The Barnes' G-Function, and related higher functions

[DreamWeaver]

Next, we consider the Barnes function at $$z=1/4$$ and $$z=3/4$$. Setting $$z=1/4$$ in the trig series gives:$$\mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{\pi k}{2}\right) \equiv $$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$$\mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \cos \pi \, \sum_{k=0}^{\infty}\frac{\log(4k+2)}{(4k+2)^2} + \cos 2\pi \, \sum_{k=0}^{\infty}\frac{\log(4k+4)}{(4k+4)^2} = $$$$- \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 2}{(2k+1)^2} + \frac{\log(2k+1)}{(2k+1)^2} \Bigg\} + \sum_{k=0}^{\infty} \Bigg\{ \frac{2\log 2 }{(k+1)^2} + \frac{\log(k+1)}{(k+1)^2} \Bigg\} = $$$$- \chi(2)\, \log 2 + \chi ' (2) + 2\, \zeta(2)\, \log 2 - \zeta ' (2)$$Where $$\chi(x)$$ is the Legendre Chi function:
$$\chi(x) = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^x}$$
By comparison with the series definition of the Riemann Zeta function, a simple calculation shows the following equivalence:$$\chi(x) = \zeta (x) - \frac{1}{2^x} \zeta(x) = (1-2^{-x}) \zeta(x)$$Differentiating both sides gives:$$\chi ' (x) = 2^{-x} \zeta (x)\, \log 2 + (1-2^{-x}) \zeta ' (x)$$Setting $$x=2$$ gives:$$\chi(2) = \frac{3}{4}\zeta(2) = \frac{\pi^2}{8}$$$$\chi ' (2) = \frac{\zeta(2)}{4}\log 2 + \frac{3}{4}\zeta ' (2) = \frac{\pi^2}{24}\log 2+ \frac{3}{4}\zeta ' (2)$$
Substituting these values back into$$- \chi(2)\, \log 2 + \chi ' (2) + 2\, \zeta(2)\, \log 2 - \zeta ' (2)$$gives$$\mathscr{S}_{\infty}\left(\tfrac{1}{4}\right) = \frac{\pi^2}{4}\log 2 - \frac{\zeta ' (2)}{4} $$
Setting $$z=1/4$$ in The Master Formula then gives:
$$\frac{3}{32} (1-\gamma - \log 2\pi) +\frac{\zeta ' (2)}{2\pi^2}
- \frac{ \text{Cl}_2(\pi/2) }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right) -\frac{1}{2\pi^2} \left[ \frac{\pi^2}{4}\log 2 - \frac{\zeta ' (2)}{4} \right]=$$$$\frac{3}{32} (1-\gamma - \log \pi) - \frac{7}{32}\log 2 +\frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right)$$Employing the reflection formula gives the case where $$z=3/4$$.

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

Barnes' function at z=1/4 and z=3/4:
$$\log G\left( \tfrac{1}{4} \right) = \frac{3}{32} (1-\gamma - \log \pi) - \frac{7}{32}\log 2 +\frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi}
-\frac{3}{4} \log \Gamma \left( \tfrac{1}{4} \right)$$
$$\log G\left( \tfrac{3}{4} \right) = \frac{3}{32} (1-\gamma) - \frac{11}{32}\log 2\pi +\frac{5 \, \zeta ' (2)}{8\pi^2}
+ \frac{ G }{4\pi}
+\frac{1}{4} \log \Gamma \left( \tfrac{1}{4} \right)$$
$$G\left( \tfrac{1}{4} \right) = \frac{e^{3(1-\gamma)/32}}{2^{7/32}\, \pi^{3/32}\, \Gamma \left( \tfrac{1}{4} \right)^{3/4} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2}
- \frac{ G }{4\pi} \right) $$
$$G\left( \tfrac{3}{4} \right) = \frac{e^{3(1-\gamma)/32\, } \Gamma \left( \tfrac{1}{4} \right)^{1/4} }{ (2\pi)^{11/32} }\, \, \text{exp} \left( \frac{5 \, \zeta ' (2)}{8\pi^2}
+ \frac{ G }{4\pi} \right)$$

 

[DreamWeaver]

