MHB A conjecture on a Generalized Barnes' function - can anyone help?

[DreamWeaver]

This is NOT a tutorial, so any and all contributions are very much welcome... :DI've recently been working on the Barnes' function - see tutorial in Math Notes board - and been trying to generalize some of my results to higher order Barnes' functions (intimately connected with the Multiple Gamma functions, $$\Gamma_n(z)$$ ). If we define the Generalized Barnes' function

by$$G_1(z) = \frac{1}{\Gamma(z)}$$

$$G_2(z) = \frac{1}{\Gamma_2(z)} = G(z) \quad [ \text{the regular Barnes function} ]$$

$$G_3(z) = \frac{1}{\Gamma_3(z)}$$etc, and$$G_n(1) = 1$$

$$G_{n+1}(1+z)=G_n(z) G_{n+1}(z)$$
My main aim has been to find an infinite product representation for the Triple Barnes' Function, $$G_3(1+z)$$, and even higher order Multiple Barnes' functions. The following results are known:$$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$$$\Rightarrow$$$$G_1(1+z)=\frac{1}{\Gamma(1+z)} =\frac{1}{z\Gamma(z)}= e^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$and$$G_2(1+z)=G(1+z) = $$

$$(2\pi)^{z/2} \text{exp}\left(- \frac{z+z^2(1+\gamma )}{2} \right) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)$$
In the Barnes' function tutorial, I presented the following Taylor series$$\log \Gamma(1+z)=-\gamma z +\sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \quad \Rightarrow$$$$\log G_1(1+z)=\gamma z - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k $$Similarly$$\log G_2(1+z)=\log G(1+z)= \frac{z}{2}\log(2\pi) - \left( \frac{z+z^2(1+\gamma )}{2} \right) + \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} $$
For various different reasons, this has led me to conjecture a family of polynomials, which I've come to call the Complementary Barnes' Function Polynomials (Barnes' Polynomials for short), $$\gamma_m(z) = \text{Polynomial of degree at most} \, m$$The central idea is this: in light of the previous two series, I've been lead to consider:The Barnes' Polynomial Conjecture:$$\log G_m(1+z)= \gamma_m(z) +(-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1}$$Or, equivalently, by exponentiating both sides$$G_m(1+z)= \text{exp} (\gamma_m(z)) \text{exp} \Bigg\{ (-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1} \Bigg\}$$Looking at the Taylor series results for $$G_1(1+z)$$ and $$G_2(1+z)$$ above, this implies that$$\gamma_1(z)= \gamma z$$$$\gamma_2(z)= \frac{z}{2}\log(2\pi) - \left( \frac{z+z^2(1+\gamma )}{2} \right)$$These are consistent with the idea that$$G_m(1+z)= \text{exp} (\gamma_m(z)) \text{exp} \Bigg\{ (-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1} \Bigg\}$$Since this implies that$$G_1(1+z)=e^{\gamma z} \text{exp} \Bigg\{ - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \Bigg\}$$and$$G_2(1+z)=$$

$$\text{exp} \left(\frac{z}{2}\log 2\pi- \frac{z+z^2(1+\gamma )}{2} \right) \Bigg\{ -
\sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} \Bigg\}$$Conversely,$$G_1(1+z)= e^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k} = $$

$$\text{exp}(\gamma_1(z)) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$and$$G_2(1+z)=G(1+z) = $$

$$(2\pi)^{z/2} \text{exp}\left(- \frac{z+z^2(1+\gamma )}{2} \right) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)=$$$$\text{exp}(\gamma_2(z)) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)$$And therein lies the problem, as I've not been able to find a way of evaluating $$\gamma_{m \ge 3}(z)$$, or proving that$$\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}= \text{exp} \Bigg\{ - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \Bigg\}
$$and$$\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)= \text{exp} \Bigg\{ \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} \Bigg\}$$Any ideas, folks...?

 

[chisigma]

Re: A conjecture on a Genralized Barnes' function - can anyone help?

