The Barnes' G-Function, and related higher functions

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In summary, the Barnes' G-function is a mathematical function used to generalize factorials and gamma functions. It has many applications in number theory, combinatorics, and statistical mechanics. The Barnes' G-function has several related higher functions, such as the multi-factorial function and the double gamma function, which extend its usefulness in various fields of mathematics. These functions have been studied extensively and have been shown to have connections to other areas of mathematics, making them an important tool in mathematical research.
  • #1
DreamWeaver
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In this tutorial we develop a number of important properties of the Barne's G-function - a close relative of the Double Gamma function (N.B. not to be confused with the digamma function!) - as well as highlight numerous intimate connections between this function, and a number of other closely-related 'higher' functions, such as the Polylogarithm, Clausen function, Gamma function, Polygamma function, Hurwitz Zeta function, the Inverse Tangent Integral, and more besides. In the process, we derive a number of closed form expressions for various arguments of the Clausen functions, an alternative closed form for the Inverse Tangent Integral, and a few lesser-known series results.

As with my other tutorials, I'll aim to prove the vast majority of relations and identities used herein - to make it mostly self-contained - but there will be a few exceptions, the first of which are the following canonical product forms for the Gamma function, respectively Barnes' G-function:\(\displaystyle (01) \quad \frac{1}{\Gamma(z)}=z e^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right) e^{-z/k}\)

\(\displaystyle (02) \quad G(1+z)=(2\pi)^{z/2} \text{exp} \left(-\frac{z+z^2(1+\gamma)}{2} \right) \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^k \text{exp} \left(\frac{z^2}{2k}-z\right)\)We will use these definitions shortly, to find various (equivalent) reflection formulae for the Barne's Function, but firstly, a more basic definition of this function is required.

For \(\displaystyle z \in \mathbb{R}\) (or \(\displaystyle \mathbb{C}\)) and \(\displaystyle n \in \mathbb{N}^{+}\), the Multiple Gamma Function \(\displaystyle \Gamma_n(z)\) is uniquely defined by the following:\(\displaystyle (03) \quad \Gamma_{n+1}(1+z)=\frac{\Gamma_{n+1}(z)}{\Gamma_n(z)}\)

\(\displaystyle (04) \quad \Gamma_1(z) \equiv \Gamma(z)\)

\(\displaystyle (05) \quad \Gamma_n(1)=1\)

\(\displaystyle (06) \quad z > 0 \Rightarrow (-1)^{n+1}\frac{d^{n+1}}{dz^{n+1}}\log\Gamma_n(z) \ge 0\)That last property is the condition of convexity, and without it, the former 3 properties would NOT define a unique set of unambiguous functions. The reasons for this are beyond the scope of this tutorial, so I'm affraid you'll just have to take my word for it. Or not... (Heidy)The Barnes' G-function can be considered as a second order extension to the factorial function x!, in much the same way as the Euler Gamma function is a first order extension:

\(\displaystyle (07) \quad \Gamma(1+x)= x \, \Gamma(x)\)

\(\displaystyle (08) \quad G(1+x)=\Gamma (x) \, G(x)\)Referring back to eqn (03) above, the connection between the Barnes' function and Double Gamma function becomes clear, since, by the definitions given:

\(\displaystyle G(z) =\frac{1}{\Gamma_2(z)}\)
----------------------------------------The vast majority of the results that follow will depend on the following parametric evaluation of the Loggamma Integral (the assumption here is that \(\displaystyle 0 < z \le 1\), although these restrictions can be relaxed):\(\displaystyle (09) \quad \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z)
\)

We prove integral (07) somewhat indirectly, by first considering the logarithmic difference:

\(\displaystyle z\log \Gamma(z)-\log G(1+z)\)Inserting the canonical product forms for the Gamma function and Barnes' function into this difference - in logarithmic form - we get:

\(\displaystyle z\log \Gamma(z)-\log G(1+z)=-z \log\left(\frac{1}{\Gamma (z)}\right)-\log G(1+z)=\)\(\displaystyle -z \left[ \log z+\gamma z +\sum_{k=1}^{\infty} \Bigg\{ \log\left(1+\frac{z}{k} \right) -\frac{z}{k} \Bigg\} \right]\)

\(\displaystyle -\left[ \frac{z}{2}\log 2\pi -\frac{z}{2}-\frac{z^2}{2} -\frac{z^2 \gamma}{2} + \sum_{k=1}^{\infty} \Bigg\{k\log\left(1+\frac{z}{k}\right) +\frac{z^2}{2k} -z \Bigg\} \right]\)After a little reordering of terms we get the following series expression:\(\displaystyle (10) \quad \sum_{k=1}^{\infty} \Bigg\{ (k+z)\log \left(1+\frac{z}{k}\right)-\frac{z^2}{2k}-z \Bigg\}=\)

\(\displaystyle -z\log z-\frac{z}{2}\log 2\pi +\frac{z}{2} +\frac{z^2}{2}- \frac{z^2 \gamma}{2}- z\log\Gamma(z) +\log G(1+z)\)Next, we once again take the logarithm of the Gamma function infinite product, and then integrate:\(\displaystyle \int_0^z\log\Gamma(x)\,dx=-\int_0^z \log\left(\frac{1}{\Gamma(x)}\right)\,dx=\)\(\displaystyle -(z\log z-z)-\frac{z^2 \gamma}{2}- \sum_{k=1}^{\infty} \Bigg\{ (k+z)\log \left(1+\frac{z}{k}\right)-\frac{z^2}{2k}-z \Bigg\}\)Equating this with (10) above, after some simplification, gives the desired evaluation:\(\displaystyle \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z)\)
More to follow shortly... (Heidy)

Questions/comments should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-barnes-g-function-related-higher-functions-7863.html
 
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  • #2
This is a commentary thread for http://mathhelpboards.com/math-notes-49/barnes-g-function-related-higher-functions-7861.html

Any comment/question regarding the thread should be noted here rather than the tutorial thread.
 
  • #3
To derive our first reflection formula for the Barnes' Function, we perform an integration by parts on the Loggamma Integral:\(\displaystyle \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z)=\)

\(\displaystyle x\log\Gamma(x) \Bigg|_0^z-\int_0^zx\psi_0(x)\,dx\)to obtain

\(\displaystyle (11) \quad \int_0^zx\psi_0(x)\,dx=\log G(1+z)- \frac{z}{2}\log 2\pi-\frac{z(1-z)}{2}\)Next, we use the definition of the Digamma function

\(\displaystyle \psi_0(x)=\frac{d}{dx}\log\Gamma(x)\)in conjunction with the Euler reflection formula for the Gamma function to obtain the reflection formula for the Digamma function:\(\displaystyle \Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin \pi x} \, \Rightarrow\)\(\displaystyle \frac{d}{dx}\left[ \log \Gamma(x) +\log \Gamma(1-x) \right]= \frac{d}{dx} \left[\log \pi-\log(\sin \pi x)\right] \, \Rightarrow\)\(\displaystyle \psi_0(x)=\psi_0(1-x)-\pi\cot\pi x\)Applying the Digamma reflection formula to our Digamma integral we get:\(\displaystyle \int_0^zx\psi_0(x)\,dx=\int_0^z\left( \psi_0(1-x) -\pi\cot \pi x\right)\,dx=\)

\(\displaystyle \int_0^zx\psi_0(1-x)\,dx-Lc(z)\)Where \(\displaystyle Lc(z)\) is simply a short-hand notation for the cotangent integral

\(\displaystyle Lc(z)=\int_0^z \pi x\cot \pi x\,dx\)To summarise so far, we have\(\displaystyle (12) \quad \int_0^zx\psi_0(1-x)\,dx-Lc(z)=\log G(1+z)- \frac{z}{2}\log 2\pi-\frac{z(1-z)}{2}\)Applying the reflection substitution \(\displaystyle x=1-y\) on that last Digamma integral gives:\(\displaystyle \int_0^zx\psi_0(1-x)\,dx=-\int_1^{1-z}(1-y)\psi_0(y)\,dy=\)

\(\displaystyle -\int_1^{1-z}\left[\frac{d}{dy} \log \Gamma(y) \right]\,dy+\int_1^{1-z}y\psi_0(y)\,dy=\)\(\displaystyle -\log \Gamma(1-z)+\int_1^{1-z}y\psi_0(y)\,dy\)Referring back to (11) above, we see that this last integral can be expressed in terms of the Barnes' G-function

\(\displaystyle -\log \Gamma(1-z)+\left[ \log G(1+y)- \frac{y}{2}\log 2\pi-\frac{y(1-y)}{2} \right]_1^{1-z}=\)\(\displaystyle -\log \Gamma(1-z)+ \left[ \log G(2-z) + \frac{z}{2}\log 2\pi -\frac{z(1-z)}{2} - \log G(2) \right]\)From the definition of the Barnes' function,

\(\displaystyle G(1+z)=\Gamma(z) G(z)\)

and

\(\displaystyle \Gamma(1)=1\)We note that \(\displaystyle G(1)=G(2)=G(3)=1\), hence \(\displaystyle \log G(2)=0\). Furthermore,

\(\displaystyle \log G(2-z)=\log\Gamma(1-z)+\log G(1-z) \, \Rightarrow\)\(\displaystyle \int_0^zx\psi_0(1-x)\,dx=\log G(1-z)+\frac{z}{2}\log 2\pi - \frac{z(1-z)}{2}\)Inserting this back into (12) gives us the first reflection formula for the Barnes function:\(\displaystyle (13) \quad \log \left( \frac{G(1-z)}{G(1+z)} \right)= -z\log 2\pi+ \int_0^z \pi x\cot \pi x\,dx\)
 
  • #4
Uh, I know it's not my job to make commentary topics for tutorial threads, but couldn't resist (or wait) to note some issued regarding the thread. I do hope the author or any moderator will put a link to this commentary in the original thread.

