MHB Trigonometric series related to the Hurwitz Zeta function

DreamWeaver
Messages
297
Reaction score
0
This thread is dedicated to exploring the trigonometric series shown below.

This is NOT a tutorial, so all and any contributions would be very much welcome... (Heidy)$$\mathscr{S}_{\infty}(z)= \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos(2\pi kz)$$
This series can be expressed in terms of the Trigamma function, derivatives of the Hurwitz Zeta function, derivatives of the Riemann Zeta function, logarithms, and various mathematical constants. For certain arguments - eg $$z=1/8$$ - the Dirichlet Beta function and it's derivative also occur.I'll leave the special cases $$z=1/2,\, 1/3,\, 2/3,\, 1/4,\,$$ and $$3/4$$ - perhaps someone else would like to have a go? - and tackle the slightly trickier case z=1/5 (in the next post, that is - to follow shortly).
 
Physics news on Phys.org
Here, we consider the following special case:$$\mathscr{S}_{\infty} \left( \tfrac{1}{5} \right)= \sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos \left( \frac{2\pi k}{5} \right)$$The first thing to note is that, due to the trigonometric term, this series is somewhat periodic - after a fashion - with period 5. So, we write out the first 5 terms, and consider each of these five terms to be the initial term in one of FIVE series:$$\frac{\log 1}{1^2}\cos\left( \tfrac{2\pi}{5} \right)+ \frac{\log 2}{2^2}\cos\left(

\tfrac{4\pi}{5} \right)+ \frac{\log 3}{3^2}\cos\left( \tfrac{6\pi}{5} \right)+ \frac{\log

4}{4^2}\cos\left( \tfrac{8\pi}{5} \right)+ \frac{\log 5}{5^2}\cos\left( \tfrac{10\pi}{5}

\right)+ \, \cdots $$Notice that the last term involves a term equivalent to $$\cos (2\pi)$$. The 10th, 15th, 20th, etc terms follow this pattern. Likewise the 1st, 6th, 11th, etc all have a factor of \cos $$(2\pi/5)$$. Furthermore, it is a simple matter to show that:$$\cos \left(\frac{8\pi}{5}\right) = \cos \left(\frac{2\pi}{5}\right)$$$$\cos \left(\frac{6\pi}{5}\right) = \cos \left(\frac{4\pi}{5}\right)$$So, we can collect these five groups of congruent terms together, and write them as a sum of 5 series:
$$\cos\left( \frac{2\pi}{5} \right) \left[ \frac{\log 1}{1^2}+ \frac{\log 6}{6^2}+ \frac{\log 11}{11^2}+ \, \cdots \right] +$$$$\cos\left( \frac{4\pi}{5} \right) \left[ \frac{\log 2}{2^2}+ \frac{\log 7}{7^2}+ \frac{\log 12}{12^2}+ \, \cdots \right] +$$$$\cos\left( \frac{4\pi}{5} \right) \left[ \frac{\log 3}{3^2}+ \frac{\log 8}{8^2}+ \frac{\log 13}{13^2}+ \, \cdots \right] +$$$$\cos\left( \frac{2\pi}{5} \right) \left[ \frac{\log 4}{4^2}+ \frac{\log 9}{9^2}+ \frac{\log 14}{14^2}+ \, \cdots \right] +$$$$\cos\left( 2\pi \right) \left[ \frac{\log 5}{5^2}+ \frac{\log 10}{10^2}+ \frac{\log 15}{15^2}+ \, \cdots \right]$$
This can be written as:$$\cos\left( \frac{2\pi}{5} \right) \, \sum_{k=0}^{\infty} \frac{\log(5k+1)}{(5k+1)^2} +
\cos\left( \frac{4\pi}{5} \right) \, \sum_{k=0}^{\infty} \frac{\log(5k+2)}{(5k+2)^2} +$$$$\cos\left( \frac{4\pi}{5} \right) \, \sum_{k=0}^{\infty} \frac{\log(5k+3)}{(5k+3)^2} +
\cos\left( \frac{2\pi}{5} \right) \, \sum_{k=0}^{\infty} \frac{\log(5k+4)}{(5k+4)^2} + \sum_{k=1}^{\infty} \frac{\log(5k)}{(5k)^2}$$
Note that the fifth series starts at $$k=1$$, rather than $$k=0$$... For each and every of the first four series, we can apply the following split:$$\sum_{k=0}^{\infty} \frac{\log(5k+p)}{(5k+p)^2} = \sum_{k=0}^{\infty} \frac{\log\left[5\,
(k+\frac{p}{5}) \right]}{5^2 \left[ k+\frac{p}{5} \right]^2} =$$$$\frac{\log 5}{25} \, \sum_{k=1}^{\infty} \frac{1}{ \left(k+ \tfrac{p}{5} \right)^2 } + \frac{1}{25}\, \sum_{k=1}^{\infty} \frac{\log \left(k+ \tfrac{p}{5} \right) }{ \left(k+ \tfrac{p}{5} \right)^2 } $$Next, use the series definition for the Trigamma function:$$\psi_1(z)=\sum_{k=0}^{\infty}\frac{1}{(k+z)^2}$$to get$$\frac{\psi_1\left( \tfrac{p}{5} \right) \log 5}{25} + \frac{1}{25}\, \sum_{k=1}^{\infty} \frac{\log \left(k+ \tfrac{p}{5} \right) }{ \left(k+ \tfrac{p}{5} \right)^2 } $$