This is NOT a tutorial, so any and all contributions are very much welcome... :DI've recently been working on the Barnes' function - see tutorial in Math Notes board - and been trying to generalize some of my results to higher order Barnes' functions (intimately connected with the Multiple Gamma functions, $$\Gamma_n(z)$$ ). If we define the Generalized Barnes' function

by$$G_1(z) = \frac{1}{\Gamma(z)}$$

$$G_2(z) = \frac{1}{\Gamma_2(z)} = G(z) \quad [ \text{the regular Barnes function} ]$$

$$G_3(z) = \frac{1}{\Gamma_3(z)}$$etc, and$$G_n(1) = 1$$

$$G_{n+1}(1+z)=G_n(z) G_{n+1}(z)$$
My main aim has been to find an infinite product representation for the Triple Barnes' Function, $$G_3(1+z)$$, and even higher order Multiple Barnes' functions. The following results are known:$$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$$$\Rightarrow$$$$G_1(1+z)=\frac{1}{\Gamma(1+z)} =\frac{1}{z\Gamma(z)}= e^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$and$$G_2(1+z)=G(1+z) = $$

$$(2\pi)^{z/2} \text{exp}\left(- \frac{z+z^2(1+\gamma )}{2} \right) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)$$
In the Barnes' function tutorial, I presented the following Taylor series$$\log \Gamma(1+z)=-\gamma z +\sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \quad \Rightarrow$$$$\log G_1(1+z)=\gamma z - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k $$Similarly$$\log G_2(1+z)=\log G(1+z)= \frac{z}{2}\log(2\pi) - \left( \frac{z+z^2(1+\gamma )}{2} \right) + \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} $$
For various different reasons, this has led me to conjecture a family of polynomials, which I've come to call the Complementary Barnes' Function Polynomials (Barnes' Polynomials for short), $$\gamma_m(z) = \text{Polynomial of degree at most} \, m$$The central idea is this: in light of the previous two series, I've been lead to consider:The Barnes' Polynomial Conjecture:$$\log G_m(1+z)= \gamma_m(z) +(-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1}$$Or, equivalently, by exponentiating both sides$$G_m(1+z)= \text{exp} (\gamma_m(z)) \text{exp} \Bigg\{ (-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1} \Bigg\}$$Looking at the Taylor series results for $$G_1(1+z)$$ and $$G_2(1+z)$$ above, this implies that$$\gamma_1(z)= \gamma z$$$$\gamma_2(z)= \frac{z}{2}\log(2\pi) - \left( \frac{z+z^2(1+\gamma )}{2} \right)$$These are consistent with the idea that$$G_m(1+z)= \text{exp} (\gamma_m(z)) \text{exp} \Bigg\{ (-1)^m \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+m-1}z^{k+m-1} \Bigg\}$$Since this implies that$$G_1(1+z)=e^{\gamma z} \text{exp} \Bigg\{ - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \Bigg\}$$and$$G_2(1+z)=$$

$$\text{exp} \left(\frac{z}{2}\log 2\pi- \frac{z+z^2(1+\gamma )}{2} \right) \Bigg\{ -
\sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} \Bigg\}$$Conversely,$$G_1(1+z)= e^{\gamma z}\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k} = $$

$$\text{exp}(\gamma_1(z)) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$and$$G_2(1+z)=G(1+z) = $$

$$(2\pi)^{z/2} \text{exp}\left(- \frac{z+z^2(1+\gamma )}{2} \right) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)=$$$$\text{exp}(\gamma_2(z)) \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)$$And therein lies the problem, as I've not been able to find a way of evaluating $$\gamma_{m \ge 3}(z)$$, or proving that$$\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}= \text{exp} \Bigg\{ - \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}z^k \Bigg\}
$$and$$\prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^k \text{exp} \left(\frac{z^2}{2k}-z \right)= \text{exp} \Bigg\{ \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k+1}z^{k+1} \Bigg\}$$Any ideas, folks...?

