Barnes G-function#Reflection formula 1.0

The Barnes G-function satisfies the functional equation

:$G(z+1)=\backslash Gamma(z)\backslash ,\; G(z)$

with normalisation G(1) = 1. Note the similarity between the functional equation of the Barnes G-function and that of the Euler gamma function:

:$\backslash Gamma(z+1)=z\; \backslash ,\; \backslash Gamma(z)\; .$

The functional equation implies that G takes the following values at integer arguments:

:

(in particular, $\backslash ,G(0)=0,\; G(1)=1$)

and thus

:$G(n)=\backslash frac\{(\backslash Gamma(n))^\{n-1\}\}\{K(n)\}$

where $\backslash ,\backslash Gamma(x)$ denotes the gamma function and K denotes the K-function. The functional equation uniquely defines the Barnes G-function if the convexity condition,

:$(\backslash forall\; x\; \backslash geq\; 1)\; \backslash ,\; \backslash frac\{\backslash mathrm\{d\}^3\}\{\backslash mathrm\{d\}x^3\}\backslash log(G(x))\backslash geq\; 0$

is added.M. F. Vignéras, L'équation fonctionelle de la fonction zêta de Selberg du groupe mudulaire SL$(2,\backslash mathbb\{Z\})$, Astérisque 61, 235–249 (1979). Additionally, the Barnes G-function satisfies the duplication formula,{{cite journal | url=https://koreascience.kr/article/JAKO199611919482150.page | title=A duplication formula for the double gamma function $Gamma_2$ | journal=Bulletin of the Korean Mathematical Society | year=1996 | volume=33 | issue=2 | pages=289–294 | last1=Park | first1=Junesang }}

:$G(x)G\backslash left(x+\backslash frac\{1\}\{2\}\backslash right)^\{2\}G(x+1)=e^\{\backslash frac\{1\}\{4\}\}A^\{-3\}2^\{-2x^\{2\}+3x-\backslash frac\{11\}\{12\}\}\backslash pi^\{x-\backslash frac\{1\}\{2\}\}G\backslash left(2x\backslash right)$,

where $A$ is the Glaisher–Kinkelin constant.

Similar to the Bohr–Mollerup theorem for the gamma function, for a constant $c>0$, we have for $f(x)=cG(x)${{Cite book |last=Marichal |first=Jean Luc |title=A Generalization of Bohr-Mollerup's Theorem for Higher Order Convex Functions |publisher=Springer |url=https://orbi.uliege.be/bitstream/2268/294009/1/Marichal-Zena%C3%AFdi2022_Book_AGeneralizationOfBohr-Mollerup.pdf |pages=218}}

$f(x+1)=\backslash Gamma(x)f(x)$

and for $x>0$

$f(x+n)\backslash sim\; \backslash Gamma(x)^nn^\{\{x\backslash choose\; 2\}\}f(n)$

as $n\backslash to\backslash infty$.

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):

:$\backslash log\; G(1-z)\; =\; \backslash log\; G(1+z)-z\backslash log\; 2\backslash pi+\; \backslash int\_0^z\; \backslash pi\; x\; \backslash cot\; \backslash pi\; x\; \backslash ,\; dx.$

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:

:$2\backslash pi\; \backslash log\backslash left(\; \backslash frac\{G(1-z)\}\{G(1+z)\}\; \backslash right)=\; 2\backslash pi\; z\backslash log\backslash left(\backslash frac\{\backslash sin\backslash pi\; z\}\{\backslash pi\}\; \backslash right)\; +\; \backslash operatorname\{Cl\}\_2(2\backslash pi\; z)$

The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation $\backslash operatorname\{Lc\}(z)$ for the log-cotangent integral, and using the fact that $\backslash ,(d/dx)\; \backslash log(\backslash sin\backslash pi\; x)=\backslash pi\backslash cot\backslash pi\; x$, an integration by parts gives

:$\backslash begin\{align\}$

\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{align}

Performing the integral substitution $\backslash ,\; y=2\backslash pi\; x\; \backslash Rightarrow\; dx=dy/(2\backslash pi)$ gives

