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] 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(-
z+z2(1+\gamma) | |
2 |
\right)
infty | |
\prod | |
k=1 |
\left\{\left(1+
z | |
k |
\right)k\exp\left(
z2 | |
2k |
-z\right)\right\}
where
\gamma
The integral representation, which may be deduced from the relation to the double gamma function, is
logG(1+z)=
z | |
2 |
log(2\pi)
| |||||
+\int | \left[ | ||||
0 |
1-e-zt | + | ||||||
|
z2 | |
2 |
e-t-
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.
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}
n-2 | |
0&ifn=0,-1,-2,...\ \prod | |
i=0 |
i!&ifn=1,2,...\end{cases}
(in particular,
G(0)=0,G(1)=1
G(n)= | (\Gamma(n))n-1 |
K(n) |
where
\Gamma(x)
(\forallx\geq1)
d3 | |
dx3 |
log(G(x))\geq0
is added.[2] Additionally, the Barnes G-function satisfies the duplication formula,[3]
G(x)G\left(x+ | 1 |
2 |
\right)2
| ||||
G(x+1)=e |
A-3
| ||||||||||
2 |
| ||||
\pi |
G\left(2x\right)
where
A
Similar to the Bohr–Mollerup theorem for the gamma function, for a constant
c>0
f(x)=cG(x)
f(x+1)=\Gamma(x)f(x)
and for
x>0
f(x+n)\sim\Gamma(x)nn{x\choose
as
n\toinfty
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):
logG(1-z)=logG(1+z)-zlog2\pi+
z | |
\int | |
0 |
\pix\cot\pixdx.
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\pilog\left(
G(1-z) | |
G(1+z) |
\right)=2\pizlog\left(
\sin\piz | |
\pi |
\right)+\operatorname{Cl}2(2\piz)
The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation
\operatorname{Lc}(z)
(d/dx)log(\sin\pix)=\pi\cot\pix
\begin{align} \operatorname{Lc}(z)&=
z\pi | |
\int | |
0 |
x\cot\pixdx\\ &=zlog(\sin\pi
zlog(\sin | |
z)-\int | |
0 |
\pix)dx\\ &=zlog(\sin\pi
z[log(2\sin | |
z)-\int | |
0 |
\pix)-log2]dx\\ &=zlog(2\sin\pi
zlog(2\sin | |
z)-\int | |
0 |
\pix)dx. \end{align}
Performing the integral substitution
y=2\pix ⇒ dx=dy/(2\pi)
zlog(2\sin\piz)-
1 | |
2\pi |
2\piz | |
\int | |
0 |
log\left(2\sin
y | |
2 |
\right)dy.
The Clausen function – of second order – has the integral representation
\operatorname{Cl}2(\theta)=
\theta | |
-\int | |
0 |
log|2\sin
x | |
2 |
|dx.
However, within the interval
0<\theta<2\pi
\operatorname{Lc}(z)=zlog(2\sin\piz)+
1 | |
2\pi |
\operatorname{Cl}2(2\piz).
Thus, after a slight rearrangement of terms, the proof is complete:
2\pilog\left(
G(1-z) | |
G(1+z) |
\right)=2\pizlog\left(
\sin\piz | |
\pi |
\right)+\operatorname{Cl}2(2\piz).\Box
Using the relation
G(1+z)=\Gamma(z)G(z)
2\pi
log\left(
G(1-z) | |
G(z) |
\right)=zlog\left(
\sin\piz | \right)+log\Gamma(z)+ | |
\pi |
1 | |
2\pi |
\operatorname{Cl}2(2\piz)
Adamchik (2003) has given an equivalent form of the reflection formula, but with a different proof.[5]
Replacing z with − z in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials):
log\left(
| ||||||
|
\right)=log\Gamma\left(
1 | |
2 |
-z\right)+B1(z)log2\pi+
1 | |
2 |
log2+\pi
z | |
\int | |
0 |
B1(x)\tan\pixdx
By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained:
logG(1+z)=
z | |
2 |
log2\pi-\left(
z+(1+\gamma)z2 | |
2 |
\right)+
infty | |
\sum | |
k=2 |
| ||||
(-1) |
zk+1.
It is valid for
0<z<1
\zeta(x)
infty | |
\zeta(s)=\sum | |
n=1 |
1 | |
ns |
.
