# The Barnes' G-Function, and related higher functions

#### DreamWeaver

##### Well-known member
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... 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:

$$\displaystyle G(z) =\frac{1}{\Gamma_2(z)}$$

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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... http://mathhelpboards.com/commentar...g-function-related-higher-functions-7863.html

#### DreamWeaver

##### Well-known member
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:

$$\displaystyle (13) \quad \log \left( \frac{G(1-z)}{G(1+z)} \right)= -z\log 2\pi+ \int_0^z \pi x\cot \pi x\,dx$$

#### DreamWeaver

##### Well-known member
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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#### DreamWeaver

##### Well-known member
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:

$$\displaystyle \text{Cl}_2(2\pi z)=$$

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

#### DreamWeaver

##### Well-known member
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:

$$\displaystyle (15) \quad \text{Cl}_2(2\varphi)=2\text{Cl}_2(\varphi)-2\text{Cl}_2(\pi-\varphi)$$

#### DreamWeaver

##### Well-known member
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:

$$\displaystyle \text{Cl}_2\left(\frac{2\pi}{3}\right)=\frac{2}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)$$

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The following values of the sine function are widely-known:

$$\displaystyle \sin \frac{\pi}{2}=1$$

$$\displaystyle \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$$

$$\displaystyle \sin \frac{2\pi}{3}=\frac{\sqrt{3}}{2}$$

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

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

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

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

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

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

$$\displaystyle \sin \frac{\pi}{6}=\frac{1}{2}$$

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

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

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

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

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

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

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

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

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

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

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

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

$$\displaystyle \sin \frac{11\pi}{12}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$

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Explicit evaluations of the Clausen function to follow shortly, internet connection permitting...   Last edited:

#### DreamWeaver

##### Well-known member
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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#### DreamWeaver

##### Well-known member
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:

$$\displaystyle \varphi_0(z) = \frac{ G'(z) }{G(z)} \Rightarrow$$

$$\displaystyle G'(z) = G(z) \varphi_0(z) = G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\}$$

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

$$\displaystyle G''(z)=\frac{d}{dz} G(z) \varphi_0(z) = G'(z) \varphi_0(z) + G(z) \varphi_1(z)=$$

$$\displaystyle G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\}^2 + G(z) \Bigg\{ -1+\psi_0(z)+(z-1)\psi_1(z) \Bigg\}$$

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

$$\displaystyle G'''(z) = \frac{d}{dz} \Bigg\{ G'(z) \varphi_0(z) + G(z) \varphi_1(z) \Bigg\}=$$

$$\displaystyle G''(z) \varphi_0(z) + 2\, G'(z) \varphi_1(z) + G(z) \varphi_2(z)=$$

$$\displaystyle G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\}^3 +$$

$$\displaystyle 3 \, G(z) \Bigg\{ \frac{1}{2}(1+\log 2\pi )-z+(z-1)\psi_0(z) \Bigg\} \Bigg\{ -1+\psi_0(z)+(z-1)\psi_1(z) \Bigg\} +$$

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

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More to follow shortly... My intenet connection has been a very bad mammal today... Last edited:

#### DreamWeaver

##### Well-known member
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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#### DreamWeaver

##### Well-known member
On a recent thread in the Analysis board - see here - 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:

$$\displaystyle \log G(z) =$$

$$\displaystyle \frac{z(1-z)(1-\gamma - \log 2\pi)}{2} +\frac{\zeta ' (2)}{2\pi^2} - \frac{ \text{Cl}_2(2\pi z) }{4\pi} +(z-1)\log \Gamma(z) -\frac{ \mathscr{S}_{\infty}(z) }{2\pi^2}$$

#### DreamWeaver

##### Well-known member
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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#### DreamWeaver

##### Well-known member
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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#### DreamWeaver

##### Well-known member
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:

$$\displaystyle \log G\left( \tfrac{1}{4} \right) = \frac{3}{32} (1-\gamma - \log \pi) - \frac{7}{32}\log 2 +\frac{5 \, \zeta ' (2)}{8\pi^2} - \frac{ G }{4\pi} -\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right)$$

$$\displaystyle \log G\left( \tfrac{3}{4} \right) = \frac{3}{32} (1-\gamma) - \frac{11}{32}\log 2\pi +\frac{5 \, \zeta ' (2)}{8\pi^2} + \frac{ G }{4\pi} +\frac{1}{4} \log \Gamma \left( \tfrac{1}{4} \right)$$

$$\displaystyle G\left( \tfrac{1}{4} \right) = \frac{e^{3(1-\gamma)/32}}{2^{7/32}\, \pi^{3/32}\, \Gamma \left( \tfrac{1}{4} \right)^{3/4} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2} - \frac{ G }{4\pi} \right)$$

$$\displaystyle G\left( \tfrac{3}{4} \right) = \frac{e^{3(1-\gamma)/32\, } \Gamma \left( \tfrac{1}{4} \right)^{1/4} }{ (2\pi)^{11/32} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2} + \frac{ G }{4\pi} \right)$$

#### DreamWeaver

##### Well-known member
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...? lol

With 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:

$$\displaystyle \log G\left(\tfrac{1}{5}\right) =$$

$$\displaystyle \frac{2}{25}(1-\gamma-\log 2\pi) - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 +$$

$$\displaystyle \frac{13\, \zeta ' (2)}{25\pi^2} - \frac{ \text{Cl}_2(2\pi /5) }{4\pi} -\frac{4}{5} \log G\left( \tfrac{1}{5} \right) +$$

$$\displaystyle \frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}$$

$$\displaystyle \log G\left(\tfrac{4}{5}\right) =$$

$$\displaystyle \frac{2}{25}(1-\gamma) +\frac{11}{50}\log 2 +\frac{3}{25} \log \pi - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 +$$

$$\displaystyle \frac{13\, \zeta ' (2)}{25\pi^2} + \frac{ \text{Cl}_2(2\pi /5) }{4\pi} - \frac{1}{10}\log(5-\sqrt{5} ) +\frac{1}{5} \log G\left( \tfrac{1}{5} \right) +$$

$$\displaystyle \frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}$$

#### DreamWeaver

##### Well-known member
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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