Barnes G-function



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. It can be written in terms of the double gamma function.

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


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

where $$\, \gamma $$ is the Euler–Mascheroni constant, exp(x) = ex is the exponential function, and Π denotes multiplication (capital pi notation).

The integral representation, which may be deduced from the relation to the double gamma function, is

\log G(1+z) = \frac{z}{2}\log(2\pi) +\int_0^\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 an entire function, G is of order two, and of infinite type. This can be deduced from the asymptotic expansion given below.

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:


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

(in particular, $$\,G(0)=0, G(1)=1$$) 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 Barnes G-function if the convexity condition,


 * $$(\forall x \geq 1) \, \frac{\mathrm{d}^3}{\mathrm{d}x^3}\log(G(x))\geq 0$$

is added. Additionally, the Barnes G-function satisfies the duplication formula,


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

where $$A$$ is the Glaisher–Kinkelin constant.

Characterisation
Similar to the Bohr–Mollerup theorem for the gamma function, for a constant $$c>0$$, we have for $$f(x)=cG(x)$$

$$f(x+1)=\Gamma(x)f(x)$$

and for $$x>0$$

$$f(x+n)\sim \Gamma(x)^nn^f(n)$$

as $$n\to\infty$$.

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


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

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


 * $$\begin{align}

\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{align}$$

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


 * $$z\log(2\sin \pi z)-\frac{1}{2\pi}\int_0^{2\pi z}\log\left(2\sin \frac{y}{2} \right)\,dy.$$

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


 * $$\operatorname{Cl}_2(\theta) = -\int_0^{\theta}\log\Bigg|2\sin \frac{x}{2} \Bigg|\,dx.$$

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 integral, the following relation clearly holds:


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

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


 * $$2\pi \log\left( \frac{G(1-z)}{G(1+z)} \right)= 2\pi z\log\left(\frac{\sin\pi z}{\pi} \right)+\operatorname{Cl}_2(2\pi z)\, . \, \Box $$

Using the relation $$\, G(1+z)=\Gamma(z)\, G(z) $$ 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}\operatorname{Cl}_2(2\pi z) $$

Adamchik (2003) has given 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):


 * $$\log\left( \frac{ G\left(\frac{1}{2}+z\right) }{ G\left(\frac{1}{2}-z\right) } \right) = \log \Gamma \left(\frac{1}{2}-z \right) + B_1(z) \log 2\pi+\frac{1}{2}\log 2+\pi \int_0^z B_1(x) \tan \pi x \,dx$$

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:


 * $$ \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}. $$

Exponentiating both sides of the Taylor expansion gives:


 * $$\begin{align} 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{align}$$

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


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

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.

Absolute value
It holds true that $$G(\overline z)=\overline{G(z)}$$, thus $$|G(z)|^2=G(z)G(\overline z)$$. From this relation and by the above presented Weierstrass product form one can show that

$$ This relation is valid for arbitrary $$x\in\mathbb{R}\setminus\{0,-1,-2,\dots\}$$, and $$y\in\mathbb{R}$$. If $$x=0$$, then the below formula is valid instead:
 * 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}^\infty\left(1+\frac{y^2}{(x+k)^2}\right)^{k+1}\exp\left(-\frac{y^2}{k}\right)}.

$$ for arbitrary real y.
 * G(iy)|=y\exp\left(y^2\frac{1+\gamma}{2}\right)\sqrt{\prod_{k=1}^\infty\left(1+\frac{y^2}{k^2}\right)^{k+1}\exp\left(-\frac{y^2}{k}\right)}

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


 * $$\begin{align}

\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{align}$$

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 log-gamma integral
The parametric log-gamma can be evaluated in terms of the Barnes G-function:


 * $$ \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z) $$

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 G-function and gamma function gives:



\begin{align} & 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{align} $$

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



\begin{align} & \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{align} $$

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



\begin{align} & \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{align} $$

Equating the two evaluations completes the proof:


 * $$ \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z)$$

And since $$\, G(1+z)=\Gamma(z)\, G(z) $$ then,


 * $$ \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi -(1-z)\log\Gamma(z) -\log G(z)\, .$$