Inmathematics, theBarnes G-functionG(z) is afunction that is an extension ofsuperfactorials to thecomplex numbers. It is related to thegamma function, theK-function and theGlaisher–Kinkelin constant, and was named aftermathematicianErnest William Barnes.^[1] It can be written in terms of thedouble gamma function.

Formally, the BarnesG-function is defined in the followingWeierstrass product form:

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

where${\displaystyle \gamma }$ is theEuler–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 thedouble gamma function, is

${\displaystyle \log G(1+z)=\frac{z}{2}\log (2\pi )+{\int }_0^{\mathrm{\infty }}\frac{dt}{t}\left[\frac{1-e^{-zt}}{4{\sinh }^2\frac{t}{2}}+\frac{z^2}{2}{e}^{-t}-\frac{z}{t}\right]}$

As anentire function,G is of order two, and of infinite type. This can be deduced from the asymptotic expansion given below.

The BarnesG-function satisfies thefunctional equation

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

with normalisationG(1) = 1. Note the similarity between the functional equation of the Barnes G-function and that of the Eulergamma function:

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

The functional equation implies thatG takes the following values atinteger arguments:

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

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

where${\displaystyle \mathrm{\Gamma }(x)}$ denotes thegamma function andK denotes theK-function. The functional equation uniquely defines the Barnes G-function if the convexity condition,

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

is added.^[2] Additionally, the Barnes G-function satisfies the duplication formula,^[3]

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

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

Similar to theBohr–Mollerup theorem for thegamma function, for a constant${\displaystyle c>0}$, we have for${\displaystyle f(x)=cG(x)}$^[4]

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

and for${\displaystyle x>0}$

${\displaystyle f(x+n)\sim \mathrm{\Gamma }(x{)}^n{n}^{(\genfrac{}{}{0ex}{}{x}{2})}f(n)}$

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

Thedifference equation for the G-function, in conjunction with thefunctional equation for thegamma function, can be used to obtain the followingreflection formula for the Barnes G-function (originally proved byHermann Kinkelin):

${\displaystyle \log G(1-z)=\log G(1+z)-z\log 2\pi +{\int }_0^z\pi x\cot \pi xdx.}$

The log-tangent integral on the right-hand side can be evaluated in terms of theClausen function (of order 2), as is shown below:

${\displaystyle 2\pi \log \left(\frac{G(1-z)}{G(1+z)}\right)=2\pi z\log \left(\frac{\sin \pi z}{\pi }\right)+{\mathrm{Cl}}_2(2\pi z)}$

The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation${\displaystyle \mathrm{Lc}(z)}$ 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

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

${\displaystyle z\log (2\sin \pi z)-\frac{1}{2\pi }{\int }_0^{2\pi z}\log \left(2\sin \frac{y}{2}\right)dy.}$

TheClausen function – of second order – has the integral representation

${\displaystyle {\mathrm{Cl}}_2(\theta )=-{\int }_0^{\theta }\log |2\sin \frac{x}{2}|dx.}$

However, within the interval, theabsolute value sign within theintegrand 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:

${\displaystyle \mathrm{Lc}(z)=z\log (2\sin \pi z)+\frac{1}{2\pi }{\mathrm{Cl}}_2(2\pi z).}$

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

${\displaystyle 2\pi \log \left(\frac{G(1-z)}{G(1+z)}\right)=2\pi z\log \left(\frac{\sin \pi z}{\pi }\right)+{\mathrm{Cl}}_2(2\pi z).◻}$

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

${\displaystyle \log \left(\frac{G(1-z)}{G(z)}\right)=z\log \left(\frac{\sin \pi z}{\pi }\right)+\log \mathrm{\Gamma }(z)+\frac{1}{2\pi }{\mathrm{Cl}}_2(2\pi z)}$

Adamchik (2003) has given an equivalent form of thereflection formula, but with a different proof.^[5]

Replacingz with⁠1/2⁠ − z in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involvingBernoulli polynomials):

${\displaystyle \log \left(\frac{G\left(\frac{1}{2}+z\right)}{G\left(\frac{1}{2}-z\right)}\right)=\log \mathrm{\Gamma }\left(\frac{1}{2}-z\right)+B_1(z)\log 2\pi +\frac{1}{2}\log 2+\pi {\int }_0^z{B}_1(x)\tan \pi xdx}$

ByTaylor's theorem, and considering the logarithmicderivatives of the Barnes function, the following series expansion can be obtained:

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

It is valid for. Here,${\displaystyle \zeta (x)}$ is theRiemann zeta function:

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

Exponentiating both sides of the Taylor expansion gives:

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

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

Like the gamma function, the G-function also has a multiplication formula:^[6]

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

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

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

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

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

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

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:

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

for arbitrary realy.

Thelogarithm ofG(z + 1) has the following asymptotic expansion, as established by Barnes:

Here the${\displaystyle B_k}$ are theBernoulli numbers and${\displaystyle A}$ is theGlaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes^[7] theBernoulli 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.

The parametric log-gamma can be evaluated in terms of the Barnes G-function:^[5]

${\displaystyle {\int }_0^z\log \mathrm{\Gamma }(x)dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log \mathrm{\Gamma }(z)-\log G(1+z)}$

• Askey, R.A.; Roy, R. (2010),"Barnes G-function", inOlver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.),NIST Handbook of Mathematical Functions, Cambridge University Press,ISBN 978-0-521-19225-5,MR 2723248.

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