That last series can be expressed as the derivative of the Hurwitz Zeta function:$$\zeta(s,a)=\sum_{k=0}^{\infty}\frac{1}{(k+a)^s} \quad \Rightarrow$$$$\frac{d}{ds} \zeta(s,a) = \zeta'(s,a) = - \sum_{k=0}^{\infty}\frac{\log (k+a) }{(k+a)^s}$$
Applying this to our generic series above, we get
$$\frac{\psi_1\left( \tfrac{p}{5} \right) \log 5}{25} - \frac{ \zeta ' \left(2, \tfrac{p}{5} \right) }{25} $$Applying this to the first four series gives:
$$\frac{ \cos (2\pi/5) }{25} \Bigg[
\pi^2 \csc^2 \left( \tfrac{\pi}{5} \right) \, \log 5
- \zeta ' \left( 2, \tfrac{1}{5} \right)
- \zeta ' \left( 2, \tfrac{4}{5} \right)
\Bigg] + $$$$\frac{ \cos (4\pi/5) }{25} \Bigg[
\pi^2 \csc^2 \left( \tfrac{2\pi}{5} \right) \, \log 5
- \zeta ' \left( 2, \tfrac{2}{5} \right)
- \zeta ' \left( 2, \tfrac{3}{5} \right)
\Bigg] + $$$$+ \sum_{k=1}^{\infty} \frac{\log(5k)}{(5k)^2}$$
Where the Cosecant terms appear as a result of the Reflection formula for the Trigamma function:$$\psi_1(z)+\psi_1(1-z) = \pi^2\csc \pi z$$
Part 2 to follow shortly... (Heidy)
 
