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:

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

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

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

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

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

The functional equation implies that G takes the following values at integer arguments:

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

and thus

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

where $\mathrm{\Gamma }(x)$ denotes the Gamma function and K denotes the K-function. The functional equation uniquely defines the G function if the convexity condition: $\frac{d^3}{d{x}^3}G(x)\ge 0$ 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):

$\log G(1-z)=\log G(1+z)-z\log 2\pi +\underset{0}{\overset{z}{\int }}\pi x\cot \pi xdx.$

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

$Lc(z)=\underset{0}{\overset{z}{\int }}\pi x\log (\sin \pi x)dx=z\log (\sin \pi z)-\underset{0}{\overset{z}{\int }}\log (\sin \pi x)dx=$

$z\log (\sin \pi z)-\underset{0}{\overset{z}{\int }}[\log (2\sin \pi x)-\log 2]dx=$

$z\log (2\sin \pi z)-\underset{0}{\overset{z}{\int }}\log (2\sin \pi x)dx$

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

$z\log (2\sin \pi z)-\frac{1}{2\pi }\underset{0}{\overset{2\pi z}{\int }}\log (2\sin \pi \frac{y}{2})dy$

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

${\text{Cl}}_2(\theta )=-\underset{0}{\overset{\theta }{\int }}\log |2\sin \frac{x}{2}|dx$

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:

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

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

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

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):

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

$\log \mathrm{\Gamma }(\frac{1}{2}-z)+B_1(z)\log 2\pi -\frac{1}{2}\log 2+\pi \underset{0}{\overset{z}{\int }}{B}_1(x)\tan \pi xdx$

Taylor series expansion

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

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

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

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

Exponentiating both sides of the Taylor expansion gives:

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

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

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

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

Multiplication formula

Like the Gamma function, the G-function also has a multiplication formula:

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

where $K(n)$ is a constant given by:

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

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:

$\log G(z+1)=$

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

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 $\left|z\right|$ 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):

$\underset{0}{\overset{z}{\int }}\log \mathrm{\Gamma }(x)dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log \mathrm{\Gamma }(z)-\log G(1+z)$

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

$z\log \mathrm{\Gamma }(z)-\log G(1+z)$

Where

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

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

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

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

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

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

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

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

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

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

$\underset{0}{\overset{z}{\int }}\log \mathrm{\Gamma }(x)dx=-\underset{0}{\overset{z}{\int }}\log \left(\frac{1}{\mathrm{\Gamma }(x)}\right)dx=$

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

Equating the two evaluations completes the proof:

$\underset{0}{\overset{z}{\int }}\log \mathrm{\Gamma }(x)dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log \mathrm{\Gamma }(z)-\log G(1+z).◻$

References

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