 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 $G(n)=sf(n-2)$ for $n\geq 1$ , where $sf$ is the superfactorial.

and thus

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

where $\,\Gamma (x)$ denotes the gamma function and $K$ denotes the K-function. In general, $K(z)G(z)=e^{(z-1)\ln \Gamma (z)}$ for all complex $z$ .

The functional equation $G(z+1)=\Gamma (z)\,G(z)$ 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.

02Characterisation

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 $n\to \infty$ .

03Reflection 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) when $0<z<1$ , 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 log-tangent 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:

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

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):

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

04Taylor 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:

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

05Multiplication 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.

06Absolute 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.

07Asymptotic 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.

08Relation to the log-gamma integral

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

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

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

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:

$z\log \Gamma (z)-\log G(1+z)$

where

${\frac {1}{\Gamma (z)}}=ze^{\gamma z}\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)e^{-z/k}\right\}$

and $\,\gamma$ is the Euler-Mascheroni constant.

Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives:

${\begin{aligned}&z\log \Gamma (z)-\log G(1+z)=-z\log \left({\frac {1}{\Gamma (z)}}\right)-\log G(1+z)\\[5pt]={}&{-z}\left[\log z+\gamma z+\sum _{k=1}^{\infty }{\Bigg \{}\log \left(1+{\frac {z}{k}}\right)-{\frac {z}{k}}{\Bigg \}}\right]\\[5pt]&{}-\left[{\frac {z}{2}}\log 2\pi -{\frac {z}{2}}-{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}+\sum _{k=1}^{\infty }{\Bigg \{}k\log \left(1+{\frac {z}{k}}\right)+{\frac {z^{2}}{2k}}-z{\Bigg \}}\right]\end{aligned}}$

A little simplification and re-ordering of terms gives the series expansion:

${\begin{aligned}&\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}\\[5pt]={}&{-z}\log z-{\frac {z}{2}}\log 2\pi +{\frac {z}{2}}+{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}-z\log \Gamma (z)+\log G(1+z)\end{aligned}}$

Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval $\,[0,\,z]$ to obtain:

${\begin{aligned}&\int _{0}^{z}\log \Gamma (x)\,dx=-\int _{0}^{z}\log \left({\frac {1}{\Gamma (x)}}\right)\,dx\\[5pt]={}&{-(z\log z-z)}-{\frac {z^{2}\gamma }{2}}-\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}\end{aligned}}$

Equating the two evaluations completes the proof:

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

And since $\,G(1+z)=\Gamma (z)\,G(z)$ then,

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

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

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

where

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

with Taylor series

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

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Sources and credits

This article is adapted from the Wikipedia article “Barnes G-function”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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