# Barnes G-function

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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.[1] 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:

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

where $\, \gamma \,$ is the Euler–Mascheroni constant, exp(x) = ex, 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:

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

(in particular, $\,G(0)=G(1)=1\,$) 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.[2]

## 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:

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

## Multiplication formula

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

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

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 [4] 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

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

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

Taking the logarithm of the Weierstrass product forms of the Barnes 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

1. ^ E.W.Barnes, "The theory of the G-function", Quarterly Journ. Pure and Appl. Math. 31 (1900), 264–314.
2. ^ M. F. Vignéras, L'équation fonctionelle de la fonction zêta de Selberg du groupe mudulaire SL$(2,\mathbb{Z})$, Astérisque 61, 235–249 (1979).
3. ^ I. Vardi, Determinants of Laplacians and multiple gamma functions, SIAM J. Math. Anal. 19, 493–507 (1988).
4. ^ E.T.Whittaker and G.N.Watson, "A course of modern analysis", CUP.