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About Barnes G-function About Barnes G-function In mathematics, the Barnes G-function 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.
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In mathematics, the Barnes G-function 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.
Article In mathematics, the Barnes G-function
G
(
z
)
{\displaystyle 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
π
)
z
/
2
exp
(
−
z
+
z
2
(
1
+
γ
)
2
)
∏
k
=
1
∞
{
(
1
+
z
k
)
k
exp
(
z
2
2
k
−
z
)
}
{\displaystyle 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\}}
where
γ
≈
0.5772
{\displaystyle \gamma \approx 0.5772}
is the Euler–Mascheroni constant,
exp
(
x
)
=
e
x
{\displaystyle \exp(x)=e^{x}}
is the exponential function, and
Π
{\displaystyle \Pi }
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
)
=
z
2
log
(
2
π
)
+
∫
0
∞
d
t
t
[
1
−
e
−
z
t
4
sinh
2
t
2
+
z
2
2
e
−
t
−
z
t
]
{\displaystyle \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]}
As an entire function,
G
{\displaystyle G}
is of order two, and of infinite type. This can be deduced from the asymptotic expansion given below.
Contents Functional equation and integer arguments The Barnes G-function satisfies the functional equation
G
(
z
+
1
)
=
Γ
(
z
)
G
(
z
)
{\displaystyle G(z+1)=\Gamma (z)\,G(z)}
with normalization
G
(
1
)
=
1
{\displaystyle G(1)=1}
. Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function:
Γ
(
z
+
1
)
Characterisation Similar to the Bohr–Mollerup theorem for the gamma function, for a constant
c
>
0
{\displaystyle c>0}
we have for
f
(
x
)
=
c
G
(
x
)
{\displaystyle f(x)=cG(x)}
f
(
x
+
1
)
=
Γ
(
x
)
f
(
x
)
{\displaystyle f(x+1)=\Gamma (x)f(x)}
and for
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
π
+
∫
0
z
π
x
cot
π
x
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
)
=
z
2
log
2
π
−
(
z
+
(
1
+
γ
)
z
2
2
)
+
∑
k
=
2
∞
(
Like the gamma function, the G-function also has a multiplication formula:
G
(
n
z
)
=
K
(
n
)
n
n
2
z
2
/
2
−
n
z
(
2
π
)
−
n
2
−
n
2
z
∏
i
=
0
n
Absolute value It holds true that
G
(
z
¯
)
=
G
(
z
)
¯
{\displaystyle G({\overline {z}})={\overline {G(z)}}}
, thus
|
G
(
z
)
|
2
=
G
(
z
)
G
(
z
¯
)
{\displaystyle |G(z)|^{2}=G(z)G({\overline {z}})}
. From this relation and by the above presented Weierstrass product form one can show that
Asymptotic expansion The logarithm of G(z + 1) has the following asymptotic expansion, as established by Barnes:
log
G
(
z
+
1
)
=
z
2
2
log
z
−
3
z
2
4
+
z
2
log
2
π
−
1
12
log
z
+
(
Relation to the log-gamma integral The parametric log-gamma can be evaluated in terms of the Barnes G-function:
∫
0
z
log
Γ
(
x
)
d
x
=
z
(
1
−
z
)
2
+
z
2
log
2
π
+
(
z
−
1
)
log
Γ
Article from Wikipedia (CC BY-SA 4.0), where it is maintained by volunteer editors.
{\displaystyle \Gamma (z+1)=z\,\Gamma (z).}
The functional equation implies that
takes the following values at integer arguments:
{\displaystyle 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}}}
{\displaystyle G(0)=0,G(1)=1}
{\displaystyle G(n)=sf(n-2)}
{\displaystyle G(n)={\frac {(\Gamma (n))^{n-1}}{K(n)}}}
{\displaystyle \,\Gamma (x)}
denotes the gamma function and
denotes the K-function. In general,
{\displaystyle K(z)G(z)=e^{(z-1)\ln \Gamma (z)}}
{\displaystyle G(z+1)=\Gamma (z)\,G(z)}
uniquely defines the Barnes G-function if the convexity condition,
{\displaystyle (\forall x\geq 1)\,{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}\log(G(x))\geq 0}
is added. Additionally, the Barnes G-function satisfies the duplication formula,
{\displaystyle 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)}
is the Glaisher–Kinkelin constant.
