The Barnes G function along part of the real axis
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
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
{\displaystyle \,\gamma }
is the Euler–Mascheroni constant , exp (x ) = e x , and ∏ is capital pi notation .
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 normalisation 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
)
=
z
Γ
(
z
)
.
{\displaystyle \Gamma (z+1)=z\,\Gamma (z).}
The functional equation implies that G takes the following values at integer arguments:
G
(
n
)
=
{
0
if
n
=
0
,
−
1
,
−
2
,
…
∏
i
=
0
n
−
2
i
!
if
n
=
1
,
2
,
…
{\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}}}
(in particular,
G
(
0
)
=
0
,
G
(
1
)
=
1
{\displaystyle \,G(0)=0,G(1)=1}
)
and thus
G
(
n
)
=
(
Γ
(
n
)
)
n
−
1
K
(
n
)
{\displaystyle G(n)={\frac {(\Gamma (n))^{n-1}}{K(n)}}}
where
Γ
(
x
)
{\displaystyle \,\Gamma (x)}
denotes the gamma function and K denotes the K-function . The functional equation uniquely defines the G function if the convexity condition:
d
3
d
x
3
G
(
x
)
≥
0
{\displaystyle \,{\frac {d^{3}}{dx^{3}}}G(x)\geq 0}
is added.[ 2]
Value at 1/2
G
(
1
2
)
=
2
1
24
e
3
2
ζ
′
(
−
1
)
π
−
1
4
.
{\displaystyle G\left({\frac {1}{2}}\right)=2^{\frac {1}{24}}e^{{\frac {3}{2}}\zeta '(-1)}\pi ^{-{\frac {1}{4}}}.}
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
d
x
.
{\displaystyle \log G(1-z)=\log G(1+z)-z\log 2\pi +\int _{0}^{z}\pi x\cot \pi x\,dx.}
The logtangent integral on the right-hand side can be evaluated in terms of the Clausen function (of order 2), as is shown below:
2
π
log
(
G
(
1
−
z
)
G
(
1
+
z
)
)
=
2
π
z
log
(
sin
π
z
π
)
+
Cl
2
(
2
π
z
)
{\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
Lc
(
z
)
{\displaystyle \operatorname {Lc} (z)}
for the logcotangent integral, and using the fact that
(
d
/
d
x
)
log
(
sin
π
x
)
=
π
cot
π
x
{\displaystyle \,(d/dx)\log(\sin \pi x)=\pi \cot \pi x}
, an integration by parts gives
Lc
(
z
)
=
∫
0
z
π
x
cot
π
x
d
x
=
z
log
(
sin
π
z
)
−
∫
0
z
log
(
sin
π
x
)
d
x
=
z
log
(
sin
π
z
)
−
∫
0
z
[
log
(
2
sin
π
x
)
−
log
2
]
d
x
=
z
log
(
2
sin
π
z
)
−
∫
0
z
log
(
2
sin
π
x
)
d
x
.
{\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
y
=
2
π
x
⇒
d
x
=
d
y
/
(
2
π
)
{\displaystyle \,y=2\pi x\Rightarrow dx=dy/(2\pi )}
gives
z
log
(
2
sin
π
z
)
−
1
2
π
∫
0
2
π
z
log
(
2
sin
y
2
)
d
y
.
{\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
Cl
2
(
θ
)
=
−
∫
0
θ
log
|
2
sin
x
2
|
d
x
.
{\displaystyle \operatorname {Cl} _{2}(\theta )=-\int _{0}^{\theta }\log {\Bigg |}2\sin {\frac {x}{2}}{\Bigg |}\,dx.}
However, within the interval
0
<
θ
<
2
π
{\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 logtangent integral, the following relation clearly holds:
Lc
(
z
)
=
z
log
(
2
sin
π
z
)
+
1
2
π
Cl
2
(
2
π
z
)
.
{\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:
2
π
log
(
G
(
1
−
z
)
G
(
1
+
z
)
)
=
2
π
z
log
(
sin
π
z
π
)
+
Cl
2
(
2
π
z
)
.
