Barnes G-function

The Barnes G-function satisfies the functional equation

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

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

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

The functional equation implies that 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 K denotes the K-function. The functional equation uniquely defines the G function if the convexity condition: ${\displaystyle \,{\frac {d^{3}}{dx^{3}}}G(x)\geq 0}$ is added.[2]

${\displaystyle G\left({\frac {1}{2}}\right)=2^{\frac {1}{24}}e^{{\frac {3}{2}}\zeta '(-1)}\pi ^{-{\frac {1}{4}}}.}$

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

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

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

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

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:

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

Ref: see Adamchik below for 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):

${\displaystyle \log \left({\frac {G\left({\frac {1}{2}}+z\right)}{G\left({\frac {1}{2}}-z\right)}}\right)=}$

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

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:

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

Like the gamma function, the G-function also has a multiplication formula:[3]

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

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.

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 [4] 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.

The parametric Loggamma can be evaluated in terms of the Barnes G-function (Ref: this result is found in Adamchik below, but stated without proof):

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

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 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,

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

• Askey, R.A.; Roy, R. (2010), "Barnes G-function", in Olver, 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

• Adamchik, Viktor S. (2003). "Contributions to the Theory of the Barnes function". arXiv:math/0308086.