Now, first things first, excellent thread DW. As functional analysis is one of my interests, I was able to digest the first post in the thread. Not sure about the others though...

In the first post, DW introduced the multiple gamma and indicated that it can be defined for \(\displaystyle z \in \mathbb{C}\) too. I don't contradict it, but the 3 conditions are not sufficient to do it.

In the same post, DW also indicated the property of log-convexity. Now, at this point, I would like to note that something similar can be defined for gamma function, also called Bohr-Mollerups theorem, which is pretty well-known. Now this reminds me of an exercise I posted on MMF a long time ago, less than a year, anyway, which is on finding another analytic continuation of Gamma that is not log-convex.

Balarka
.
 
  • #5
Thanks Balara for starting the commentary thread. I wanted to thank DW. Further I think there is a typo under the integral representation EQ(09) where you said

We prove integral (07) somewhat indirectly ...

I think you meant (09) .

Keep it up , you are doing well.
 
  • #6
To obtain a second reflection formula for the Barnes' function, we replace \(\displaystyle z\) with \(\displaystyle \frac{1}{2}-z\) in (13) and use \(\displaystyle G(1+z)=\Gamma(z) G(z)\) to obtain\(\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)=-\left(\frac{1}{2}-z\right)\log

2\pi+\int_0^{1/2-z}\pi x\cot \pi x\,dx\)Next, we consider the cotangent integral, and apply the substitution \(\displaystyle x=\frac{1}{2}-y\):\(\displaystyle Lc\left(\frac{1}{2}-z\right)=\int_0^{1/2-z}\pi x\cot \pi x\,dx=\pi\int_{1/2}^z

\left(\frac{1}{2}-z\right) \tan \pi x\,dx=\)\(\displaystyle \pi\int_0^z \left(\frac{1}{2}-z\right) \tan \pi x\,dx - \pi\int_0^{1/2}

\left(\frac{1}{2}-z\right) \tan \pi x\,dx\)We keep the leftmost integral as is, and evaluate the second by means of integration by parts\(\displaystyle \frac{d}{dx}\log(\cos \pi x)=-\pi \tan \pi x \, \Rightarrow\)\(\displaystyle \pi\int_0^{1/2} \left(\frac{1}{2}-z\right) \tan \pi x\,dx=\)\(\displaystyle - \left(\frac{1}{2}-z\right) \log(\cos \pi x) \Bigg|_0^{1/2} +\int_0^{1/2}\log(\cos \pi x)\,dx= \int_0^{1/2}\log(\cos \pi x)\,dx\)Applying the substitution \(\displaystyle y=\pi x\) on the logcosine integral, in conjunction with the classic result

\(\displaystyle \int_0^{\pi/2}\log(\sin x)\,dx= \int_0^{\pi/2}\log(\cos x)\,dx= -\frac{\pi}{2}\log

2\)We obtain

\(\displaystyle \int_0^{1/2}\log(\cos \pi x)\,dx=-\frac{1}{2}\log 2\)
Hence
\(\displaystyle \log\left( \frac{ G\left(\frac{1}{2}+z\right) }{ G\left(\frac{1}{2}-z\right) } \right) =\)\(\displaystyle \log
\Gamma \left(\frac{1}{2}-z \right) -\left(\frac{1}{2}-z\right)\log 2\pi-\frac{1}{2}\log

2+\pi \int_0^z\left(x-\frac{1}{2}\right) \tan \pi x \,dx\)Finally, using the definition \(\displaystyle B_1(x)=x-\frac{1}{2}\) - for the first Bernoulli Polynomial - we arrive at our second (equivalent) reflection formula for the Barnes' function:
\(\displaystyle (14) \quad \log\left( \frac{ G\left(\frac{1}{2}+z\right) }{ G\left(\frac{1}{2}-z\right) } \right) =\)\(\displaystyle
\log \Gamma \left(\frac{1}{2}-z \right) + B_1(z) \log 2\pi-\frac{1}{2}\log 2+\pi \int_0^z B_1(x) \tan \pi x \,dx\)
 
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  • #7
From the first of the two reflection formulae - (13) above - we now obtain an explicit evaluation for the Clausen function, in terms of logarithms of the Barnes' function and Gamma function. Namely, we wish to prove the following identity (for \(\displaystyle 0 < z < 1)\):\(\displaystyle (15) \quad \text{Cl}_2(2\pi z)=\)

\(\displaystyle 2\pi \log\left( \frac{G(1-z)}{G(1+z)} \right)+2\pi z \log\left( \frac{\pi}{\sin \pi z} \right)=\)

\(\displaystyle 2\pi \log\left( \frac{G(1-z)}{G(z)} \right)-2\pi \log \Gamma(z)+2\pi z \log\left( \frac{\pi}{\sin \pi z} \right)\)The Clausen Function \(\displaystyle \text{Cl}_2(\theta)\) has (Fourier) series definition\(\displaystyle \text{Cl}_2(\theta)=\sum_{k=1}^{\infty} \frac{\sin k\theta}{k^2}\)and the integral representation\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta} \log \Bigg| 2\sin \frac{x}{2} \Bigg| \,dx\)For \(\displaystyle 0 \le \theta < 2\pi\), the absolute value sign within the integral above can be dropped.

The series definition of \(\displaystyle \text{Cl}_2(\theta)\) makes it abundantly clear that \(\displaystyle \text{Cl}_2(\pi)=\text{Cl}_2(0)=0\),

since \(\displaystyle \sin k\pi = \sin \pi = \sin 0 = 0\) for \(\displaystyle k \in \mathbb{Z}\).Starting off with the first of our reflection formulae - (13) above - we perform an integration by parts on the cotangent integral:\(\displaystyle \log \left( \frac{G(1-z)}{G(1+z)} \right)=-z\log 2\pi+ Lc(z)\)\(\displaystyle Lc(z)=\int_0^z\pi x \cot \pi x\,dx= \int_0^z x\left[\frac{d}{dx}\log(\sin \pi x)\right]\,dx=
\)\(\displaystyle z\log(\sin \pi z)-\int_0^z\log (\sin \pi x)\,dx=\)\(\displaystyle z\log(\sin \pi z)+z\log 2-\int_0^z\log (2 \sin \pi x)\,dx\)Next, we apply the substitution \(\displaystyle y=2\pi x\) on the logsine integral to obtain\(\displaystyle z\log(2\sin \pi z)-\frac{1}{2\pi}\int_0^{2\pi z}\log \left(2 \sin \frac{y}{2} \right)\,dy=\)\(\displaystyle z\log(2\sin \pi z)+\frac{1}{2\pi}\text{Cl}_2(2\pi z)\)Inserting this back into (13) we arrive at, after a little simplification, (15), and the proof is complete:\(\displaystyle \text{Cl}_2(2\pi z)=\)

\(\displaystyle 2\pi \log\left( \frac{G(1-z)}{G(1+z)} \right)+2\pi z \log\left( \frac{\pi}{\sin \pi z} \right)\)
 
  • #8
I just wanted to mention another way to derive the first reflection formula.In a http://mathhelpboards.com/challenge-questions-puzzles-28/loggamma-integral-6723.html#post34305 I showed that you can express $\log G(1+z)$ as

$$ \log G(1+z) = \frac{z}{2} \log(2 \pi) - \frac{1}{2} z(z+1) - \frac{1}{2}\gamma z^{2} +\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \zeta(k) z^{k+1}$$Then

$$\log G(1-z) = -\frac{z}{2} \log(2 \pi) + \frac{1}{2} z(1-z) - \frac{1}{2}\gamma z^{2} - \sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \zeta(k) z^{k+1}$$

So

$$ \log \frac{G(1+z)}{G(1-z)} = z \log(2 \pi) -z + 2 \sum_{k=1}^{\infty} \frac{\zeta(2k)}{2k+1} z^{2k+1} = z \log(2 \pi) + z - 2 \zeta(0) z + 2 \sum_{k=0}^{\infty} \frac{\zeta(2k)}{2k+1} z^{2k+1} $$

$$= z \log(2 \pi) + 2 \sum_{k=0}^{\infty} \zeta(2k) \int_{0}^{z} x^{2k} \ dx = z \log(2 \pi) + 2 \int_{0}^{z} \sum_{k=0}^{\infty} \zeta(2k) x^{2k} \ dx $$

$$ = z \log(2 \pi) - \pi \int_{0}^{z} x \cot (\pi x) \ dx $$
 
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  • #9
In light of (15), we are clearly ready to evaluate various arguments of the Clausen function, in terms of the Barnes' G-function. However, that being said, we can reap a slightly richer harvest if we also prove the following duplication formula for the Clausen function:\(\displaystyle \text{Cl}_2(2\varphi)=2\text{Cl}_2(\varphi)-2\text{Cl}_2(\pi-\varphi)
\)