:$z\backslash log(2\backslash sin\; \backslash pi\; z)-\backslash frac\{1\}\{2\backslash pi\}\backslash int\_0^\{2\backslash pi\; z\}\backslash log\backslash left(2\backslash sin\; \backslash frac\{y\}\{2\}\; \backslash right)\backslash ,dy.$

The Clausen function – of second order – has the integral representation

:$\backslash operatorname\{Cl\}\_2(\backslash theta)\; =\; -\backslash int\_0^\{\backslash theta\}\backslash log\backslash Bigg|2\backslash sin\; \backslash frac\{x\}\{2\}\; \backslash Bigg|\backslash ,dx.$

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:

:$\backslash operatorname\{Lc\}(z)=z\backslash log(2\backslash sin\; \backslash pi\; z)+\backslash frac\{1\}\{2\backslash pi\}\; \backslash operatorname\{Cl\}\_2(2\backslash pi\; z).$

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

:$2\backslash pi\; \backslash log\backslash left(\; \backslash frac\{G(1-z)\}\{G(1+z)\}\; \backslash right)=\; 2\backslash pi\; z\backslash log\backslash left(\backslash frac\{\backslash sin\backslash pi\; z\}\{\backslash pi\}\; \backslash right)+\backslash operatorname\{Cl\}\_2(2\backslash pi\; z)\backslash ,\; .\; \backslash ,\; \backslash Box$

Using the relation $\backslash ,\; G(1+z)=\backslash Gamma(z)\backslash ,\; G(z)$ and dividing the reflection formula by a factor of $\backslash ,\; 2\backslash pi$ gives the equivalent form:

:$\backslash log\backslash left(\; \backslash frac\{G(1-z)\}\{G(z)\}\; \backslash right)=\; z\backslash log\backslash left(\backslash frac\{\backslash sin\backslash pi\; z\}\{\backslash pi\}$

\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.{{cite arXiv|last=Adamchik|first=Viktor S.|title=Contributions to the Theory of the Barnes function|year=2003|eprint=math/0308086}}

Replacing z with {{Sfrac|1|2}} − z in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving Bernoulli polynomials):

:$\backslash log\backslash left(\; \backslash frac\{\; G\backslash left(\backslash frac\{1\}\{2\}+z\backslash right)\; \}\{\; G\backslash left(\backslash frac\{1\}\{2\}-z\backslash right)\; \}\; \backslash right)\; =\; \backslash log\; \backslash Gamma\; \backslash left(\backslash frac\{1\}\{2\}-z\; \backslash right)\; +\; B\_1(z)\; \backslash log\; 2\backslash pi+\backslash frac\{1\}\{2\}\backslash log\; 2+\backslash pi\; \backslash int\_0^z\; B\_1(x)\; \backslash tan\; \backslash pi\; x\; \backslash ,dx$

By Taylor's theorem, and considering the logarithmic derivatives of the Barnes function, the following series expansion can be obtained:

:$\backslash log\; G(1+z)\; =\; \backslash frac\{z\}\{2\}\backslash log\; 2\backslash pi\; -\backslash left(\; \backslash frac\{z+(1+\backslash gamma)z^2\}\{2\}\; \backslash right)\; +\; \backslash sum\_\{k=2\}^\{\backslash infty\}(-1)^k\backslash frac\{\backslash zeta(k)\}\{k+1\}z^\{k+1\}.$

It is valid for . Here, $\backslash ,\; \backslash zeta(x)$ is the Riemann zeta function:

:$\backslash zeta(s)=\backslash sum\_\{n=1\}^\{\backslash infty\}\backslash frac\{1\}\{n^s\}.$

Exponentiating both sides of the Taylor expansion gives:

:

&=(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{align}

Comparing this with the Weierstrass product form of the Barnes function gives the following relation:

:$\backslash exp\; \backslash left[\backslash sum\_\{k=2\}^\backslash infty\; (-1)^k\backslash frac\{\backslash zeta(k)\}\{k+1\}z^\{k+1\}\; \backslash right]\; =\; \backslash prod\_\{k=1\}^\{\backslash infty\}\; \backslash left\backslash \{\; \backslash left(1+\backslash frac\{z\}\{k\}\backslash right)^k\; \backslash exp\; \backslash left(\backslash frac\{z^2\}\{2k\}-z\backslash right)\; \backslash right\backslash \}$

Like the gamma function, the G-function also has a multiplication formula:I. Vardi, Determinants of Laplacians and multiple gamma functions, SIAM J. Math. Anal. 19, 493–507 (1988).