◻
{\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)\,.\,\Box }
Using the relation
G
(
1
+
z
)
=
Γ
(
z
)
G
(
z
)
{\displaystyle \,G(1+z)=\Gamma (z)\,G(z)}
and dividing the reflection formula by a factor of
2
π
{\displaystyle \,2\pi }
gives the equivalent form:
log
(
G
(
1
−
z
)
G
(
z
)
)
=
z
log
(
sin
π
z
π
)
+
log
Γ
(
z
)
+
1
2
π
Cl
2
(
2
π
z
)
{\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)}
Ref: see Adamchik below for an equivalent form of the reflection formula , but with a different proof.
Replacing z with (1/2) − z'' in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials ):
log
(
G
(
1
2
+
z
)
G
(
1
2
−
z
)
)
=
{\displaystyle \log \left({\frac {G\left({\frac {1}{2}}+z\right)}{G\left({\frac {1}{2}}-z\right)}}\right)=}
log
Γ
(
1
2
−
z
)
+
B
1
(
z
)
log
2
π
+
1
2
log
2
+
π
∫
0
z
B
1
(
x
)
tan
π
x
d
x
{\displaystyle \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}
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
∞
(
−
1
)
k
ζ
(
k
)
k
+
1
z
k
+
1
.
{\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}.}
It is valid for
0
<
z
<
1
{\displaystyle \,0<z<1}
. Here,
ζ
(
x
)
{\displaystyle \,\zeta (x)}
is the Riemann Zeta function :
ζ
(
s
)
=
∑
n
=
1
∞
1
n
s
.
{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.}
Exponentiating both sides of the Taylor expansion gives:
G
(
1
+
z
)
=
exp
[
z
2
log
2
π
−
(
z
+
(
1
+
γ
)
z
2
2
)
+
∑
k
=
2
∞
(
−
1
)
k
ζ
(
k
)
k
+
1
z
k
+
1
]
=
(
2
π
)
z
/
2
exp
[
−
z
+
(
1
+
γ
)
z
2
2
]
exp
[
∑
k
=
2
∞
(
−
1
)
k
ζ
(
k
)
k
+
1
z
k
+
1
]
.
{\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:
exp
[
∑
k
=
2
∞
(
−
1
)
k
ζ
(
k
)
k
+
1
z
k
+
1
]
=
∏
k
=
1
∞
{
(
1
+
z
k
)
k
exp
(
z
2
2
k
−
z
)
}
{\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\}}
Like the gamma function, the G-function also has a multiplication formula:[ 3]
G
(
n
z
)
=
K
(
n
)
n
n
2
z
2
/
2
−
n
z
(
2
π
)
−
n
2
−
n
2
z
∏
i
=
0
n
−
1
∏
j
=
0
n
−
1
G
(
z
+
i
+
j
n
)
{\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)}
where
K
(
n
)
{\displaystyle K(n)}
is a constant given by:
K
(
n
)
=
e
−
(
n
2
−
1
)
ζ
′
(
−
1
)
⋅
n
5
12
⋅
(
2
π
)
(
n
−
1
)
/
2
=
(
A
e
−
1
12
)
n
2
−
1
⋅
n
5
12
⋅
(
2
π
)
(
n
−
1
)
/
2
.
{\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}.}
Here
ζ
′
{\displaystyle \zeta ^{\prime }}
is the derivative of the Riemann zeta function and
A
{\displaystyle A}
is the Glaisher–Kinkelin constant .
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
+
(
1
12
−
log
A
)
+
∑
k
=
1
N
B
2
k
+
2
4
k
(
k
+
1
)
z
2
k
+
O
(
1
z
2
N
+
2
)
.