By the integral representation of the Clausen function, within the range \(\displaystyle 0 < \phi < \pi\), we may write\(\displaystyle \text{Cl}_2(2\phi)=-\int_0^{2 \phi}\log \left( 2\sin \frac{x}{2} \right)\,dx=\)\(\displaystyle -\int_0^{2 \phi}\log \left[ \left(2\sin \frac{x}{4}\right) \left(2\cos \frac{x}{4}\right) \right]\,dx\)Since \(\displaystyle \sin 2x=2 \sin x \cos x\)Setting \(\displaystyle x=2y\) in that last integral gives\(\displaystyle -2\int_0^{\phi}\log \left[ \left(2\sin \frac{x}{2}\right) \left(2\cos \frac{x}{2}\right) \right]\,dx=\)\(\displaystyle -2\int_0^{\phi}\log\left(2\sin \frac{x}{2} \right)\,dx
-2\int_0^{\phi}\log\left(2\cos \frac{x}{2} \right)\,dx=\)\(\displaystyle 2\text{Cl}_2(\phi)-2\int_0^{\phi}\log\left(2\cos \frac{x}{2} \right)\,dx\)Finally, set \(\displaystyle x=\pi -y\) in the logcosine integral to obtain\(\displaystyle \text{Cl}_2(2\phi)=2\text{Cl}_2(\phi)-2\int_0^{\phi}\log\left(2\cos \frac{x}{2} \right)\,dx=\)\(\displaystyle 2\text{Cl}_2(\phi)+2\int_{\pi}^{\pi-\phi}\log\left(2\sin \frac{y}{2} \right)\,dy=\)\(\displaystyle 2\text{Cl}_2(\phi)-2\text{Cl}_2(\pi-\phi)+2\text{Cl}_2(\pi)=2\text{Cl}_2(\phi)-2\text{Cl}_2(\pi-\phi)\)We now have the duplication formula for the Clausen function:\(\displaystyle (15) \quad \text{Cl}_2(2\varphi)=2\text{Cl}_2(\varphi)-2\text{Cl}_2(\pi-\varphi)\)
 
  • #10
Henceforth, we define the Catalan constant in the usual way:\(\displaystyle G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}\)Setting \(\displaystyle \theta =\frac{\pi}{2}\) in the series definition of \(\displaystyle \text{Cl}_2(\theta)\) gives\(\displaystyle \text{Cl}_2\left(\frac{\pi}{2}\right)= \sum_{k=1}^{\infty}\frac{\sin(\pi k/2)}{k^2}=G\)Alternatively, setting \(\displaystyle \theta=\frac{\pi}{4}\) in the Clausen function duplication formula

gives\(\displaystyle \text{Cl}_2\left(\frac{\pi}{2}\right)=2\text{Cl}_2\left(\frac{\pi}{4}\right)-2\text{Cl}_2\left(\frac{3\pi}{4}\right)=G\)Hence\(\displaystyle \text{Cl}_2\left(\frac{\pi}{4}\right)-\text{Cl}_2\left(\frac{3\pi}{4}\right)=\frac{G}{2}\)Setting \(\displaystyle \theta = \frac{\pi}{3}\) in the Clausen function duplication formula also gives the following useful identity:\(\displaystyle \text{Cl}_2\left(\frac{2\pi}{3}\right)=\frac{2}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)\)
----------------------------------------------------The following values of the sine function are widely-known:\(\displaystyle \sin \frac{\pi}{2}=1\)

\(\displaystyle \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

\(\displaystyle \sin \frac{2\pi}{3}=\frac{\sqrt{3}}{2}\)

\(\displaystyle \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)

\(\displaystyle \sin \frac{3\pi}{4}=\frac{1}{\sqrt{2}}\)

\(\displaystyle \sin \frac{\pi}{5}=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\)

\(\displaystyle \sin \frac{2\pi}{5}=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\)

\(\displaystyle \sin \frac{3\pi}{5}=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\)

\(\displaystyle \sin \frac{4\pi}{5}=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}
\)

\(\displaystyle \sin \frac{\pi}{6}=\frac{1}{2}\)

\(\displaystyle \sin \frac{5\pi}{6}=\frac{1}{2}\)

\(\displaystyle \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\)

\(\displaystyle \sin \frac{3\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\)

\(\displaystyle \sin \frac{5\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\)

\(\displaystyle \sin \frac{7\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\)

\(\displaystyle \sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}\)

\(\displaystyle \sin \frac{3\pi}{10}=\frac{\sqrt{5}+1}{4}\)

\(\displaystyle \sin \frac{7\pi}{10}=\frac{\sqrt{5}+1}{4}\)

\(\displaystyle \sin \frac{9\pi}{10}=\frac{\sqrt{5}-1}{4}\)

\(\displaystyle \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)

\(\displaystyle \sin \frac{5\pi}{12}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)

\(\displaystyle \sin \frac{7\pi}{12}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)

\(\displaystyle \sin \frac{11\pi}{12}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
----------------------------------------------------
Explicit evaluations of the Clausen function to follow shortly, internet connection permitting... (Hug) (Hug) (Hug)
 
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  • #11
In light of the above, we have the following explicit evaluations of the Clausen function:\(\displaystyle (16) \quad \text{Cl}_2\left(\frac{\pi}{3}\right)=3\pi \log\left(

\frac{G\left(\frac{2}{3}\right)}{ G\left(\frac{1}{3}\right)} \right)-3\pi \log

\Gamma\left(\frac{1}{3}\right)+\pi \log \left(\frac{ 2\pi }{\sqrt{3}}\right)\)
\(\displaystyle (17) \quad \text{Cl}_2\left(\frac{2\pi}{3}\right)=2\pi \log\left(

\frac{G\left(\frac{2}{3}\right)}{ G\left(\frac{1}{3}\right)} \right)-2\pi \log

\Gamma\left(\frac{1}{3}\right)+\frac{2\pi}{3} \log \left(\frac{ 2\pi

}{\sqrt{3}}\right)\)
\(\displaystyle (18) \quad \text{Cl}_2\left(\frac{\pi}{4}\right)=
2\pi\log \left( \frac{G\left(\frac{7}{8}\right)}{G\left(\frac{1}{8}\right)} \right) -2\pi

\log \Gamma\left(\frac{1}{8}\right)+\frac{\pi}{4}\log \left( \frac{2\pi}{\sqrt{2-\sqrt{2}}}

\right)\)
\(\displaystyle (19) \quad \text{Cl}_2\left(\frac{3\pi}{4}\right)=
2\pi\log \left( \frac{G\left(\frac{5}{8}\right)}{G\left(\frac{3}{8}\right)} \right) -2\pi

\log \Gamma\left(\frac{3}{8}\right)+\frac{3\pi}{4}\log \left( \frac{2\pi}{\sqrt{2+\sqrt{2}}}

\right)\)
\(\displaystyle (20) \quad \text{Cl}_2\left(\frac{\pi}{6}\right)=
2\pi\log \left( \frac{G\left(\frac{11}{12}\right)}{G\left(\frac{1}{12}\right)} \right) -2\pi

\log \Gamma\left(\frac{1}{12}\right)+\frac{\pi}{6}\log \left( \frac{2\pi \sqrt{2}

}{\sqrt{3}-1} \right)\)
\(\displaystyle (21) \quad \text{Cl}_2\left(\frac{5\pi}{6}\right)=
2\pi\log \left( \frac{G\left(\frac{7}{12}\right)}{G\left(\frac{5}{12}\right)} \right) -2\pi

\log \Gamma\left(\frac{5}{12}\right)+\frac{5\pi}{6}\log \left( \frac{2\pi \sqrt{2}

}{\sqrt{3}+1} \right)\)
 
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  • #12
mathbalarka said:
Uh, I know it's not my job to make commentary topics for tutorial threads, but couldn't resist (or wait) to note some issued regarding the thread. I do hope the author or any moderator will put a link to this commentary in the original thread...

Yes, thank you for creating the commentary thread. I added the link to this thread. :D
 
  • #13
Thanks for the feedback, everyone... And a special thanks to Mathbalarka for setting up this thread. :D@ Random Variable...

That's a very nice proof of the reflection formula - you are a series Jedi... In my notes, some of which will be added shortly, I've done the same thing but in reverse, namely proved the reflection formula from the loggamma integral, and then converted that into the Zeta series.@ MathBalarka...

Once again, your help and feedback is very much appreciated! :D

Re logarithmic convexity, there is an extension to the Bohr-Mollerup theorem that can be applied to the Multiple Gamma function, but it's a bit beyond me.

In terms of the Multiple Gamma function being extended to \(\displaystyle \mathbb{C}\), I might be mistaken here, but it's my understanding that, since the double Gamma function is essentially defined as a function of the Gamma function, which itself can be extended to \(\displaystyle \mathbb{C}\), the the Double Gamma function inherits a \(\displaystyle \mathbb{C}\)-extension from the Gamma function.

More explicitly, as I understand it, the Weierstrauss infinite product for the Gamma function is used to define the Gamma function for \(\displaystyle z \in \mathbb{C}, Re(z) \ne \mathbb{Z}^{-}\), and the same holds for the Infinite product for the Barnes G-function...?