:$$

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 $K(n)$ is a constant given by:

:$K(n)=\; e^\{-(n^2-1)\backslash zeta^\backslash prime(-1)\}\; \backslash cdot$

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 $\backslash zeta^\backslash prime$ is the derivative of the Riemann zeta function and $A$ is the Glaisher–Kinkelin constant.

It holds true that $G(\backslash overline\; z)=\backslash overline\{G(z)\}$, thus $|G(z)|^2=G(z)G(\backslash overline\; z)$. From this relation and by the above presented Weierstrass product form one can show that

:$$

|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 $x\backslash in\backslash mathbb\{R\}\backslash setminus\backslash \{0,-1,-2,\backslash dots\backslash \}$, and $y\backslash in\backslash mathbb\{R\}$. If $x=0$, then the below formula is valid instead:

:$$

|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)}

for arbitrary real y.

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

:$\backslash begin\{align\}$

\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{align}

Here the $B\_k$ are the Bernoulli numbers and $A$ is the Glaisher–Kinkelin constant. (Note that somewhat confusingly at the time of Barnes E. T. Whittaker and G. N. Watson, "A Course of Modern Analysis", CUP. the Bernoulli number $B\_\{2k\}$ would have been written as $(-1)^\{k+1\}\; B\_k$, but this convention is no longer current.) This expansion is valid for $z$ in any sector not containing the negative real axis with $|z|$ large.

The parametric log-gamma can be evaluated in terms of the Barnes G-function:

:$\backslash int\_0^z\; \backslash log\; \backslash Gamma(x)\backslash ,dx=\backslash frac\{z(1-z)\}\{2\}+\backslash frac\{z\}\{2\}\backslash log\; 2\backslash pi\; +z\backslash log\backslash Gamma(z)\; -\backslash log\; G(1+z)$

{{Collapse top|title=A proof of the formula}}

The proof is somewhat indirect, and involves first considering the logarithmic difference of the gamma function and Barnes G-function:

:$z\backslash log\; \backslash Gamma(z)-\backslash log\; G(1+z)$

where

:$\backslash frac\{1\}\{\backslash Gamma(z)\}=\; z\; e^\{\backslash gamma\; z\}\; \backslash prod\_\{k=1\}^\backslash infty\; \backslash left\backslash \{\; \backslash left(1+\backslash frac\{z\}\{k\}\backslash right)e^\{-z/k\}\; \backslash right\backslash \}$

and $\backslash ,\backslash gamma$ is the Euler–Mascheroni constant.

Taking the logarithm of the Weierstrass product forms of the Barnes G-function and gamma function gives:

:$$

\begin{align}

& 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{align}

A little simplification and re-ordering of terms gives the series expansion:

:$$

\begin{align}

& \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{align}

Finally, take the logarithm of the Weierstrass product form of the gamma function, and integrate over the interval $\backslash ,\; [0,\backslash ,z]$ to obtain:

:$$

\begin{align}

& \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{align}

Equating the two evaluations completes the proof:

:$\backslash int\_0^z\; \backslash log\; \backslash Gamma(x)\backslash ,dx=\backslash frac\{z(1-z)\}\{2\}+\backslash frac\{z\}\{2\}\backslash log\; 2\backslash pi\; +z\backslash log\backslash Gamma(z)\; -\backslash log\; G(1+z)$

And since $\backslash ,\; G(1+z)=\backslash Gamma(z)\backslash ,\; G(z)$ then,

:$\backslash int\_0^z\; \backslash log\; \backslash Gamma(x)\backslash ,dx=\backslash frac\{z(1-z)\}\{2\}+\backslash frac\{z\}\{2\}\backslash log\; 2\backslash pi\; -(1-z)\backslash log\backslash Gamma(z)\; -\backslash log\; G(z)\backslash ,\; .$

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