alt=Plot of the Barnes G function G(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D|thumb|Plot of the Barnes G aka double gamma function G(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D
thumb|The Barnes G function along part of the real axis
In mathematics, the Barnes G-function <math>G(z)</math> 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:
:<math> 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\}</math>
where <math>\, \gamma </math> is the Euler–Mascheroni constant, exp(x) = e<sup>x</sup> is the exponential function, and <math>\Pi</math> denotes multiplication (capital pi notation).
The integral representation, which may be deduced from the relation to the double gamma function, is
:<math>
\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]
</math>
As an entire function, <math>G</math> 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
:<math> G(z+1)=\Gamma(z)\, G(z) </math>
with normalization <math>G(1)=1</math>. Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function:
:<math> \Gamma(z+1)=z \, \Gamma(z) .</math>
The functional equation implies that <math>G</math> takes the following values at integer arguments:
:<math>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}</math>
In particular, <math>G(0)=0, G(1) = 1</math> and <math>G(n) = sf(n-2)</math> for <math>n \geq 1</math>, where <math>sf</math> is the superfactorial.
and thus
:<math>G(n)=\frac{(\Gamma(n))^{n-1{K(n)}</math>
where <math>\,\Gamma(x)</math> denotes the gamma function and <math>K</math> denotes the K-function. In general,<math display="block">K(z) G(z)=e^{(z-1) \ln \Gamma(z)}</math>for all complex <math>z</math>.
The functional equation <math> G(z+1)=\Gamma(z)\, G(z) </math> uniquely defines the Barnes G-function if the convexity condition,
:<math>(\forall x \geq 1) \, \frac{\mathrm{d}^3}{\mathrm{d}x^3}\log(G(x))\geq 0</math>
is added. Additionally, the Barnes G-function satisfies the duplication formula,
:<math>G(x)G\left(x+\frac{1}{2}\right)^{2}G(x+1)=e^{\frac{1}{4A^{-3}2^{-2x^{2}+3x-\frac{11}{12\pi^{x-\frac{1}{2G\left(2x\right)</math>,
where <math>A</math> is the Glaisher–Kinkelin constant.
Characterisation
Similar to the Bohr–Mollerup theorem for the gamma function, for a constant <math>c>0</math> we have for <math>f(x)=cG(x)</math>
<math>f(x+1)=\Gamma(x)f(x)</math>
and for <math>x>0</math>
<math>f(x+n)\sim \Gamma(x)^nn^f(n)</math>
as <math>n\to\infty</math>.
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):
:<math> \log G(1-z) = \log G(1+z)-z\log 2\pi+ \int_0^z \pi x \cot \pi x \, dx.</math>
The log-tangent integral on the right-hand side can be evaluated in terms of the Clausen function (of order 2) when <math>0 < z < 1</math>, as is shown below:
Replacing <math>z </math> with <math>1/2 - z </math> in the previous reflection formula gives, after some simplification, the equivalent formula shown below
(involving Bernoulli polynomials):
:<math>\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</math>
Taylor series expansion
By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained:
:<math>\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}.</math>
It is valid for <math>\, 0 < z < 1 </math>. Here, <math>\, \zeta(x) </math> is the Riemann zeta function:
:<math> \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}. </math>
Exponentiating both sides of the Taylor expansion gives:
:<math>\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}</math>
Comparing this with the Weierstrass product form of the Barnes function gives the following relation:
:<math>\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\}</math>
Multiplication formula
Like the gamma function, the G-function also has a multiplication formula:
:<math>
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)
</math>
where <math>K(n)</math> is a constant given by:
:<math> 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}.</math>
Here <math>\zeta^\prime</math> is the derivative of the Riemann zeta function and <math>A</math> is the Glaisher–Kinkelin constant.
Absolute value
It holds true that <math>G(\overline z)=\overline{G(z)}</math>, thus <math>|G(z)|^2=G(z)G(\overline z)</math>. From this relation and by the above presented Weierstrass product form one can show that
:<math>
|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)}.
</math>
This relation is valid for arbitrary <math>x\in\mathbb{R}\setminus\{0,-1,-2,\dots\}</math>, and <math>y\in\mathbb{R}</math>. If <math>x=0</math>, then the below formula is valid instead:
:<math>
|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)}
</math>
for arbitrary real y.
Asymptotic expansion
The logarithm of G(z + 1) has the following asymptotic expansion, as established by Barnes:
:<math>\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}</math>
Here the <math>B_k</math> are the Bernoulli numbers and <math>A</math> is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes the Bernoulli number <math>B_{2k}</math> would have been written as <math>(-1)^{k+1} B_k </math>, but this convention is no longer current.) This expansion is valid for <math>z </math> in any sector not containing the negative real axis with <math>|z|</math> large.
Relation to the log-gamma integral
The parametric log-gamma can be evaluated in terms of the Barnes G-function:
:<math> \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +(z-1)\log\Gamma(z) -\log G(z) </math>
The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function:
:<math>z\log \Gamma(z)-\log G(1+z)</math>
where
:<math>\frac{1}{\Gamma(z)}= z e^{\gamma z} \prod_{k=1}^\infty \left\{ \left(1+\frac{z}{k}\right)e^{-z/k} \right\}</math>
and <math>\,\gamma</math> is the Euler–Mascheroni constant.
Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives:
:<math>
\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}
</math>
A little simplification and re-ordering of terms gives the series expansion:
:<math>
\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}
</math>
Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval <math>\, [0,\,z]</math> to obtain:
:<math>
\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}
</math>
Equating the two evaluations completes the proof:
:<math> \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)</math>
And since <math>\, G(1+z)=\Gamma(z)\, G(z) </math> then,
:<math> \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)\, .</math>
Taking the logarithm of both sides introduces the analog of the Digamma function <math>\psi(x)</math>,
<math>
\varphi(x) \equiv \frac{d}{dx}\log G(x),
</math>
where
:<math>
\varphi(x) =(x-1)[\psi(x)-1]+\varphi(1),\quad \varphi(1)=\frac{\ln(2\pi)-1}{2}
</math>
with Taylor series
:<math>
\varphi(x)=\varphi(1)-(\gamma+1)(x-1)+\sum_{k\ge 2}(-1)^k\zeta(k)(x-1)^k .
</math>