In...

http://www.math.ucla.edu/~wdduke/preprints/special-jntb.pdf

... is written... $\displaystyle \gamma_{3} (z) = [\zeta^{\ '}(-1) - \frac{\ln 2 \pi}{4} + \frac{7}{24}]\ z + (\frac{\gamma + \ln 2 \pi}{4} + \frac{1}{8})\ z^{2} - (\frac{\gamma}{6} + \frac{\pi^{2}}{36} + \frac{1}{4})\ z^{3}$

Kind regards

$\chi$ $\sigma$

 

[polygamma]

Re: A conjecture on a Genralized Barnes' function - can anyone help?

$ \displaystyle \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k}$$\displaystyle \log \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right) e^{-z/k} = \sum_{k=1}^{\infty} \Big[ \log \left( 1 + \frac{z}{k} \right) - \frac{z}{k} \Big] $

$ = \displaystyle \sum_{k=1}^{\infty} \Big( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \frac{z^{n}}{k^{n}} - \frac{z}{k} \Big) = \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{n} \frac{z^{n}}{k^{n}}$

$ = \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^{n-1} z^{n}}{n} \sum_{k=1}^{\infty} \frac{1}{k^{n}} = \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \zeta(n)}{n} z^{n} $I showed the equivalence of the second one in that thread about the log gamma integral.

 

[DreamWeaver]

Thank you very much... Both of you. (Hug)

With our combined skills, it seems like an infinite product representation for $$\log G_3(1+z)$$ and $$\log G_4(1+z)$$ shouldn't be far off.

In addition, although I've not yet posted the details, assuming the Barnes' Polynomial hypothesis above, then I have already worked out reflection formulae for $$\log G_3(1+z)$$, $$\log G_4(1+z)$$, $$\log G_5(1+z)$$, and $$\log G_6(1+z)$$... These and others will follow shortly.

Watch this space! (Heidy) (Heidy) (Heidy)
ps. Sorry RV... I must seem a bit dim at times, but I was a tad tipsy when I read your previous reply - the other day - re the Barnes function. Will take another look...

 

[DreamWeaver]

A reflection formula for the Triple Barnes Function - PART 1:From the Taylor series expansion for the Barnes function $$\log G_2(1+z)$$, we have$$\int_0^z \log G_2(1+x) \,dx = $$$$\int_0^z \Bigg\{ \frac{x}{2}\log 2\pi- \left( \frac{x+x^2(1+\gamma) }{2} \right) + \sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+1)} x^{k+1} \Bigg\} \,dx=$$$$\frac{z^2}{4}\log 2\pi-\frac{z^2}{4}-\frac{z^3}{6}(1+\gamma)+ \sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+1)(k+2)} x^{k+2}=$$$$\frac{z^2}{4}\log 2\pi-\frac{z^2}{4}-\frac{z^3}{6}(1+\gamma)+$$$$z \left[ \sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+1)} x^{k+1} \right] -\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+2)} x^{k+2}=$$
$$\frac{z^2}{4}\log 2\pi-\frac{z^2}{4}-\frac{z^3}{6}(1+\gamma) -\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+2)} x^{k+2}$$$$+ z \left[ \log G_2(1+z) - \frac{z}{2}\log 2\pi + \left( \frac{z+z^2(1+\gamma) }{2} \right) \right] =$$$$\frac{z^2}{4}(1-\log 2\pi ) +\frac{z^3}{3}(1+\gamma) +z \log G_2(1+z) $$$$-\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{(k+2)} x^{k+2}$$
So, assuming that $$\log G_m(1+z) = \gamma_m(z)+ (-1)^m \sum_{k=2}^{\infty} (-1)^k\frac{\zeta(k)}{(k+m-1)} z^{k+m-1}$$This becomes$$\int_0^z \log G_2(1+x) \,dx = $$$$\frac{z^2}{4}(1-\log 2\pi ) +\frac{z^3}{3}(1+\gamma) -\gamma_3(z) +z \log G_2(1+z) +$$$$ \log G_3(1+z) $$

