Stable polynomial

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. Up to elementary factors, it is a special case of the double gamma function.

Formally, the Barnes G-function is defined in the following Weierstrass product form:

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

where $\,\gamma \,$ is the Euler–Mascheroni constant, exp(x) = e^x, and ∏ is capital pi notation.

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:

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

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 G function if the convexity condition: $\,{\frac {d^{3}}{dx^{3}}}G(x)\geq 0\,$ is added.

Reflection formula 1.0

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

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

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

$Lc(z)=\int _{0}^{z}\pi x\log(\sin \pi x)\,dx=z\log(\sin \pi z)-\int _{0}^{z}\log(\sin \pi x)\,dx=$

$z\log(\sin \pi z)-\int _{0}^{z}{\Bigg [}\log(2\sin \pi x)-\log 2{\Bigg ]}\,dx=$

$z\log(2\sin \pi z)-\int _{0}^{z}\log(2\sin \pi x)\,dx$

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

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

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

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

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 itegral, the following relation clearly holds:

$Lc(z)=z\log(2\sin \pi z)+{\frac {1}{2\pi }}\,{\text{Cl}}_{2}(2\pi z)$

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

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

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:

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

Ref: see Adamchik below for an equivalent form of the reflection formula, but with a different proof.

Reflection formula 2.0

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:

$\zeta (x)=\sum _{k=1}^{\infty }{\frac {1}{k^{x}}}$

Exponentiating both sides of the Taylor expansion gives:

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

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

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

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

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.

Asymptotic expansion

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

$\log G(z+1)=$

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

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 Loggamma integral

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):

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

The proof is somewhat indirect, and involves first considering the logarithmic difference of the Gamma function and Barnes G-function:

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

Where

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

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

Taking the logarithm of the Weierstrass product forms of the Barnes function and Gamma function gives:

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

$-z\left[\log z+\gamma z+\sum _{k=1}^{\infty }{\Bigg \{}\log \left(1+{\frac {z}{k}}\right)-{\frac {z}{k}}{\Bigg \}}\right]$

$-\left[{\frac {z}{2}}\log 2\pi -{\frac {z}{2}}-{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}+\sum _{k=1}^{\infty }{\Bigg \{}k\log \left(1+{\frac {z}{k}}\right)+{\frac {z^{2}}{2k}}-z{\Bigg \}}\right]$

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

$\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}=$

$-z\log z-{\frac {z}{2}}\log 2\pi +{\frac {z}{2}}+{\frac {z^{2}}{2}}-{\frac {z^{2}\gamma }{2}}-z\log \Gamma (z)+\log G(1+z)$

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

$\int _{0}^{z}\log \Gamma (x)\,dx=-\int _{0}^{z}\log \left({\frac {1}{\Gamma (x)}}\right)\,dx=$

$-(z\log z-z)-{\frac {z^{2}\gamma }{2}}-\sum _{k=1}^{\infty }{\Bigg \{}(k+z)\log \left(1+{\frac {z}{k}}\right)-{\frac {z^{2}}{2k}}-z{\Bigg \}}$

Equating the two evaluations completes the proof:

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

References

• Template:Dlmf

• Template:Cite web