As I say though, my understanding is a little sketchy here, so if you'd like to write a bit about the uniqueness, analyticity, etc of the Barnes' G-function, I could add it to the tutorial (with full and unequivocal credit to you, of course). Totally up to you, mind, but I think it would make a great addition... (Hug)
 
  • #14
DreamWeaver said:
In terms of the Multiple Gamma function being extended to $\mathbb{C}$, I might be mistaken here, but it's my understanding that, since the double Gamma function is essentially defined as a function of the Gamma function, which itself can be extended to $\mathbb{C}$, the the Double Gamma function inherits a $\mathbb{C}$-extension from the Gamma function.

Your understanding is correct. Let me clarify my points here :

What we are basically referring to extensions are really analytic continuations, otherwise it doesn't make much sense. For example, there are lots of extensions of Riemann $\zeta$, even some beth-extensions, but that is completely useless.

Now, this analytic continuation is a strange beast. What I tried to point out in a couple of posts back is that the conditions to define a certain analytic function might not be sufficient to define an continuation, especially unique ones (i.e., analytic), in certain domain. This applies to both double/multiple or the usual gamma function. Also, bear in mind that gamma cannot be extended to whole $\mathbb{C}$ since it attains branch points around the negative integers.

DreamWeaver said:
More explicitly, as I understand it, the Weierstrauss infinite product for the Gamma function is used to define the Gamma function for $z \in \mathbb{C}, Re(z) \neq \mathbb{Z}^-$

In general, yes, but there are other continuations too. For example, take the functional equation of zeta function. In fact, cobbling up a bunch of reflection formulas, alongside with the basic properties might have some luck at specific points.

DreamWeaver said:
As I say though, my understanding is a little sketchy here, so if you'd like to write a bit about the uniqueness, analyticity, etc of the Barnes' G-function, I could add it to the tutorial (with full and unequivocal credit to you, of course). Totally up to you, mind, but I think it would make a great addition... (Hug)

I can try, but I don't see I can have time anywhere soon. I can't even complete my zeta tutorial (I am collaborating a paper with a few people, that is the main reason to it.) I'll definitely try, though.
 
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  • #15
@ mathbalarka

I'm sure you meant to say that the gamma function has poles at the negative integers (and the origin), not branch points.

And fortunately an analytic continuation is unique. So it's technically incorrect to speak of analytically continuing the gamma function in a different way.

I think what you're referring to is extending the factorials to values besides positive integers in a different way. But that's not a process of analytic continuation, and is thus not unique. And also the Barnes G function is an entire function, and that infinite product representation defines it everywhere.
 
  • #16
Random Variable" said:
I'm sure you meant to say that the gamma function has poles at the negative integers (and the origin), not branch points.

Yes, that is a typo.

Random Variable said:
And fortunately an analytic continuation is unique.

I never contradicted that.

Random Variable said:
And also the Barnes G function is an entire function, and that infinite product representation defines it everywhere.

Yes, but as I said, there are other elegant ways of continuing it.
 
  • #17
In your post from a couple of days ago, you misspoke and mentioned finding an analytic continuation of the gamma function that is not log convex. Clearly what you meant is finding an extension of the factorial that is not the gamma function, similar to how you can find an extension of the super factorial that is not the Barnes G-function.
 
  • #18
Next, we consider the derivatives of the Barnes' function. In the process, we introduce what I will henceforth call the Double Polygamma functions - the logarithmic derivatives of the Barnes' function - which are analogous to the regular Polygamma functions. These functions might well be entirely new, but I honestly don't know (I've certainly not encountered them elsewhere).

Define the Double Polygamma functions by:\(\displaystyle (22) \quad \varphi_{-1}(z)=\log G(z) \)\(\displaystyle (23) \quad \varphi_0(z)=\frac{d}{dz}\log G(z)=\frac{G'(z)}{G(z)} \)\(\displaystyle (24) \quad \varphi_{m \ge 1}(z)=\frac{d^{m+1}}{dz^{m+1}}\log G(z), \quad m= 1, 2, 3, \cdots \)Perhaps unsurprisingly, the Double Polygamma functions can be expressed as linear combinations of regular Polygamma functions.We start off by considering the parametric Loggamma integral, in the alternate form:\(\displaystyle (25) \quad \log G(1+z)= \frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi+\int_0^zx\psi_0(x)\,dx \)From the first fundamental theorem of Calculus, differentiating both sides with respect to the parameter \(\displaystyle z \) gives:\(\displaystyle \varphi_0(1+z)=\frac{G'(1+z)}{G(1+z)}=\frac{1}{2}(1+\log 2\pi )-z+z\psi_0(z) \)Conversely, since \(\displaystyle G(1+z)=\Gamma(z) G(z) \)we have\(\displaystyle \varphi_0(1+z)=\frac{d}{dz} \Gamma(z) G(z)= \Gamma '(z) G(z) +\Gamma (z) G'(z) \)Plugging this back into (25), and using \(\displaystyle \psi_0(z)=\frac{d}{dx}\log \Gamma(z)=\frac{\Gamma '(z)}{\Gamma(z)} \)gives\(\displaystyle (26) \quad \varphi_0(z)= \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \)Further differentiations yield\(\displaystyle (27) \quad \varphi_1(z)=-1+\psi_0(z)+(z-1)\psi_1(z) \)\(\displaystyle (28) \quad \varphi_2(z)= 2\psi_1(z)+(z-1)\psi_2(z) \)\(\displaystyle (29) \quad \varphi_3(z)=3\psi_2(z)+(z-1)\psi_3(z) \)\(\displaystyle (30) \quad \varphi_4(z)=4\psi_3(z)+(z-1)\psi_4(z) \)\(\displaystyle (31) \quad \varphi_5(z)=5\psi_4(z)+(z-1)\psi_5(z) \)
Much like the Gamma function, the higher order derivatives of the Barnes function quickly escalate in complexity, but at the very least, we now have an effective algorithm for evaluating them - first, in terms of Double Polygamma functions, and then, in terms of regular Polygammas:\(\displaystyle \varphi_0(z) = \frac{ G'(z) }{G(z)} \Rightarrow \)\(\displaystyle G'(z) = G(z) \varphi_0(z) = G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\} \)----------------------------------------
\(\displaystyle G''(z)=\frac{d}{dz} G(z) \varphi_0(z) = G'(z) \varphi_0(z) + G(z) \varphi_1(z)= \)\(\displaystyle G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\}^2 + G(z) \Bigg\{ -1+\psi_0(z)+(z-1)\psi_1(z) \Bigg\} \)----------------------------------------
\(\displaystyle G'''(z) = \frac{d}{dz} \Bigg\{ G'(z) \varphi_0(z) + G(z) \varphi_1(z) \Bigg\}= \)\(\displaystyle G''(z) \varphi_0(z) + 2\, G'(z) \varphi_1(z) + G(z) \varphi_2(z)= \)\(\displaystyle G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\}^3 +\)\(\displaystyle 3 \, G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\} \Bigg\{ -1+\psi_0(z)+(z-1)\psi_1(z) \Bigg\} +\)

\(\displaystyle G(z) \Bigg\{ 2\psi_1(z)+(z-1)\psi_2(z) \Bigg\}\)

----------------------------------------

More to follow shortly... My intenet connection has been a very bad mammal today... (Heidy)
 
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  • #19
Thanks MathBalarka! That helps... A lot :D

And please DON'T worry about my suggestion of writing a few bits about the Barnes' function for this 'ere tutorial... You clearly have bigger fish to fry, and not much time to fry them in.

If I can be a tad cheeky though, would you mind PMing me a link when you get your paper published? Sounds right up my street... :eek::eek::eek:

All the best!

Gethin
 
  • #20
@ Random Variable...

I hope you don't mind, but it (belatedly) strikes me that, in your Zeta-series proof of the Barnes' function reflection formula, you must (?) have either assumed or used something equivalent to the reflection formula or the parametric Loggamma integral evaluation... Perhaps?

Just wanted to check that you weren't using a modicum of circular reasoning, is all...

Would you mind explaining how you connected the series

\(\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1}\)

with the Barnes' function...?Thanks! :D
 
  • #21
In the commentary thread for this tutorial, Random Variable derived the Reflection formula for the Barnes' function via a Zeta-function series. Here, having already proved the reflection formula by other means, we use the Double Polygamma functions to derive the very same Zeta-series representation for the Barnes' function.