Last edited:
For the fifth/last series, we write:$$\sum_{k=1}^{\infty} \frac{\log(5k)}{(5k)^2} = \frac{1}{5^2} \sum_{k=1}^{\infty} \frac{(\log 5+\log k)}{k^2}= $$$$\frac{1}{5^2} \left[ \zeta(2)\, \log 5 + \sum_{k=1}^{\infty} \frac{\log k}{k^2} \right]$$Differentiating the Riemann Zeta function, we get$$\frac{d}{dx} \zeta(x) =\frac{d}{dx} \, \sum_{k=1}^{\infty} \frac{1}{k^x} = - \sum_{k=1}^{\infty} \frac{\log k}{k^x}$$Hence the fifth series is given by:$$\sum_{k=1}^{\infty} \frac{\log(5k)}{(5k)^2} = \frac{1}{25} \left[ \zeta(2)\, \log 5 - \zeta ' (2) \right] $$And the full evaluation - so far - is:$$\mathscr{S}_{\infty} \left( \tfrac{1}{5} \right)=$$$$\frac{ \cos (2\pi/5) }{25} \Bigg[
\pi^2 \csc^2 \left( \tfrac{\pi}{5} \right) \, \log 5
- \zeta ' \left( 2, \tfrac{1}{5} \right)
- \zeta ' \left( 2, \tfrac{4}{5} \right)
\Bigg] + $$$$\frac{ \cos (4\pi/5) }{25} \Bigg[
\pi^2 \csc^2 \left( \tfrac{2\pi}{5} \right) \, \log 5
- \zeta ' \left( 2, \tfrac{2}{5} \right)
- \zeta ' \left( 2, \tfrac{3}{5} \right)
\Bigg] + $$$$+ \frac{1}{25} \left[ \zeta(2)\, \log 5 - \zeta ' (2) \right] $$
Finally, we use the following special values:$$\zeta(2) = \frac{\pi^2}{6}$$$$\cos(2\pi/5) = \frac{\sqrt{5}-1}{2}$$$$\cos(4\pi/5) = -\frac{\sqrt{5}+1}{2}$$$$\sin(\pi/5) = \frac{1}{2} \sqrt{ \frac{5-\sqrt{5} }{2} }$$$$\sin(2\pi/5) = \frac{1}{2} \sqrt{ \frac{5+\sqrt{5} }{2} }$$

After marginal simplification, we now have the final evaluation:$$\sum_{k=1}^{\infty}\frac{\log k}{k^2}\, \cos \left( \frac{2\pi k}{5} \right) = \mathscr{S}_{\infty} \left( \frac{1}{5} \right)=$$$$\frac{(\sqrt{5}-1)}{50} \, \left[ \frac{8\pi^2}{ (5-\sqrt{5}) }\, \log 5 - \zeta ' \left( 2, \tfrac{1}{5} \right)
- \zeta ' \left( 2, \tfrac{4}{5} \right) \right]$$$$-\frac{(\sqrt{5}+1)}{50} \, \left[ \frac{8\pi^2}{ (5+\sqrt{5}) }\, \log 5 - \zeta ' \left( 2, \tfrac{2}{5} \right)
- \zeta ' \left( 2, \tfrac{3}{5} \right) \right]$$$$+\frac{\pi^2}{150}\, \log 5 - \frac{\zeta ' (2)}{25}$$
 
Last edited:
The 8-case - PART 1:This next one is far trickier, but the result is quite beautiful, so it's worth the extra effort.