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:$$\mathscr{S}_{\infty}\left(\tfrac{1}{5}\right) = \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{2\pi k}{5}\right) \equiv $$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):$$\mathscr{S}_{\infty}\left(\tfrac{1}{5}\right) \equiv $$$$\cos\left(\frac{2\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+1)}{(5k+1)^2}+
\cos\left(\frac{4\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+2)}{(5k+2)^2}+$$$$\cos\left(\frac{6\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+3)}{(5k+3)^2}+
\cos\left(\frac{8\pi}{5}\right)\, \sum_{k=0}^{\infty}\frac{\log(5k+4)}{(5k+4)^2}+$$$$\cos\left(2\pi \right)\, \sum_{k=0}^{\infty}\frac{\log(5k+5)}{(5k+5)^2} $$Using a little basic trigonometry - $$\cos(2\pi-\theta) = \cos \theta$$ - and a few special values of the Cosine (see here: Cosine: Specific values (subsection 03/02)), we can re-write this as:$$\cos\left( \frac{2\pi}{5} \right)\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log(5k+1)}{(5k+1)^2}+ \frac{\log(5k+4)}{(5k+4)^2} \Bigg\} +$$$$\cos\left( \frac{4\pi}{5} \right)\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log(5k+2)}{(5k+2)^2}+ \frac{\log(5k+3)}{(5k+3)^2} \Bigg\} +$$$$\frac{1}{5^2} \sum_{k=0}^{\infty} \Bigg\{ \frac{\log 5 }{(k+1)^2 } + \frac{\log (k+1) }{(k+1)^2 } \Bigg\} = $$

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

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

$$\frac{ (\sqrt{5}+1) }{100} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\} -
\frac{ (\sqrt{5}-1) }{100} \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\}
$$
Finally, setting $$z=1/5$$ in The Master Formula gives:$$\frac{2}{25}(1-\gamma-\log 2\pi) - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + $$$$\frac{13\, \zeta ' (2)}{25\pi^2} - \frac{ \text{Cl}_2(2\pi /5) }{4\pi} -\frac{4}{5} \log G\left( \tfrac{1}{5} \right) + $$$$\frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
$$
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:
$$\log G\left(\tfrac{1}{5}\right) = $$$$\frac{2}{25}(1-\gamma-\log 2\pi) - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + $$$$\frac{13\, \zeta ' (2)}{25\pi^2} - \frac{ \text{Cl}_2(2\pi /5) }{4\pi} -\frac{4}{5} \log G\left( \tfrac{1}{5} \right) + $$$$\frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
$$$$\log G\left(\tfrac{4}{5}\right) = $$$$\frac{2}{25}(1-\gamma) +\frac{11}{50}\log 2 +\frac{3}{25} \log \pi - \frac{1}{50}\, \left( \frac{1}{6} +\frac{16}{(\sqrt{5}+1)^3} - \frac{16}{(\sqrt{5}-1)^3} \right)\, \log 5 + $$$$\frac{13\, \zeta ' (2)}{25\pi^2} + \frac{ \text{Cl}_2(2\pi /5) }{4\pi} - \frac{1}{10}\log(5-\sqrt{5} ) +\frac{1}{5} \log G\left( \tfrac{1}{5} \right) + $$$$\frac{ (\sqrt{5}-1) }{200 \pi^2 } \, \Bigg\{ \zeta ' \left(2, \tfrac{1}{5} \right) + \zeta ' \left(2, \tfrac{4}{5} \right) \Bigg\} - \frac{ (\sqrt{5}+1) }{200\, \pi^2} \, \Bigg\{ \zeta ' \left(2, \tfrac{2}{5} \right) - \zeta ' \left(2, \tfrac{3}{5} \right) \Bigg\}
$$

 

[DreamWeaver]

