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.

[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: where

is the Euler–Mascheroni constant, exp(x) = ex 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 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 with normalisation G(1) = 1.

Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function: The functional equation implies that G takes the following values at integer arguments: (in particular,

The functional equation uniquely defines the Barnes G-function if the convexity condition, is added.

[2] Additionally, the Barnes G-function satisfies the duplication formula,[3] where

Similar to the Bohr–Mollerup theorem for the gamma function, for a constant

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): 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: The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation

gives The Clausen function – of second order – has the integral representation However, within the interval

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

and dividing the reflection formula by a factor of

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

[5] Replacing z with ⁠1/2⁠ − z in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials): By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained: It is valid for

is the Riemann zeta function: Exponentiating both sides of the Taylor expansion gives: Comparing this with the Weierstrass product form of the Barnes function gives the following relation: Like the gamma function, the G-function also has a multiplication formula:[6] where

is the derivative of the Riemann zeta function and

From this relation and by the above presented Weierstrass product form one can show that This relation is valid for arbitrary

, then the below formula is valid instead: for arbitrary real y.

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

(Note that somewhat confusingly at the time of Barnes [7] the Bernoulli number

, but this convention is no longer current.)

in any sector not containing the negative real axis with

The parametric log-gamma can be evaluated in terms of the Barnes G-function:[5] The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function: where and

Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives: A little simplification and re-ordering of terms gives the series expansion: Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval

to obtain: Equating the two evaluations completes the proof: And since