Английская Википедия:Barnes G-function

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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

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

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:

<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^x 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

<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, G is of order two, and of infinite type. This can be deduced from the asymptotic expansion given below.The Barnes G function along part of the real axis

Functional equation and integer arguments

The Barnes G-function satisfies the functional equation

<math> G(z+1)=\Gamma(z)\, G(z) </math>

with normalisation G(1) = 1. 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 G 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 thus

<math>G(n)=\frac{(\Gamma(n))^{n-1}}{K(n)}</math>

where <math>\,\Gamma(x)</math> denotes the gamma function and K denotes the K-function. The functional equation 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}{4}}A^{-3}2^{-2x^{2}+3x-\frac{11}{12}}\pi^{x-\frac{1}{2}}G\left(2x\right)</math>

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^Шаблон:X\choose 2f(n)</math>

as <math>n\to\infty</math>.

Value at 1/2

<math>\begin{align}G\left(\tfrac{1}{2}\right) &= 2^{\frac{1}{24}} \, e^{\frac32 \zeta'(-1)} \, \pi^{-\frac14}\\

&= 2^{\frac{1}{24}} \, e^{\frac{1}{8}} \, \pi^{-\frac{1}{4}} \, A^{-\frac{3}{2}},\end{align}</math>

where <math>A</math> is the Glaisher–Kinkelin constant.Шаблон:Citation neededШаблон:Importance inline

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):

<math> \log G(1-z) = \log G(1+z)-z\log 2\pi+ \int_0^z \pi x \cot \pi x \, dx.</math>

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

<math>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)</math>

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

<math>\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}</math>

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

<math>z\log(2\sin \pi z)-\frac{1}{2\pi}\int_0^{2\pi z}\log\left(2\sin \frac{y}{2} \right)\,dy.</math>

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

<math>\operatorname{Cl}_2(\theta) = -\int_0^{\theta}\log\Bigg|2\sin \frac{x}{2} \Bigg|\,dx.</math>

However, within the interval <math>\, 0 < \theta < 2\pi </math>, 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:

<math>\operatorname{Lc}(z)=z\log(2\sin \pi z)+\frac{1}{2\pi} \operatorname{Cl}_2(2\pi z).</math>

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

<math>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 </math>

Using the relation <math>\, G(1+z)=\Gamma(z)\, G(z) </math> and dividing the reflection formula by a factor of <math>\, 2\pi </math> gives the equivalent form:

<math> \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) </math>

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):

<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 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):

<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>

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>

References

• Шаблон:Dlmf

• Шаблон:Cite arXiv