I'll skip the two remaining arguments for the 5-case and move on to the 6-case:$$\mathscr{S}_{\infty}\left(\tfrac{1}{6}\right) = \sum_{k=1}^{\infty} \frac{\log
k}{k^2}\cos\left(\frac{\pi}{3}\right) \equiv$$
$$\cos\left( \frac{\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+1)}{(6k+1)^2} +
\cos\left( \frac{2\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+2)}{(6k+2)^2}$$$$\cos\left( \pi \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+3)}{(6k+3)^2} +
\cos\left( \frac{4\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+4)}{(6k+4)^2}$$$$\cos\left( \frac{5\pi}{3} \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+5)}{(6k+5)^2} +
\cos\left( 2\pi \right) \, \sum_{k=0}^{\infty} \frac{\log (6k+6)}{(6k+6)^2} = $$
$$\frac{1}{2} \, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log (6k+1)}{(6k+1)^2} -
\frac{\log (6k+2)}{(6k+2)^2} -
\frac{\log (6k+4)}{(6k+4)^2} +
\frac{\log (6k+5)}{(6k+5)^2}
\Bigg\} +$$$$+\frac{1}{9} \, \sum_{k=0}^{\infty} \Bigg\{
- \frac{\log 3}{(2k+1)^2} - \frac{\log (2k+1)}{(2k+1)^2}
+ \frac{\log 3}{(2k+2)^2} + \frac{\log (2k+2)}{(2k+2)^2}
\Bigg\}=
$$
$$\frac{\log 6}{72} \, \Bigg\{
\psi_1\left( \tfrac{1}{6} \right) + \psi_1\left( \tfrac{5}{6} \right)
- \psi_1\left( \tfrac{1}{3} \right) - \psi_1\left( \tfrac{2}{3} \right)
\Bigg\} +$$$$\frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} +$$$$\frac{1}{9}\, \Bigg[
\eta ' (2) - \eta(2)\, \log 3
\Bigg] $$
By the reflection formula for the Trigamma function,$$\psi_1\left( \tfrac{1}{6} \right) + \psi_1\left( \tfrac{5}{6} \right)
- \psi_1\left( \tfrac{1}{3} \right) - \psi_1\left( \tfrac{2}{3} \right) =$$$$\pi^2 \left( \csc^2(\pi/6) - \csc^2(\pi/3) \right)= \frac{8 \pi^2}{3} $$$$\Rightarrow$$
$$\frac{\pi^2}{27}\log 6 + \frac{1}{9}\, \Bigg[ \frac{\pi^2}{12}\log 2 + \frac{\zeta ' (2)}{2} -\frac{\pi^2}{12}\log 3 \Bigg] +$$$$\frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} =$$
$$\frac{\pi^2}{9}\log 2 + \frac{5\pi^2}{108}\log 3+ \frac{\zeta ' (2)}{18} +$$$$\frac{1}{72}\, \Bigg\{
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)
\Bigg\} $$For the sake of brevity, I'll use the shorthand notation$$\kappa_1 (6) =
\zeta ' \left(2, \tfrac{1}{3} \right) + \zeta ' \left(2, \tfrac{2}{3} \right) -
\zeta ' \left(2, \tfrac{1}{6} \right) - \zeta ' \left(2, \tfrac{5}{6} \right)$$

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

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

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

$$ G\left( \tfrac{1}{6} \right) = \frac{e^{5(1-\gamma)/72}}{2^{1/8}\, 3^{5/216}\, \pi^{5/72}\, \Gamma\left( \tfrac{1}{6} \right)^{5/6} }\, \, \text{exp}
\left( \frac{17 \, \zeta ' (2)}{36\pi^2} -\frac{\text{Cl}_2(\pi/3)}{4\pi} - \frac{\kappa_1(6)}{144\pi^2} \right) $$
$$ G\left( \tfrac{5}{6} \right) = \frac{e^{5(1-\gamma)/72}\, \Gamma\left( \tfrac{1}{6} \right)^{1/6} }{2^{7/24}\, 3^{5/216}\, \pi^{17/72}\, }\, \, \text{exp}
\left( \frac{17 \, \zeta ' (2)}{36\pi^2} +\frac{\text{Cl}_2(\pi/3)}{4\pi} - \frac{\kappa_1(6)}{144\pi^2} \right) $$

 

[DreamWeaver]

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:$$G(n\, z) = $$$$n^{n^2z^2/2-nz+5/12}(2\pi)^{(n-1)(1-nz)/2} \text{exp} \Bigg((1-n^2)\, \zeta ' (-1)\Bigg)\, \times$$$$\prod_{i=0}^{n-1}\, \prod_{j=0}^{n-1} G\left(z+\frac{i+j}{n}\right)$$For $$n\in \mathbb{N}$$

 

[polygamma]

I haven't thought about it much, but perhaps it can be derived by using the multiplication formula for the Hurwitz zeta function.

 

[DreamWeaver]

I haven't thought about it much, but perhaps it can be derived by using the multiplication formula for the Hurwitz zeta function.

I think you might be right there, RV... There's nothing scientific about it, obviously, but I 'feel' like I'm not far off it. At least that's the hope: just a bit more gruntwork and I'll accidentally stumble across it when least expected.

It's quite a formidable formula, though... (Headbang)I suspect the relation $$\log G(1+z)-z\log \Gamma(z)= \zeta ' (-1)-\zeta ' (-1,z)$$ might also offer a decent starting point.