{\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}}}
Here the
B
k
{\displaystyle B_{k}}
are the Bernoulli numbers and
A
{\displaystyle A}
is the Glaisher–Kinkelin constant . (Note that somewhat confusingly at the time of Barnes [ 4] the Bernoulli number
B
2
k
{\displaystyle B_{2k}}
would have been written as
(
−
1
)
k
+
1
B
k
{\displaystyle (-1)^{k+1}B_{k}}
, but this convention is no longer current.) This expansion is valid for
z
{\displaystyle z}
in any sector not containing the negative real axis with
|
z
|
{\displaystyle |z|}
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):
∫
0
z
log
Γ
(
x
)
d
x
=
z
(
1
−
z
)
2
+
z
2
log
2
π
+
z
log
Γ
(
z
)
−
log
G
(
1
+
z
)
{\displaystyle \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)}
The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function:
z
log
Γ
(
z
)
−
log
G
(
1
+
z
)
{\displaystyle z\log \Gamma (z)-\log G(1+z)}
where
1
Γ
(
z
)
=
z
e
γ
z
∏
k
=
1
∞
{
(
1
+
z
k
)
e
−
z
/
k
}
{\displaystyle {\frac {1}{\Gamma (z)}}=ze^{\gamma z}\prod _{k=1}^{\infty }\left\{\left(1+{\frac {z}{k}}\right)e^{-z/k}\right\}}
and
γ
{\displaystyle \,\gamma }
is the Euler–Mascheroni constant .
Taking the logarithm of the Weierstrass product forms of the Barnes function and gamma function gives:
z
log
Γ
(
z
)
−
log
G
(
1
+
z
)
=
−
z
log
(
1
Γ
(
z
)
)
−
log
G
(
1
+
z
)
=
−
z
[
log
z
+
γ
z
+
∑
k
=
1
∞
{
log
(
1
+
z
k
)
−
z
k
}
]
−
[
z
2
log
2
π
−
z
2
−
z
2
2
−
z
2
γ
2
+
∑
k
=
1
∞
{
k
log
(
1
+
z
k
)
+
z
2
2
k
−
z
}
]
{\displaystyle {\begin{aligned}&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{aligned}}}
A little simplification and re-ordering of terms gives the series expansion:
∑
k
=
1
∞
{
(
k
+
z
)
log
(
1
+
z
k
)
−
z
2
2
k
−
z
}
=
−
z
log
z
−
z
2
log
2
π
+
z
2
+
z
2
2
−
z
2
γ
2
−
z
log
Γ
(
z
)
+
log
G
(
1
+
z
)
{\displaystyle {\begin{aligned}&\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{aligned}}}
Finally, take the logarithm of the Weierstrass product form of the gamma function , and integrate over the interval
[
0
,
z
]
{\displaystyle \,[0,\,z]}
to obtain:
∫
0
z
log
Γ
(
x
)
d
x
=
−
∫
0
z
log
(
1
Γ
(
x
)
)
d
x
=
−
(
z
log
z
−
z
)
−
z
2
γ
2
−
∑
k
=
1
∞
{
(
k
+
z
)
log
(
1
+
z
k
)
−
z
2
2
k
−
z
}
{\displaystyle {\begin{aligned}&\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{aligned}}}
Equating the two evaluations completes the proof:
∫
0
z
log
Γ
(
x
)
d
x
=
z
(
1
−
z
)
2
+
z
2
log
2
π
+
z
log
Γ
(
z
)
−
log
G
(
1
+
z
)
{\displaystyle \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)}
And since
G
(
1
+
z
)
=
Γ
(
z
)
G
(
z
)
{\displaystyle \,G(1+z)=\Gamma (z)\,G(z)}
then,
∫
0
z
log
Γ
(
x
)
d
x
=
z
(
1
−
z
)
2
+
z
2
log
2
π
−
(
1
−
z
)
log
Γ
(
z
)
−
log
G
(
z
)
.
{\displaystyle \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)\,.}
References
^ E. W. Barnes, "The theory of the G-function", Quarterly Journ. Pure and Appl. Math. 31 (1900), 264–314.
^ M. F. Vignéras, L'équation fonctionelle de la fonction zêta de Selberg du groupe mudulaire SL
(
2
,
Z
)
{\displaystyle (2,\mathbb {Z} )}
, Astérisque 61 , 235–249 (1979).
^ I. Vardi, Determinants of Laplacians and multiple gamma functions , SIAM J. Math. Anal. 19 , 493–507 (1988).
^ E. T. Whittaker and G. N. Watson , "A Course of Modern Analysis ", CUP.
Askey, R.A.; Roy, R. (2010), "Barnes G-function" , in Olver, Frank W. J. ; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.), NIST Handbook of Mathematical Functions , Cambridge University Press, ISBN 978-0-521-19225-5 , MR 2723248 .
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