By Taylor's Theorem,\(\displaystyle f(a+x)= \sum_{k=0}^{\infty}\frac{x^k}{k!}f^{(k)}(x)=\)

\(\displaystyle f(a) + \frac{x}{1!}f^{(1)}(x)+ \frac{x^2}{2!}f^{(2)}(x)+ \frac{x^3}{3!}f^{(3)}(x) + \cdots \)Setting \(\displaystyle a=1\) and \(\displaystyle x=z\) in the above - with the understanding that \(\displaystyle 0 < z < 1\) - we get the following Taylor series for the logarithm of the Barnes function:\(\displaystyle \log G(1+z)=\log G(1) + \sum_{k=1}^{\infty}\frac{z^k}{k!}\varphi_{k-1}(1)\)For reasons that will become clear in a moment, we write the first two terms in the above series separately, and use \(\displaystyle \log G(1)=\log (1)=0\) to get\(\displaystyle \log G(1+z)=\frac{z}{1!} \varphi_0(1)+\frac{z^2}{2!} \varphi_1(1) + \sum_{k=3}^{\infty}\frac{z^k}{k!} \varphi_{k-1}(1)\)By appealing to the earlier evaluations of the Double Polygamma functions, and using the classic Polygamma function value \(\displaystyle \psi_0(1)=-\gamma\), where \(\displaystyle \gamma\) is the Euler-Mascheroni constant, we evaluate the first two Double Polygamma functions above to obtain:\(\displaystyle \log G(1+z)=z \left(-\frac{1}{2} +\frac{1}{2} \log 2\pi \right) +\frac{(-1-\gamma) z^2}{2} + \sum_{k=3}^{\infty}\frac{z^k}{k!}\varphi_{k-1}(1)=
\)\(\displaystyle \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{k=3}^{\infty}\frac{z^k}{k!}\varphi_{k-1}(1)\)Next, we change the summation index (and thereby lower limit of summation) via the substitution \(\displaystyle n=k-1\):\(\displaystyle \log G(1+z)= \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{n=2}^{\infty}\frac{z^{n+1}}{(n+1)!}\varphi_{n}(1)\)On the tutorial concerning the evaluation of certain types of Logarithmic Integrals, I proved the following property of the (regular) Polygamma function:\(\displaystyle \psi_{m \ge 1}(1)=(-1)^{m+1}m! \, \zeta(m+1)\)Whereas here, a closer look at the Double Polygamma functions at \(\displaystyle z=1\) gives\(\displaystyle \varphi_{m \ge 2}(1)=m\psi_{m-1}(1)=m [(-1)^m(m-1)! \, \zeta(m)]=(-1)^mm! \, \zeta(m)\)Applying this relation to our Taylor series gives the final result:
\(\displaystyle \log G(1+z)= \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1}\)
Alternatively, we can rewrite this as\(\displaystyle \log G(z) = -\log \Gamma(z)+ \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1}\)and then differentiate both sides repeatedly to obtain Zeta series for the Double Polygamma functions:\(\displaystyle \varphi_0(z)=-\frac{1}{2}(1-\log 2\pi)-z(1+\gamma) -\psi_0(z) +\sum_{k=2}^{\infty}(-1)^k \, \zeta(k)z^k\)\(\displaystyle \varphi_1(z)=-1-\gamma-\psi_1(z) + \sum_{k=2}^{\infty}(-1)^kk\, \zeta(k)z^{k-1}\)\(\displaystyle \varphi_2(z)= - \psi_2(z) + \sum_{k=2}^{\infty}(-1)^k k(k-1) \, \zeta(k)z^{k-2}\)\(\displaystyle \varphi_3(z)= - \psi_3(z) + \sum_{k=2}^{\infty}(-1)^kk(k-1)(k-2) \, \zeta(k) z^{k-3}\)\(\displaystyle \varphi_4(z)= - \psi_4(z) + \sum_{k=2}^{\infty}(-1)^kk(k-1)(k-2)(k-3) \zeta(k) z^{k-4}\)
Notice that the coefficient \(\displaystyle (k-2)\) means that the first term in the last two Taylor series vanish. Likewise, the second term in the last series vanishes due to the coefficient \(\displaystyle (k-3)\). Hence we can write these as\(\displaystyle \varphi_3(z)= - \psi_3(z) + \sum_{k=3}^{\infty}(-1)^k k(k-1)(k-2) \zeta(k) z^{k-3}\)and\(\displaystyle \varphi_4(z)= - \psi_4(z) + \sum_{k=4}^{\infty}(-1)^kk(k-1)(k-2)(k-3) \zeta(k) z^{k-4}\)respectively.
 
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  • #22
http://mathhelpboards.com/challenge-questions-puzzles-28/loggamma-integral-6723.html#post34305

The only thing I assumed was the infinite product representation.
 
  • #23
Random Variable said:
http://mathhelpboards.com/challenge-questions-puzzles-28/loggamma-integral-6723.html#post34305

The only thing I assumed was the infinite product representation.

Nice! (Muscle)

Would you mind showing how, as and when you have time...?

You're much, much better with series than I am, so I'd really like to see that.

[Sorry to be a pest! lol :eek: ]
 
  • #24
Ummm... didn't you click on the link?
 
  • #25
Doh! Massive face-palm moment... (Sleepy)

Thanks fella, but I just thought you were using rainbow text...

[runs and hides]
 
  • #26
Random Variable said:
If I can be a tad cheeky though, would you mind PMing me a link when you get your paper published? Sounds right up my street...

Sure. I will let you know when the pre-print releases on arXiv.
 
  • #27
mathbalarka said:
Sure. I will let you know when the pre-print releases on arXiv.

Thank you! (Hug)

Can't wait...
 
  • #28
On a recent thread in the Analysis board - see http://mathhelpboards.com/analysis-50/trigonometric-series-related-hurwitz-zeta-function-8097.html - I introduced the following trigonometric series:\(\displaystyle \mathscr{S}_{\infty}(z)=\sum_{k=1}^{\infty}\frac{ \log k }{k^2} \cos 2\pi kz\)The motivation behind exploring this series was simple enough; by evaluating this series for small, rational \(\displaystyle z\), where \(\displaystyle 0 < z < 1 \, \in \mathbb{Q}\), it is then possible to evaluate small rational arguments of the Barnes function (with a little help from the reflection formula evaluated in terms of the Clausen function). To arrive at this series, we integrate Kummer's Fourier expansion for the loggamma function:\(\displaystyle \log \left(\frac{ \Gamma(x) }{ \sqrt{2\pi} }\right)=\)\(\displaystyle -\frac{1}{2} \log(2\sin \pi x)+\frac{1}{2}(1-2x)(\gamma+\log 2\pi) + \frac{1}{\pi} \, \sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi k x\)Integrating Kummer's expansion term by term gives:\(\displaystyle \int_0^z\log\Gamma(x)\,dx=\)\(\displaystyle \frac{z\,(\gamma +2 \log 2\pi)}{2} -\frac{z^2\, (\gamma+\log 2\pi)}{2} +\frac{ \text{Cl}_2(2\pi z) }{4\pi} +\frac{1}{2\pi^2}\, \sum_{k=1}^{\infty}\frac{\log k}{k^2} \)\(\displaystyle - \frac{1}{2\pi^2}\, \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos 2\pi kz\)Differentiating the series expansion for the Riemann Zeta function, we get\(\displaystyle \zeta^{(n)}(x) = \frac{d^n}{dx^n}\, \sum_{k=1}^{\infty}\frac{1}{k^x} = (-1)^n\, \sum_{k=1}^{\infty} \frac{(\log k)^n}{k^x}\)Hence the loggamma integral can be written as\(\displaystyle \int_0^z\log\Gamma(x)\,dx =\)\(\displaystyle \frac{z(1-z)(\gamma + \log 2\pi)}{2} + \frac{z}{2}\log 2\pi - \frac{\zeta ' (2)}{2\pi^2} +\frac{ \text{Cl}_2(2\pi z) }{4\pi} + \frac{ \mathscr{S}_{\infty}(z) }{2\pi^2}\)We've already evaluated this integral, however, and obtained\(\displaystyle \int_0^z\log\Gamma(x)\,dx =\)\(\displaystyle \frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) - \log G(1+z)\)Equating the two evaluations gives:\(\displaystyle \log G(1+z) = \)\(\displaystyle \left[ \frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) \right]\)\(\displaystyle - \left[ \frac{z(1-z)(\gamma + \log 2\pi)}{2} + \frac{z}{2}\log 2\pi - \frac{\zeta ' (2)}{2\pi^2} +\frac{ \text{Cl}_2(2\pi z) }{4\pi} + \frac{ \mathscr{S}_{\infty}(z) }{2\pi^2} \right]= \)\(\displaystyle \frac{z(1-z)(1-\gamma - \log 2\pi)}{2} +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(2\pi z) }{4\pi}
+z\log \Gamma(z) -\frac{ \mathscr{S}_{\infty}(z) }{2\pi^2}\)Finally, an application of the functional relation \(\displaystyle G(1+z) = \Gamma(z)\, G(z)\) gives the desired result:

\(\displaystyle \log G(z) = \)\(\displaystyle \frac{z(1-z)(1-\gamma - \log 2\pi)}{2} +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(2\pi z) }{4\pi}
+(z-1)\log \Gamma(z) -\frac{ \mathscr{S}_{\infty}(z) }{2\pi^2}\)
 