Consider the case for $$z=1/8$$:$$\mathscr{S}_{\infty} \left( \frac{1}{8} \right)= \sum_{k=1}^{\infty}\frac{\log k}{k^2}\, \cos \left( \frac{\pi k}{4} \right)$$The previous example had a 'pseudo- periodicity' of 5, so it is unsurprising that this series has a 'period' of 8. However, every second and sixth term vanishes, due to the coefficients $$\cos(\pi/2)$$ and $$\cos(3\pi/2)$$ respectively. Hence we may write$$\mathscr{S}_{\infty} \left( \frac{1}{8} \right) =$$$$\cos \left( \frac{\pi}{4}\right) \, \sum_{k=0}^{\infty} \frac{\log(8k+1)}{(8k+1)^2} +
\cos \left( \frac{3\pi}{4}\right) \, \sum_{k=0}^{\infty} \frac{\log(8k+3)}{(8k+3)^2} +$$$$\cos \left( \frac{5\pi}{4}\right) \, \sum_{k=0}^{\infty} \frac{\log(8k+5)}{(8k+5)^2} +
\cos \left( \frac{7\pi}{4}\right) \, \sum_{k=0}^{\infty} \frac{\log(8k+7)}{(8k+7)^2}+$$$$\cos \left( \pi \right) \, \sum_{k=0}^{\infty} \frac{\log(8k+4)}{(8k+4)^2} +
\cos \left( 2\pi \right) \, \sum_{k=0}^{\infty} \frac{\log(8k+8)}{(8k+8)^2} $$Observing that $$\cos\left( \frac{3\pi}{4} \right) = - \cos\left( \frac{\pi}{4} \right)$$$$\cos\left( \frac{5\pi}{4} \right) = - \cos\left( \frac{\pi}{4} \right)$$$$\cos\left( \frac{7\pi}{4} \right) = \cos\left( \frac{\pi}{4} \right)$$The four 'odd' series can be combined into a single series:$$\cos\left( \frac{\pi}{4} \right) \, \sum_{k=0}^{\infty} \, \Bigg\{
\frac{\log(8k+1)}{(8k+1)^2} -
\frac{\log(8k+3)}{(8k+3)^2} -
\frac{\log(8k+5)}{(8k+5)^2} +
\frac{\log(8k+7)}{(8k+7)^2}
\Bigg\}$$The key observation here is that this combined series is - albeit in differentiated form - almost that of the Dirichlet Beta function:$$\beta(x) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x} = $$$$\sum_{k=0}^{\infty} \, \Bigg\{
\frac{1}{(4k+1)^x} - \frac{1}{(4k+3)^x
} \Bigg\} \equiv $$$$\sum_{k=0}^{\infty} \, \Bigg\{
\frac{1}{(8k+1)^x }
- \frac{1}{(8k+3)^x }
+ \frac{1}{(8k+5)^x }
- \frac{1}{(8k+7)^x } \Bigg\} $$The Dirichlet Beta function alternates sign $$+\, -\, +\, -\, \cdots$$, whereas the undifferentiated equivalent to our series alternates like $$+\, -\, -\, +\, \cdots$$:$$\sum_{k=0}^{\infty} \, \Bigg\{
\frac{1}{(8k+1)^x }
- \frac{1}{(8k+3)^x }
- \frac{1}{(8k+5)^x }
+ \frac{1}{(8k+7)^x } \Bigg\} $$Notice that if we add two times the third series-term and subtract two times the fourth series term we recover the Dirichlet Beta function, plus two Hurwitz Zeta functions:$$ (01) \quad \sum_{k=0}^{\infty} \, \Bigg\{
\frac{1}{(8k+1)^x }
- \frac{1}{(8k+3)^x }
- \frac{1}{(8k+5)^x }
+ \frac{1}{(8k+7)^x } \Bigg\} $$$$+\, 2 \, \sum_{k=0}^{\infty} \frac{1}{(8k+5)^x} -
2 \, \sum_{k=0}^{\infty} \frac{1}{(8k+7)^x} \equiv $$$$ (02) \quad \sum_{k=0}^{\infty} \, \Bigg\{
\frac{1}{(8k+1)^x }
- \frac{1}{(8k+3)^x }
+ \frac{1}{(8k+5)^x }
- \frac{1}{(8k+7)^x } \Bigg\} \equiv$$$$ (03) \quad \beta(x)\, +\, 2 \, \sum_{k=0}^{\infty} \frac{1}{(8k+5)^x} -
2 \, \sum_{k=0}^{\infty} \frac{1}{(8k+7)^x} $$Next, we differentiate that last expression. Notice that, by differentiating, the sign preceding those last two Hurwitz Zeta function changes: $$\beta ' (x) \, -\, 2 \, \sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^x} +
2 \, \sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^x} $$On the other hand, the differentiated form of (02) above is:$$ - \left[ \sum_{k=0}^{\infty} \, \Bigg\{
\frac{\log (8k+1) }{(8k+1)^x }
- \frac{\log (8k+3)}{(8k+3)^x }
- \frac{\log (8k+5)}{(8k+5)^x }
+ \frac{\log (8k+7)}{(8k+7)^x } \Bigg\} \right] $$Hence:$$ \left[ \sum_{k=0}^{\infty} \, \Bigg\{
\frac{\log (8k+1) }{(8k+1)^x }
- \frac{\log (8k+3)}{(8k+3)^x }
- \frac{\log (8k+5)}{(8k+5)^x }
+ \frac{\log (8k+7)}{(8k+7)^x } \Bigg\} \right] \equiv $$ $$-\beta ' (x) \, +\, 2 \, \sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^x}
- 2 \, \sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^x} $$Applying this to the series expansion for $$\mathscr{S}_{\infty}(1/8) $$, we have$$\mathscr{S}_{\infty}\left( \frac{1}{8} \right) \equiv$$ $$\cos\left(\frac{\pi}{4}\right) \, \left[ -\beta ' (2) \, +\, 2 \, \sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^2}
- 2 \, \sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^2} \right] + $$$$\cos \left( \pi \right) \, \sum_{k=0}^{\infty} \frac{\log(8k+4)}{(8k+4)^2} +
\cos \left( 2\pi \right) \, \sum_{k=0}^{\infty} \frac{\log(8k+8)}{(8k+8)^2} $$

