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)^ \exp\left(- \frac \right) \, \prod_^\infty \left\$

where $\, \gamma$ 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
:
\log G(1+z) = \frac\log(2\pi) +\int_0^\infty\frac\left frac +\frace^ -\frac\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:

:$G(n)=\begin 0&\textn=0,-1,-2,\dots\\ \prod_^ i!&\textn=1,2,\dots\end$

(in particular, $\,G(0)=0, G(1)=1$)
and thus

:$G(n)=\frac$

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,

:$(\forall x \geq 1) \, \frac\log(G(x))\geq 0$

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

:$G(x)G\left(x+\frac\right)^G(x+1)=e^A^2^\pi^G\left(2x\right)$,

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:

:$2\pi \log\left( \frac \right)= 2\pi z\log\left(\frac \right) + \operatorname_2(2\pi z)$

The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation $\operatorname(z)$ 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
\operatorname(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

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

:$z\log(2\sin \pi z)-\frac\int_0^\log\left(2\sin \frac \right)\,dy.$

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

:$\operatorname_2(\theta) = -\int_0^\log\Bigg, 2\sin \frac \Bigg, \,dx.$

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:

:$\operatorname(z)=z\log(2\sin \pi z)+\frac \operatorname_2(2\pi z).$

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

:$2\pi \log\left( \frac \right)= 2\pi z\log\left(\frac \right)+\operatorname_2(2\pi z)\, . \, \Box$

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:

:$\log\left( \frac \right)= z\log\left(\frac \right)+\log\Gamma(z)+\frac\operatorname_2(2\pi z)$

Adamchik (2003) has given an equivalent form of the reflection formula, but with a different proof.

Replacing ''z'' with  − ''z'' in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials):

:$\log\left( \frac \right) = \log \Gamma \left(\frac-z \right) + B_1(z) \log 2\pi+\frac\log 2+\pi \int_0^z B_1(x) \tan \pi x \,dx$

Taylor series expansion

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

:$\log G(1+z) = \frac\log 2\pi -\left( \frac \right) + \sum_^(-1)^k\fracz^.$

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

:$\zeta(s)=\sum_^\frac.$

Exponentiating both sides of the Taylor expansion gives:

:\begin G(1+z) &= \exp \left \frac\log 2\pi -\left( \frac \right) + \sum_^(-1)^k\fracz^ \right\\
&=(2\pi)^\exp\left -\frac \right\exp \left sum_^(-1)^k\fracz^ \right\end

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

:\exp \left sum_^\infty (-1)^k\fracz^ \right= \prod_^ \left\

Multiplication formula

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

:$G(nz)= K(n) n^ (2\pi)^\prod_^\prod_^G\left(z+\frac\right)$

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

:$K(n)= e^ \cdot n^\cdot(2\pi)^\,=\, (Ae^)^\cdot n^\cdot (2\pi)^.$

Here $\zeta^\prime$ is the derivative of the Riemann zeta function and $A$ is the Glaisher–Kinkelin constant.

Absolute value

It holds true that $G(\overline z)=\overline$, thus $, G(z), ^2=G(z)G(\overline z)$. From this relation and by the above presented Weierstrass product form one can show that
:$, G(x+iy), =, G(x), \exp\left(y^2\frac\right)\sqrt\sqrt.$
This relation is valid for arbitrary $x\in\mathbb\setminus\$, and $y\in\mathbb$. If $x=0$, then the below formula is valid instead:
:$, G(iy), =y\exp\left(y^2\frac\right)\sqrt$
for arbitrary real ''y''.

Asymptotic expansion

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

:$\begin \log G(z+1) = & \frac \log z - \frac + \frac\log 2\pi -\frac \log z \\ & + \left(\frac-\log A \right) +\sum_^N \frac~+~O\left(\frac\right). \end$

Here the $B_k$ are the Bernoulli numbers and $A$ is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes E. T. Whittaker and G. N. Watson, "A Course of Modern Analysis", CUP. the Bernoulli number $B_$ would have been written as $(-1)^ B_k$, 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:

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

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= z e^ \prod_^\infty \left\$

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
& z\log \Gamma(z)-\log G(1+z)=-z \log\left(\frac\right)-\log G(1+z) \\ pt= & \left \log z+\gamma z +\sum_^\infty \Bigg\ \right\\ pt& -\left \frac\log 2\pi -\frac-\frac -\frac + \sum_^\infty \Bigg\ \right\end

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

:
\begin
& \sum_^\infty \Bigg\ \\ pt= & \log z-\frac\log 2\pi +\frac +\frac- \frac- z\log\Gamma(z) +\log G(1+z)
\end

Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval \, ,\,z/math> to obtain:

:
\begin
& \int_0^z\log\Gamma(x)\,dx=-\int_0^z \log\left(\frac\right)\,dx \\ pt= & -\frac- \sum_^\infty \Bigg\
\end

Equating the two evaluations completes the proof:

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

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

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

References

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{{DEFAULTSORT:Barnes G-Function
Number theory
Special functions
Gamma and related functions