  • #29
Setting \(\displaystyle z=1/2\) in the trig series above gives:\(\displaystyle \mathscr{S}_{\infty}\left(\frac{1}{2}\right)=\sum_{k=1}^{\infty}\frac{ \log k }{k^2} \cos \pi k \equiv \sum_{k=1}^{\infty} \frac{(-1)^k\, \log k }{k^2} \)This is none other than the derivative of the Eta function:\(\displaystyle \eta(x) = \left(1-2^{\, 1-x}\right)\, \zeta (x) = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^x} \quad \Rightarrow\)\(\displaystyle \eta ' (2) = - \sum_{k=1}^{\infty}\frac{(-1)^{k+1}\, \log k}{k^2} = \sum_{k=1}^{\infty}\frac{(-1)^k\, \log k}{k^2}\)Conversely,\(\displaystyle \eta ' (x) = \frac{d}{dx} \left[ \left(1-2^{\, 1-x}\right)\, \zeta (x)\right] = 2^{\, 1-x} \zeta(x)\,\log 2 + \left(1-2^{\, 1-x}\right)\, \zeta ' (x)\)Hence\(\displaystyle \eta ' (2) = \frac{\zeta (2)}{2} \, \log 2 + \frac{\zeta ' (2)}{2}\)and since \(\displaystyle \zeta(2) = \frac{\pi^2}{6}\) this becomes:\(\displaystyle \eta ' (2) = \frac{\pi^2}{12} \, \log 2 + \frac{\zeta ' (2)}{2}\)and consequently,\(\displaystyle \log G\left(\tfrac{1}{2}\right) = \)\(\displaystyle \frac{1}{8}(1-\gamma-\log 2\pi) + \frac{\zeta ' (2) }{2\pi^2}-\frac{ \text{Cl}_2(\pi) }{4\pi}-\frac{1}{2} \log\Gamma\left(\tfrac{1}{2}\right) \)\(\displaystyle - \frac{1}{2\pi^2} \left[ \frac{\pi^2}{12} \, \log 2 + \frac{\zeta ' (2)}{2} \right] = \)\(\displaystyle \frac{(1-\gamma) }{8}- \frac{1}{6}\log 2 - \frac{3}{8}\log \pi + \frac{\zeta ' (2) }{4\pi^2} \)
Since \(\displaystyle \text{Cl}_2(\pi) = 0\) and \(\displaystyle \Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi}\).

-------------------------
Barnes' function at z=1/2:
\(\displaystyle \log G\left( \tfrac{1}{2} \right) = \frac{(1-\gamma) }{8}- \frac{1}{6}\log 2 - \frac{3}{8}\log \pi + \frac{\zeta ' (2) }{4\pi^2} \)
\(\displaystyle G\left( \tfrac{1}{2} \right) = \frac{e^{(1-\gamma)/8}}{2^{1/6}\, \pi^{3/8}}\, \text{exp} \left(\frac{\zeta ' (2) }{4\pi^2}\right)\)
 
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  • #30
Next, we consider the Barnes function at \(\displaystyle z=1/3\) and \(\displaystyle z=2/3\). Setting \(\displaystyle z=1/3\) in the trig series gives:\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{3}\right) = \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{2\pi k}{3}\right) \equiv \)
\(\displaystyle \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty} \frac{\log(3k+1)}{(3k+1)^2}+
\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty} \frac{\log(3k+2)}{(3k+2)^2}+ \)\(\displaystyle \cos\left( 2\pi \right)\, \sum_{k=0}^{\infty} \frac{\log(3k+3)}{(3k+3)^2} =
\)
\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty} \frac{\log(3k+1)}{(3k+1)^2}
-\frac{1}{2}\, \sum_{k=0}^{\infty} \frac{\log(3k+2)}{(3k+2)^2}\)\(\displaystyle + \sum_{k=0}^{\infty} \frac{\log(3k+3)}{(3k+3)^2}
=\)\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log(3k+1)}{(3k+1)^2} + \frac{\log(3k+2)}{(3k+2)^2} + \frac{\log(3k+3)}{(3k+3)^2} \Bigg\}\)\(\displaystyle + \frac{3}{2} \, \sum_{k=0}^{\infty} \frac{\log(3k+3)}{(3k+3)^2}
=\) \(\displaystyle -\frac{1}{2}\, \sum_{k=1}^{\infty} \frac{\log k}{k^2} + \frac{3}{2} \, \sum_{k=0}^{\infty} \frac{\log(3k+3)}{(3k+3)^2}=\)\(\displaystyle \frac{\zeta ' (2) }{2} + \frac{3}{2} \, \sum_{k=0}^{\infty} \frac{\log(3k+3)}{(3k+3)^2}=\)\(\displaystyle \frac{\zeta ' (2) }{2} + \frac{3}{2} \, \sum_{k=0}^{\infty} \frac{1}{3^2} \Bigg\{ \frac{\log 3}{(k+1)^2} + \frac{\log(k+1)}{(k+1)^2} \Bigg\} = \)\(\displaystyle \frac{\zeta ' (2) }{2} + \frac{1}{6} \, \Bigg\{ \zeta(2) \, \log 3 - \zeta ' (2) \Bigg\} = \frac{ \zeta ' (2) }{3}+ \frac{\pi^2}{36}\, \log 3 \)For the sake of clarity, henceforth I'll refer to the following formula as The Master Formula:\(\displaystyle \log G(z) = \)\(\displaystyle \frac{z(1-z)(1-\gamma - \log 2\pi)}{2} +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(2\pi z) }{4\pi}
+(z-1)\log \Gamma(z) -\frac{ \mathscr{S}_{\infty}(z) }{2\pi^2}\)
Setting \(\displaystyle z=1/3\) in The Master Formula gives:\(\displaystyle \frac{1}{9}(1-\gamma-\log 2\pi) +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(2\pi /3) }{4\pi} - \frac{2}{3}\log\Gamma \left( \tfrac{1}{3} \right) - \frac{1}{2\pi^2} \left[ \frac{ \zeta ' (2) }{3}+ \frac{\pi^2}{36}\, \log 3 \right] = \)\(\displaystyle \frac{1}{9}(1-\gamma-\log 2\pi) -\frac{1}{72} \log 3 +\frac{\zeta ' (2)}{3\pi^2}
- \frac{ \text{Cl}_2(2\pi /3) }{4\pi} - \frac{2}{3}\log\Gamma \left( \tfrac{1}{3} \right) = \log G\left( \tfrac{1}{3} \right) \)Conversely, by the reflection formula:\(\displaystyle \log G(1-z) = \log G(z) + \log \Gamma (z) + z\log\left( \frac{\sin \pi z}{\pi} \right)+ \frac{1}{2\pi} \text{Cl}_2 (2\pi z)\)we have\(\displaystyle \log G \left( \tfrac{2}{3} \right)= \log G \left( \tfrac{1}{3} \right) + \log\Gamma \left( \tfrac{1}{3} \right) + \frac{1}{3} \log \left( \frac{ \sqrt{3} }{2\pi} \right) + \frac{ \text{Cl}_2(2\pi/3) }{2\pi}=\)
\(\displaystyle \frac{1}{9}(1-\gamma) -\frac{4}{9} \log 2\pi -\frac{11}{72} \log 3 +\frac{\zeta ' (2)}{3\pi^2}
+ \frac{ \text{Cl}_2(2\pi /3) }{4\pi} + \frac{1}{3}\log\Gamma \left( \tfrac{1}{3} \right)\)

-------------------------

Barnes' function at z=1/3 and z=2/3:
\(\displaystyle \log G\left( \tfrac{1}{3} \right) = \frac{1}{9}(1-\gamma-\log 2\pi) -\frac{1}{72} \log 3 +\frac{\zeta ' (2)}{3\pi^2}
- \frac{ \text{Cl}_2(2\pi /3) }{4\pi} - \frac{2}{3}\log\Gamma \left( \tfrac{1}{3} \right) \)
\(\displaystyle \log G\left( \tfrac{2}{3} \right) = \frac{1}{9}(1-\gamma) -\frac{4}{9} \log 2\pi -\frac{11}{72} \log 3 +\frac{\zeta ' (2)}{3\pi^2}
+ \frac{ \text{Cl}_2(2\pi /3) }{4\pi} + \frac{1}{3}\log\Gamma \left( \tfrac{1}{3} \right)\)
\(\displaystyle G\left( \tfrac{1}{3} \right) = \frac{e^{(1-\gamma)/9}}{ 3^{1/72} \, (2\pi)^{1/9} \, \Gamma \left( \tfrac{1}{3} \right)^{2/3} }
\, \, \text{exp} \left( \frac{\zeta ' (2)}{3\pi^2}
- \frac{ \text{Cl}_2(2\pi /3) }{4\pi} \right)\)
\(\displaystyle G\left( \tfrac{2}{3} \right) = \frac{e^{(1-\gamma)/9} \, \Gamma \left( \tfrac{1}{3} \right)^{1/3} }{ 3^{11/72} \, (2\pi)^{4/9}\, }
\, \, \text{exp} \left( \frac{\zeta ' (2)}{3\pi^2}
+ \frac{ \text{Cl}_2(2\pi /3) }{4\pi} \right)\)
 
