
In mathematics, the Barnes G-function
G(z)
Formally, the Barnes G-function is defined in the following Weierstrass product form:[2]
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
\Pi
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
The Barnes G-function satisfies the functional equation
G(z+1)=\Gamma(z)G(z)
with normalization
G(1)=1
\Gamma(z+1)=z\Gamma(z).
The functional equation implies that
G
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)=sf(n-2)
n\geq1
sf
and thus
| G(n)= | (\Gamma(n))n-1 |
| K(n) |
where
\Gamma(x)
K
z
The functional equation
G(z+1)=\Gamma(z)G(z)
(\forallx\geq1)
| d3 | |
| dx3 |
log(G(x))\geq0
is added.[3] Additionally, the Barnes G-function satisfies the duplication formula,[4]
| 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) when
0<z<1
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)
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.[6]
Replacing
z
1/2-z
(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:[7]
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:[9]
| z | |
| \int | |
| 0 |
log\Gamma(x)dx=
| z(1-z) | + | |
| 2 |
| z | |
| 2 |
log2\pi+(z-1)log\Gamma(z)-logG(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).
\psi(x)
\varphi(x)\equiv
| d | |
| dx |
logG(x),
where [10]
\varphi(x)=(x-1)[\psi(x)-1]+\varphi(1), \varphi(1)=
| ln(2\pi)-1 | |
| 2 |
\varphi(x)=\varphi(1)-(\gamma+1)(x-1)+\sumk\ge(-1)k\zeta(k)(x-1)k.
(2,Z)