Exponentiating both sides of the Taylor expansion gives:
\begin{align}G(1+z)&=\exp\left[
z | |
2 |
log2\pi-\left(
z+(1+\gamma)z2 | |
2 |
\right)+
infty | |
\sum | |
k=2 |
| ||||
(-1) |
zk+1\right]\\ &=(2\pi)z/2\exp\left[-
z+(1+\gamma)z2 | |
2 |
\right]\exp
infty | |
\left[\sum | |
k=2 |
| ||||
(-1) |
zk+1\right].\end{align}
Comparing this with the Weierstrass product form of the Barnes function gives the following relation:
\exp
infty | |
\left[\sum | |
k=2 |
| ||||
(-1) |
zk+1\right]=
infty | |
\prod | |
k=1 |
\left\{\left(1+
z | |
k |
\right)k\exp\left(
z2 | |
2k |
-z\right)\right\}
Like the gamma function, the G-function also has a multiplication formula:[6]
G(nz)=K(n)
n2z2/2-nz | |
n |
| |||||
(2\pi) |
n-1 | |
\prod | |
i=0 |
n-1 | ||
\prod | G\left(z+ | |
j=0 |
i+j | |
n |
\right)
where
K(n)
K(n)=
-(n2-1)\zeta\prime(-1) | |
e |
| ||||
⋅ n |
⋅ (2\pi)(n-1)/2
| ||||
= (Ae |
n2-1 | |
) |
⋅
| ||||
n |
⋅ (2\pi)(n-1)/2.
Here
\zeta\prime
A
It holds true that
G(\overlinez)=\overline{G(z)}
|G(z)|2=G(z)G(\overlinez)
| |||||
|G(x+iy)|=|G(x)|\exp\left(y | \right)\sqrt{1+ |
y2 | |
x2 |
x\inR\setminus\{0,-1,-2,...\}
y\inR
x=0
| ||||
|G(iy)|=y\exp\left(y |
| ||||
\right)\sqrt{\prod | ||||
k=1 |
\right)k+1\exp\left(-
y2 | |
k |
\right)}
The logarithm of G(z + 1) has the following asymptotic expansion, as established by Barnes:
\begin{align} logG(z+1)={}&
z2 | |
2 |
logz-
3z2 | |
4 |
+
z | |
2 |
log2\pi-
1 | |
12 |
logz\\ &{}+\left(
1 | |
12 |
-logA\right)
N | |
+\sum | |
k=1 |
B2k | ~+~O\left( | |
4k\left(k+1\right)z2k |
1 | |
z2N |
\right). \end{align}
Here the
Bk
A
B2k
(-1)k+1Bk
z
|z|
The parametric log-gamma can be evaluated in terms of the Barnes G-function:
z | |
\int | |
0 |
log\Gamma(x)dx=
z(1-z) | + | |
2 |
z | |
2 |
log2\pi+zlog\Gamma(z)-logG(1+z)
The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function:
zlog\Gamma(z)-logG(1+z)
where
1 | |
\Gamma(z) |
=ze\gamma
infty | |
\prod | |
k=1 |
\left\{\left(1+
z | |
k |
\right)e-z/k\right\}
and
\gamma
Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives:
\begin{align} &zlog\Gamma(z)-logG(1+z)=-zlog\left(
1 | |
\Gamma(z) |
\right)-logG(1+z)\\[5pt] ={}&{-z}\left[logz+\gammaz
infty | |
+\sum | |
k=1 |
\{log\left(1+
z | |
k |
\right)-
z | |
k |
\}\right]\\[5pt] &{}-\left[
z | |
2 |
log2\pi-
z | - | |
2 |
z2 | - | |
2 |
z2\gamma | |
2 |
+
infty | ||
\sum | \{klog\left(1+ | |
k=1 |
z | |
k |
\right)+
z2 | |
2k |
-z\}\right] \end{align}
A little simplification and re-ordering of terms gives the series expansion:
\begin{align} &
infty | |
\sum | |
k=1 |
\{(k+z)log\left(1+
z | \right)- | |
k |
z2 | |
2k |
-z\}\\[5pt] ={}&{-z}logz-
z | |
2 |
log2\pi+
z | + | |
2 |
z2 | |
2 |
-
z2\gamma | |
2 |
-zlog\Gamma(z)+logG(1+z) \end{align}
Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval
[0,z]
\begin{align} &
z | ||
\int | log\left( | |
0 |
1 | |
\Gamma(x) |
\right)dx\\[5pt] ={}&{-(zlogz-z)}-
z2\gamma | |
2 |
-
infty | |
\sum | |
k=1 |
\{(k+z)log\left(1+
z | \right)- | |
k |
z2 | |
2k |
-z\} \end{align}
Equating the two evaluations completes the proof:
z | |
\int | |
0 |
log\Gamma(x)dx=
z(1-z) | + | |
2 |
z | |
2 |
log2\pi+zlog\Gamma(z)-logG(1+z)
And since
G(1+z)=\Gamma(z)G(z)
z | |
\int | |
0 |
log\Gamma(x)dx=
z(1-z) | + | |
2 |
z | |
2 |
log2\pi-(1-z)log\Gamma(z)-logG(z).
(2,Z)