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  • #31
Next, we consider the Barnes function at \(\displaystyle z=1/4\) and \(\displaystyle z=3/4\). Setting \(\displaystyle z=1/4\) in the trig series gives:\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{\pi k}{2}\right) \equiv \)In the previous 3-case, we split this into 3 sums, but in this 4-case, the first and third sums vanish, due to the factors \cos(\pi/2) and \cos(3\pi/2) respectively. Consequently, we have\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \cos \pi \, \sum_{k=0}^{\infty}\frac{\log(4k+2)}{(4k+2)^2} + \cos 2\pi \, \sum_{k=0}^{\infty}\frac{\log(4k+4)}{(4k+4)^2} = \)\(\displaystyle - \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 2}{(2k+1)^2} + \frac{\log(2k+1)}{(2k+1)^2} \Bigg\} + \sum_{k=0}^{\infty} \Bigg\{ \frac{2\log 2 }{(k+1)^2} + \frac{\log(k+1)}{(k+1)^2} \Bigg\} = \)\(\displaystyle - \chi(2)\, \log 2 + \chi ' (2) + 2\, \zeta(2)\, \log 2 - \zeta ' (2)\)Where \(\displaystyle \chi(x)\) is the Legendre Chi function:
\(\displaystyle \chi(x) = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^x}\)
By comparison with the series definition of the Riemann Zeta function, a simple calculation shows the following equivalence:\(\displaystyle \chi(x) = \zeta (x) - \frac{1}{2^x} \zeta(x) = (1-2^{-x}) \zeta(x)\)Differentiating both sides gives:\(\displaystyle \chi ' (x) = 2^{-x} \zeta (x)\, \log 2 + (1-2^{-x}) \zeta ' (x)\)Setting \(\displaystyle x=2\) gives:\(\displaystyle \chi(2) = \frac{3}{4}\zeta(2) = \frac{\pi^2}{8}\)\(\displaystyle \chi ' (2) = \frac{\zeta(2)}{4}\log 2 + \frac{3}{4}\zeta ' (2) = \frac{\pi^2}{24}\log 2+ \frac{3}{4}\zeta ' (2)\)
Substituting these values back into\(\displaystyle - \chi(2)\, \log 2 + \chi ' (2) + 2\, \zeta(2)\, \log 2 - \zeta ' (2)\)gives\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \frac{\pi^2}{4}\log 2 - \frac{\zeta ' (2)}{4} \)
Setting \(\displaystyle z=1/4\) in The Master Formula then gives:
\(\displaystyle \frac{3}{32} (1-\gamma - \log 2\pi) +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(\pi/2) }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right) -\frac{1}{2\pi^2} \left[ \frac{\pi^2}{4}\log 2 - \frac{\zeta ' (2)}{4} \right]=\)\(\displaystyle \frac{3}{32} (1-\gamma - \log \pi) - \frac{7}{32}\log 2 +\frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right)\)Employing the reflection formula gives the case where \(\displaystyle z=3/4\).

-------------------------

Barnes' function at z=1/4 and z=3/4:
\(\displaystyle \log G\left( \tfrac{1}{4} \right) = \frac{3}{32} (1-\gamma - \log \pi) - \frac{7}{32}\log 2 +\frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right)\)
\(\displaystyle \log G\left( \tfrac{3}{4} \right) = \frac{3}{32} (1-\gamma) - \frac{11}{32}\log 2\pi +\frac{5 \, \zeta ' (2)}{8\pi^2}
+ \frac{ G }{4\pi}
+\frac{1}{4} \log \Gamma \left( \tfrac{1}{4} \right)\)
\(\displaystyle G\left( \tfrac{1}{4} \right) = \frac{e^{3(1-\gamma)/32}}{2^{7/32}\, \pi^{3/32}\, \Gamma \left( \tfrac{1}{4} \right)^{3/4} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi} \right) \)
\(\displaystyle G\left( \tfrac{3}{4} \right) = \frac{e^{3(1-\gamma)/32\, } \Gamma \left( \tfrac{1}{4} \right)^{1/4} }{ (2\pi)^{11/32} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2}
+ \frac{ G }{4\pi} \right)\)
 
  • #32
Right then...! Now that we've got the 'easy' [sarcasm] cases out of the way, it's time to get a little more - erm - stressed...? lolWith the 5-case, the first thing that causes difficulty - well, actually, just more gruntwork - is that to get all four arguments, z=1/5, 2/5, 3/5, and 4/5, we need two evaluations of our trig seris, and then two applications of the reflection formula. That being the case, I'll first deal with the pair z=1/5 and z=4/5.Setting z=1/5 in our series, we now need to evaluate the sum:\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{5}\right) = \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{2\pi k}{5}\right) \equiv \)As before, we split this sum into exactly the same number of sums as the denominator of our argument (5 partial sums in this case):\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{5}\right) \equiv \)\(\displaystyle \cos\left(\frac{2\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+1)}{(5k+1)^2}+
\cos\left(\frac{4\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+2)}{(5k+2)^2}+\)\(\displaystyle \cos\left(\frac{6\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+3)}{(5k+3)^2}+
\cos\left(\frac{8\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+4)}{(5k+4)^2}+\)\(\displaystyle \cos\left(2\pi \right)\, \sum_{k=0}^{\infty}\frac{\log(5k+5)}{(5k+5)^2} \)Using a little basic trigonometry - \(\displaystyle \cos(2\pi-\theta) = \cos \theta\) - and a few special values of the Cosine (see here: Cosine: Specific values (subsection 03/02)), we can re-write this as:\(\displaystyle \cos\left( \frac{2\pi}{5} \right)\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log(5k+1)}{(5k+1)^2}+ \frac{\log(5k+4)}{(5k+4)^2} \Bigg\} +\)\(\displaystyle \cos\left( \frac{4\pi}{5} \right)\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log(5k+2)}{(5k+2)^2}+ \frac{\log(5k+3)}{(5k+3)^2} \Bigg\} +\)\(\displaystyle \frac{1}{5^2} \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5 }{(k+1)^2 } + \frac{\log (k+1) }{(k+1)^2 } \Bigg\} = \)

\(\displaystyle \frac{ \cos\left( \frac{2\pi}{5} \right)}{5^2} \, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5}{(k+1/5)^2} + \frac{\log 5}{(k+4/5)^2}+ \frac{\log(k+1/5)}{(k+1/5)^2} + \frac{\log(k+4/5)}{(k+4/5)^2} \Bigg\} +\)\(\displaystyle \frac{ \cos\left( \frac{4\pi}{5} \right)}{5^2} \, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5}{(k+2/5)^2} + \frac{\log 5}{(k+3/5)^2}+ \frac{\log(k+2/5)}{(k+2/5)^2} + \frac{\log(k+3/5)}{(k+3/5)^2} \Bigg\} +\)\(\displaystyle \frac{1}{5^2} \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5 }{(k+1)^2 } + \frac{\log (k+1) }{(k+1)^2 } \Bigg\} = \)

\(\displaystyle \frac{ (\sqrt{5}-1)}{100} \, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5}{(k+1/5)^2} + \frac{\log 5}{(k+4/5)^2}+ \frac{\log(k+1/5)}{(k+1/5)^2} + \frac{\log(k+4/5)}{(k+4/5)^2} \Bigg\} +\)\(\displaystyle -\frac{ (\sqrt{5}+1) }{100} \, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5}{(k+2/5)^2} + \frac{\log 5}{(k+3/5)^2}+ \frac{\log(k+2/5)}{(k+2/5)^2} + \frac{\log(k+3/5)}{(k+3/5)^2} \Bigg\} +\)\(\displaystyle \frac{\zeta(2)}{25}\, \log 5 - \frac{\zeta ' (2)}{25} \)
Next, we use the following series definitions of the Trigamma function and Hurwitz Zeta function:\(\displaystyle \psi_1(z) = \sum_{k=0}^{\infty}\frac{1}{(k+z)^2}\)\(\displaystyle \zeta(s,a) = \sum_{k=0}^{\infty}\frac{1}{(k+a)^s}\)Differentiating the Hurwitz Zeta function with respect to it's first argument gives:\(\displaystyle \sum_{k=0}^{\infty}\frac{\log(k+s)}{(k+a)^s} = -\zeta ' (s,a)\)We will also use the reflection formula for the Trigamma function:\(\displaystyle \psi_1(z)+\psi_1(1-z) = \pi^2 \csc^2 \pi z\)Putting all of this together, we get
\(\displaystyle \frac{ (\sqrt{5}-1)}{100} \, \Bigg\{ \psi_1\left( \tfrac{1}{5} \right)\, \log 5 + \psi_1\left( \tfrac{4}{5} \right)\, \log 5 - \zeta ' \left(2, \tfrac{1}{5} \right) - \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} +\)\(\displaystyle -\frac{ (\sqrt{5}+1) }{100} \, \Bigg\{ \psi_1\left( \tfrac{2}{5} \right)\, \log 5 + \psi_1\left( \tfrac{3}{5} \right)\, \log 5 - \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\} +\)\(\displaystyle \frac{\pi^2}{150}\, \log 5 - \frac{\zeta ' (2)}{25} \)
The first pair of Trigammas can be evaluated via the reflection formula, as can the second:\(\displaystyle \psi_1\left( \tfrac{1}{5} \right)
+ \psi_1\left( \tfrac{4}{5} \right) = \pi^2\csc^2 (\pi/5) = \frac{16\pi^2}{(\sqrt{5}+1)^2} \)\(\displaystyle \psi_1\left( \tfrac{2}{5} \right)
+ \psi_1\left( \tfrac{3}{5} \right) = \pi^2\csc^2 (2\pi/5) = \frac{16\pi^2}{(\sqrt{5}-1)^2}\)\(\displaystyle \Rightarrow\)\(\displaystyle \frac{\pi^2}{25}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 - \frac{\zeta ' (2)}{25} + \)