 

[DreamWeaver]

A reflection formula for the Triple Barnes Function - PART 2:Next, we integrate the Reflection formula for the (regular) Barnes function:$$\log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) = -x \log \left( \frac{\sin \pi x}{\pi} \right) - \frac{1}{2\pi} \text{Cl}_2(2\pi x)$$$$\Rightarrow$$$$\int_0^z \log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) \,dx=$$$$ -\int_0^z x \log \left( \frac{\sin \pi x}{\pi} \right) \,dx-\frac{1}{2\pi} \int_0^z \text{Cl}_2(2\pi x)\,dx=
$$$$\frac{z^2}{2}\log 2\pi -\int_0^z x \log \left( \sin \pi x \right) \,dx-\frac{1}{2\pi} \int_0^z \text{Cl}_2(2\pi x)\,dx= $$$$\frac{z^2}{2}\log2\pi -\int_0^z x \log \left( 2 \sin \pi x \right) \,dx-\frac{1}{2\pi} \int_0^z \text{Cl}_2(2\pi x)\,dx=$$The substitution $$y=2\pi x$$ gives$$\frac{z^2}{2}\log2\pi -\frac{1}{4\pi^2} \int_0^{2\pi z} y \log \left( 2 \sin \frac{y}{2} \right) \,dy-\frac{1}{4\pi^2 } \int_0^{2\pi z} \text{Cl}_2(y)\,dy
$$
The first integral is evaluated as$$\int_0^{2\pi z} y \log \left( 2 \sin \frac{y}{2} \right) \,dy= -y \text{Cl}_2(y) \, \Bigg|_0^{2\pi z} +\int_0^{2\pi z} \text{Cl}_2(y) \, dy=$$$$-2\pi z \text{Cl}_2(2\pi z) +\int_0^{2\pi z} \text{Cl}_2(y) \, dy$$Hence we have$$\frac{z^2}{2}\log2\pi + \frac{z}{2\pi} \text{Cl}_2(2\pi z) -\frac{1}{2\pi^2 } \int_0^{2\pi z} \text{Cl}_2(y)\,dy
$$
That last integral can be evaluated by appealing to the series definitions of the Clausen functions of orders 2 and 3, as well as the Zeta function $$\zeta(3)$$:$$ \text{Cl}_2(\varphi)= \sum_{k=1}^{\infty} \frac{\sin k \varphi}{k^2} $$$$ \text{Cl}_3(\varphi)= \sum_{k=1}^{\infty} \frac{\cos k \varphi}{k^3} $$$$ \zeta(3) = \sum_{k=1}^{\infty} \frac{1}{k^3} $$$$\Rightarrow$$$$\int_0^{2\pi z} \text{Cl}_2(x)\,dx= \sum_{k=1}^{\infty} \frac{1}{k^2} \int_0^{2\pi z} \sin k x \,dx=$$$$ \sum_{k=1}^{\infty} \frac{1}{k^2} \left[ -\frac{1}{k} \cos kx \right]_0^{2\pi z}=$$$$\sum_{k=1}^{\infty} \frac{1}{k^3} - \sum_{k=1}^{\infty} \frac{\cos 2\pi kz}{k^3}= \zeta(3)- \text{Cl}_3( 2\pi z )$$
Putting all of this together, we have the integral evaluation:$$\int_0^z \log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) \,dx=$$$$\frac{z^2}{2}\log2\pi + \frac{z}{2\pi} \text{Cl}_2(2\pi z) -\frac{\zeta(3)}{2\pi^2 }+ \frac{1}{2\pi^2 } \text{Cl}_3(2\pi z)$$

 

[DreamWeaver]