{\displaystyle f(x+n)\sim \Gamma (x)^{n}n^{x \choose 2}f(n)}
{\displaystyle n\to \infty }
{\displaystyle \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) when
{\displaystyle 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)}
The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation
{\displaystyle \operatorname {Lc} (z)}
for the log-cotangent integral, and using the fact that
{\displaystyle \,(d/dx)\log(\sin \pi x)=\pi \cot \pi x}
, an integration by parts gives
{\displaystyle {\begin{aligned}\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{aligned}}}
Performing the integral substitution
{\displaystyle \,y=2\pi x\Rightarrow dx=dy/(2\pi )}
{\displaystyle z\log(2\sin \pi z)-{\frac {1}{2\pi }}\int _{0}^{2\pi z}\log \left(2\sin {\frac {y}{2}}\right)\,dy.}
The Clausen function – of second order – has the integral representation
{\displaystyle \operatorname {Cl} _{2}(\theta )=-\int _{0}^{\theta }\log {\Bigg |}2\sin {\frac {x}{2}}{\Bigg |}\,dx.}
However, within the interval
{\displaystyle \,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 log-tangent integral, the following relation clearly holds:
{\displaystyle \operatorname {Lc} (z)=z\log(2\sin \pi z)+{\frac {1}{2\pi }}\operatorname {Cl} _{2}(2\pi z).}
Thus, after a slight rearrangement of terms, the proof is complete:
{\displaystyle 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)}
{\displaystyle \,G(1+z)=\Gamma (z)\,G(z)}
and dividing the reflection formula by a factor of
gives the equivalent form:
{\displaystyle \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)}
Adamchik (2003) has given an equivalent form of the reflection formula, but with a different proof.
in the previous reflection formula gives, after some simplification, the equivalent formula shown below
(involving Bernoulli polynomials):
{\displaystyle \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}
{\displaystyle \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}.}
{\displaystyle \,\zeta (x)}
is the Riemann zeta function:
{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.}
Exponentiating both sides of the Taylor expansion gives:
{\displaystyle {\begin{aligned}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{aligned}}}
Comparing this with the Weierstrass product form of the Barnes function gives the following relation:
{\displaystyle \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\}}
{\displaystyle 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)}
{\displaystyle 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}.}
{\displaystyle \zeta ^{\prime }}
is the derivative of the Riemann zeta function and
is the Glaisher–Kinkelin constant.
{\displaystyle |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)}}.}
This relation is valid for arbitrary
{\displaystyle x\in \mathbb {R} \setminus \{0,-1,-2,\dots \}}
{\displaystyle y\in \mathbb {R} }
, then the below formula is valid instead:
{\displaystyle |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)}}}
{\displaystyle {\begin{aligned}\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{aligned}}}
are the Bernoulli numbers and
is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes the Bernoulli number
would have been written as
{\displaystyle (-1)^{k+1}B_{k}}
, but this convention is no longer current.) This expansion is valid for
in any sector not containing the negative real axis with
{\displaystyle \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)}
Taking the logarithm of both sides introduces the analog of the Digamma function
{\displaystyle \varphi (x)\equiv {\frac {d}{dx}}\log G(x),}
{\displaystyle \varphi (x)=(x-1)[\psi (x)-1]+\varphi (1),\quad \varphi (1)={\frac {\ln(2\pi )-1}{2}}}
{\displaystyle \varphi (x)=\varphi (1)-(\gamma +1)(x-1)+\sum _{k\geq 2}(-1)^{k}\zeta (k)(x-1)^{k}.}