\(\displaystyle \frac{ (\sqrt{5}+1) }{100} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\} -
\frac{ (\sqrt{5}-1) }{100} \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\}
\)
Finally, setting \(\displaystyle z=1/5\) in The Master Formula gives:\(\displaystyle \frac{2}{25}(1-\gamma-\log 2\pi) - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + \)\(\displaystyle \frac{13\, \zeta ' (2)}{25\pi^2} - \frac{ \text{Cl}_2(2\pi /5) }{4\pi} -\frac{4}{5} \log G\left( \tfrac{1}{5} \right) + \)\(\displaystyle \frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
\)
Clearly, due to the complexity of the expression above, writing this particular argument of the Barnes' function - and that of z=4/5 - in exponential form is impractical, so only the logarithmic forms are given below.
-------------------------

Barnes' function at z=1/5 and z=4/5:
\(\displaystyle \log G\left(\tfrac{1}{5}\right) = \)\(\displaystyle \frac{2}{25}(1-\gamma-\log 2\pi) - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + \)\(\displaystyle \frac{13\, \zeta ' (2)}{25\pi^2} - \frac{ \text{Cl}_2(2\pi /5) }{4\pi} -\frac{4}{5} \log G\left( \tfrac{1}{5} \right) + \)\(\displaystyle \frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
\)\(\displaystyle \log G\left(\tfrac{4}{5}\right) = \)\(\displaystyle \frac{2}{25}(1-\gamma) +\frac{11}{50}\log 2 +\frac{3}{25} \log \pi - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + \)\(\displaystyle \frac{13\, \zeta ' (2)}{25\pi^2} + \frac{ \text{Cl}_2(2\pi /5) }{4\pi} - \frac{1}{10}\log(5-\sqrt{5} ) +\frac{1}{5} \log G\left( \tfrac{1}{5} \right) + \)\(\displaystyle \frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
\)
 
  • #33
I'll skip the two remaining arguments for the 5-case and move on to the 6-case:\(\displaystyle \mathscr{S}_{\infty}\left(\tfrac{1}{6}\right) = \sum_{k=1}^{\infty} \frac{\log
k}{k^2}\cos\left(\frac{\pi}{3}\right) \equiv\)
\(\displaystyle \cos\left( \frac{\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+1)}{(6k+1)^2} +
\cos\left( \frac{2\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+2)}{(6k+2)^2}\)\(\displaystyle \cos\left( \pi \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+3)}{(6k+3)^2} +
\cos\left( \frac{4\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+4)}{(6k+4)^2}\)\(\displaystyle \cos\left( \frac{5\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+5)}{(6k+5)^2} +
\cos\left( 2\pi \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+6)}{(6k+6)^2} = \)
\(\displaystyle \frac{1}{2} \, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log (6k+1)}{(6k+1)^2} -
\frac{\log (6k+2)}{(6k+2)^2} -
\frac{\log (6k+4)}{(6k+4)^2} +
\frac{\log (6k+5)}{(6k+5)^2}
\Bigg\} +\)\(\displaystyle +\frac{1}{9} \, \sum_{k=0}^{\infty} \Bigg\{
- \frac{\log 3}{(2k+1)^2} - \frac{\log (2k+1)}{(2k+1)^2}
+ \frac{\log 3}{(2k+2)^2} + \frac{\log (2k+2)}{(2k+2)^2}
\Bigg\}=
\)
\(\displaystyle \frac{\log 6}{72} \, \Bigg\{
\psi_1\left( \tfrac{1}{6} \right) + \psi_1\left( \tfrac{5}{6} \right)
- \psi_1\left( \tfrac{1}{3} \right) - \psi_1\left( \tfrac{2}{3} \right)
\Bigg\} +\)\(\displaystyle \frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} +\)\(\displaystyle \frac{1}{9}\, \Bigg[
\eta ' (2) - \eta(2)\, \log 3
\Bigg] \)
By the reflection formula for the Trigamma function,\(\displaystyle \psi_1\left( \tfrac{1}{6} \right) + \psi_1\left( \tfrac{5}{6} \right)
- \psi_1\left( \tfrac{1}{3} \right) - \psi_1\left( \tfrac{2}{3} \right) =\)\(\displaystyle \pi^2 \left( \csc^2(\pi/6) - \csc^2(\pi/3) \right)= \frac{8 \pi^2}{3} \)\(\displaystyle \Rightarrow\)
\(\displaystyle \frac{\pi^2}{27}\log 6 + \frac{1}{9}\, \Bigg[ \frac{\pi^2}{12}\log 2 + \frac{\zeta ' (2)}{2} -\frac{\pi^2}{12}\log 3 \Bigg] +\)\(\displaystyle \frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} =\)
\(\displaystyle \frac{\pi^2}{9}\log 2 + \frac{5\pi^2}{108}\log 3+ \frac{\zeta ' (2)}{18} +\)\(\displaystyle \frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} \)For the sake of brevity, I'll use the shorthand notation\(\displaystyle \kappa_1 (6) =
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)\)

Setting z=1/6 in The Master Formula gives:\(\displaystyle \log G\left( \tfrac{1}{6} \right) = \)\(\displaystyle \frac{5}{72}(1-\gamma-\log \pi)-\frac{5}{72}\log 2 + \frac{\zeta ' (2)}{2\pi^2} -\frac{\text{Cl}_2(\pi/3)}{4\pi} -\frac{5}{6}\log \Gamma\left( \tfrac{1}{6} \right)\)\(\displaystyle -\frac{1}{2\pi^2}\, \left[ \frac{\pi^2}{9}\log 2 + \frac{5\pi^2}{108}\log 3+ \frac{\zeta ' (2)}{18} + \frac{\kappa_1(6)}{72} \right] =\)\(\displaystyle \frac{5}{72}(1-\gamma-\log \pi)-\frac{1}{8}\log 2 -\frac{5}{216}\log 3+ \frac{17 \, \zeta ' (2)}{36\pi^2} -\frac{\text{Cl}_2(\pi/3)}{4\pi}\)\(\displaystyle -\frac{5}{6}\log \Gamma\left( \tfrac{1}{6} \right) - \frac{\kappa_1(6)}{144\pi^2} \)

-------------------------

Barnes' function at z=1/6 and z=5/6:\(\displaystyle \log G\left( \tfrac{1}{6} \right) = \)\(\displaystyle \frac{5}{72}(1-\gamma-\log \pi)-\frac{1}{8}\log 2 -\frac{5}{216}\log 3+ \frac{17 \, \zeta ' (2)}{36\pi^2} -\frac{\text{Cl}_2(\pi/3)}{4\pi} \)\(\displaystyle -\frac{5}{6}\log \Gamma\left( \tfrac{1}{6} \right) - \frac{\kappa_1(6)}{144\pi^2} \)
\(\displaystyle \log G\left( \tfrac{5}{6} \right) = \)\(\displaystyle \frac{5}{72}(1-\gamma)-\frac{7}{24}\log 2 -\frac{5}{216}\log 3 - \frac{17}{72}\log \pi+ \frac{17 \, \zeta ' (2)}{36\pi^2} +\frac{\text{Cl}_2(\pi/3)}{4\pi} \)\(\displaystyle +\frac{1}{6}\log \Gamma\left( \tfrac{1}{6} \right) - \frac{\kappa_1(6)}{144\pi^2} \)

\(\displaystyle G\left( \tfrac{1}{6} \right) = \frac{e^{5(1-\gamma)/72}}{2^{1/8}\, 3^{5/216}\, \pi^{5/72}\, \Gamma\left( \tfrac{1}{6} \right)^{5/6} }\, \, \text{exp}
\left( \frac{17 \, \zeta ' (2)}{36\pi^2} -\frac{\text{Cl}_2(\pi/3)}{4\pi} - \frac{\kappa_1(6)}{144\pi^2} \right) \)
\(\displaystyle G\left( \tfrac{5}{6} \right) = \frac{e^{5(1-\gamma)/72}\, \Gamma\left( \tfrac{1}{6} \right)^{1/6} }{2^{7/24}\, 3^{5/216}\, \pi^{17/72}\, }\, \, \text{exp}
\left( \frac{17 \, \zeta ' (2)}{36\pi^2} +\frac{\text{Cl}_2(\pi/3)}{4\pi} - \frac{\kappa_1(6)}{144\pi^2} \right) \)
 
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  • #34
I'm wondering... Have any of you either managed to prove the multiplication formula for the Barnes' function, or even come across a not-insanely-difficult proof for it? I've had no luck with either.

And I stubbornly refuse to use it, because I can't prove it. I know, I know, it's silly, but... [ "I'm with stupid --> (Headbang) "].

Anyhoo, here it is:\(\displaystyle G(n\, z) = \)\(\displaystyle n^{n^2z^2/2-nz+5/12}(2\pi)^{(n-1)(1-nz)/2} \text{exp} \Bigg((1-n^2)\, \zeta ' (-1)\Bigg)\, \times\)\(\displaystyle \prod_{i=0}^{n-1}\, \prod_{j=0}^{n-1} G\left(z+\frac{i+j}{n}\right)\)For \(\displaystyle n\in \mathbb{N}\)
 
  • #35
I haven't thought about it much, but perhaps it can be derived by using the multiplication formula for the Hurwitz zeta function.
 

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