A reflection formula for the Triple Barnes function - PART 3Next, we split the integral$$\int_0^z \log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) \,dx$$into $$\int_0^z\log G_2(1+x)\,dx - \int_0^z\log G_2(1-x)\,dx \equiv $$$$\int_0^z\log G_2(1+x)\,dx + \int_0^{-z}\log G_2(1+x)\,dx $$and use$$\int_0^z \log G_2(1+x) \,dx = $$$$\frac{z^2}{4}(1-\log 2\pi ) +\frac{z^3}{3}(1+\gamma) -\gamma_3(z) +z \log G_2(1+z) +$$$$ \log G_3(1+z) $$
to obtain
$$\int_0^z \log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) \,dx=$$$$ \Bigg[ \frac{x^2}{4}(1-\log 2\pi ) +\frac{x^3}{3}(1+\gamma) -\gamma_3(x) +x \log G_2(1+x) +$$$$ \log G_3(1+x) \Bigg]_{x=z} + $$$$ \Bigg[ \frac{x^2}{4}(1-\log 2\pi ) +\frac{x^3}{3}(1+\gamma) -\gamma_3(x) +x \log G_2(1+x) +$$$$ \log G_3(1+x) \Bigg]_{x=-z} = $$
$$ \frac{z^2}{2}(1-\log 2\pi) -\gamma_3(z)-\gamma_3(-z) +z\log \left( \frac{G_2(1+z)}{G_2(1-z) } \right) + $$$$ \log G_3(1+z) + \log G_3(1-z) $$Using the Reflection formula for the Double Barnes' function this becomes$$ \frac{z^2}{2}(1-\log 2\pi) -\gamma_3(z)-\gamma_3(-z) + \log G_3(1+z) + \log G_3(1-z) $$$$-z^2 \log \left( \frac{\sin \pi z}{\pi} \right)- \frac{z}{2\pi} \text{Cl}_2(2\pi z)$$Equating this with $$\int_0^z \log \left( \frac{G_2(1+x)}{G_2(1-x)} \right) \,dx=$$$$\frac{z^2}{2}\log2\pi + \frac{z}{2\pi} \text{Cl}_2(2\pi z) -\frac{\zeta(3)}{2\pi^2 }+ \frac{1}{2\pi^2 } \text{Cl}_3(2\pi z)$$
Gives the Reflection formula:$$ (03) \quad \log G_3(1+z) + \log G_3(1-z)=$$$$\frac{z^2}{2}(2 \log2\pi -1) + \frac{z}{\pi} \text{Cl}_2(2\pi z) -\frac{\zeta(3)}{2\pi^2 }+ \frac{1}{2\pi^2 } \text{Cl}_3(2\pi z) +$$$$z^2 \log \left( \frac{\sin \pi z}{\pi} \right) + \gamma_3(z)+\gamma_3(-z) $$
(Heidy) (Heidy) (Heidy)

 

[DreamWeaver]

Reflection formula for the Triple Barnes function - PART 4
Using the value of $$\gamma_3(z)$$ that Chisigma posted above, we get$$\gamma_3(z)+\gamma_3(-z)= \frac{(1+2\gamma + 2\log 2\pi) }{4} z^2$$Plugging this into (03) gives:
$$ \log G_3(1+z) + \log G_3(1-z)=$$$$\frac{z^2}{2}(2 \log2\pi -1) + \frac{z}{\pi} \text{Cl}_2(2\pi z) -\frac{\zeta(3)}{2\pi^2 }+ \frac{1}{2\pi^2 } \text{Cl}_3(2\pi z) +$$$$z^2 \log \left( \frac{\sin \pi z}{\pi} \right) + \frac{(1+2\gamma + 2\log 2\pi) }{4} z^2 $$I'll have to work through this all again, just to make sure I haven't made any errors, but if I haven't, then setting $$z=1/2$$ and using

$$G_3(1+z) = G_2(z) \, G_3(z)$$gives$$\log G_3 \left( \frac{1}{2} \right) = $$$$\frac{3}{16 \log 2} + \frac{1}{16} \log \pi -\frac{1}{32} -\frac{7 \zeta (3) }{16 \pi^2} - \frac{1}{2} \log G_2 \left( \frac{1}{2} \right)$$Since$$\text{Cl}_2(\pi) = 0$$

and

$$\text{Cl}_3(\pi) = -\frac{3 \zeta(3)}{4}$$Furthermore, in terms of the Glaisher-Kinkelin constant A, the (regular) Barnes' function has the following known value:$$G \left( \frac{1}{2} \right) = G_2 \left( \frac{1}{2} \right) = \frac{ 2^{1/24} e^{1/8} }{ A^{3/2} \pi^{1/4 }}$$$$\Rightarrow$$$$\log G_3 \left( \frac{1}{2} \right) = $$$$\frac{3}{16 } \log 2 + \frac{1}{16} \log \pi -\frac{1}{32} -\frac{7 \zeta (3) }{16 \pi^2} - \frac{1}{2} \log \left( \frac{ 2^{1/24} e^{1/8} }{ A^{3/2} \pi^{1/4 }} \right)$$
Oh bugger! (Headbang)(Headbang)(Headbang)

It appears I've made a mistake or two somewhere, so I'll have to work through all this again tomorrow. The correct value, you see, should be:$$\frac{1}{ G_3( 1/2) } = \frac{A^{3/2} \pi^{3/16} }{ 2^{1/24} } \, \text{exp} \left( \frac{7 \, \zeta(3) }{32 \pi^2} - \frac{1}{8} \right)$$

On the plus-side, it looks like I'm not far off... (Heidy)

 

[polygamma]

I don't know where you made a mistake.

But I would like to share a way of finding the value of $G \left( \frac{1}{2} \right)$.For $\text{Re} (z) >0$, the Barnes G function has the closed form expression

$$ \log G(z+1) = \log \Gamma(z) + \log G (z) = z \log \Gamma(z) + \zeta'(-1) - \zeta(-1,z)$$

See here.So $$ \log G \left( \frac{1}{2} \right) = - \frac{ \log \pi }{4} + \zeta'(-1) - \zeta' \left( -1, \frac{1}{2} \right)$$In a challenge thread I used the Euler Maclaurin summation formula to show that $ \displaystyle \zeta'(-1) = \frac{1}{12}-\log A$.

http://mathhelpboards.com/challenge-questions-puzzles-28/euler-maclaurin-summation-formula-riemann-zeta-function-7702.html#post36441And manipulating the series definition of the Hurwitz zeta function, one finds that $ \displaystyle \zeta \left( z, \frac{1}{2} \right) = (2^z-1) \zeta(z)$.

Then

$$ \zeta' \left( z, \frac{1}{2} \right) = \log(2) 2^{z} \zeta(z) +(2^z-1) \zeta'(z)$$

$$ \implies \zeta' \left( -1, \frac{1}{2} \right) = \frac{\log 2}{2} \zeta(-1) - \frac{1}{2} \zeta'(-1) $$

So we have

$$ \log G \left(\frac{1}{2} \right) = - \frac{\log \pi}{4} + \zeta'(1) - \frac{\log 2}{2} \zeta(-1) + \frac{1}{2} \zeta'(-1)$$

$$ = - \frac{\log \pi}{4} + \frac{3}{2} \zeta'(1) - \frac{\log 2}{24} = - \frac{\log \pi}{4} + \frac{1}{8} - \frac{3}{2} \log A + \frac{\log 2}{24} $$

$$ \implies G \left( \frac{1}{2} \right) = \frac{2^{\frac{1}{24}} e^{\frac{1}{8}}}{A^{\frac{3}{2}} \pi^{\frac{1}{4}}}$$http://mathhelpboards.com/challenge-questions-puzzles-28/euler-maclaurin-summation-formula-riemann-zeta-function-7702.html#post36441

 

[DreamWeaver]

Very smooth, RV! Thanks for sharing... (Yes)