Back in a moment for part deux... (Heidy)
 
The 8-case - PART 2:Expressing the cosines in that last expression in numerical form we have:$$\mathscr{S}_{\infty}\left( \frac{1}{8} \right) \equiv$$ $$-\frac{\beta ' (2)}{ \sqrt{2} } \, +\, \sqrt{2} \, \sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^2}
- \sqrt{2} \, \sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^2} $$$$- \, \sum_{k=0}^{\infty} \frac{\log(8k+4)}{(8k+4)^2} + \, \sum_{k=0}^{\infty} \frac{\log(8k+8)}{(8k+8)^2} $$The last two (even) series can be written as$$\frac{1}{4^2} \Bigg\{
-\sum_{k=0}^{\infty} \frac{\log[4\, (2k+1)]}{(2k+1)^2}
+\sum_{k=0}^{\infty} \frac{\log[4\, (2k+2)]}{(2k+2)^2}
\Bigg\} = $$
$$\frac{\log 4}{16}\, \Bigg\{
-\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}
+\sum_{k=0}^{\infty} \frac{1}{(2k+2)^2}
\Bigg\} $$$$+ \frac{1}{16} \Bigg\{
-\sum_{k=0}^{\infty} \frac{\log (2k+1) }{(2k+1)^2}
+\sum_{k=0}^{\infty} \frac{\log (2k+2) }{(2k+2)^2}
\Bigg\} \equiv $$
$$-\frac{\eta(2)}{8}\, \log 2 + \frac{1}{16} \Bigg\{
-\sum_{k=0}^{\infty} \frac{\log (2k+1) }{(2k+1)^2}
+\sum_{k=0}^{\infty} \frac{\log (2k+2) }{(2k+2)^2}
\Bigg\} \equiv $$Where that last step is justified by the definition of the $$Eta function$$:$$\eta(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^x} = \sum_{k=0}^{\infty} \frac{1}{(2k+1)^x}
-\sum_{k=0}^{\infty} \frac{1}{(2k+2)^x}$$Differentiating both sides gives:$$\eta ' (x) = -\sum_{k=0}^{\infty} \frac{\log (2k+1) }{(2k+1)^x}
-\sum_{k=0}^{\infty} \frac{\log (2k+2) }{(2k+2)^x}$$Hence, the sum of our two even series is given by:$$-\frac{\eta(2)}{8}\, \log 2 + \frac{\eta ' (2) }{16}$$We can re-express this partial evaluation in terms of the Riemann Zeta function, since:$$\eta(x) = \left(1-2^{\, 1-x}\right)\, \zeta(x) \quad \Rightarrow$$$$\eta'(x) = 2^{\, 1-x}\zeta(x)\, \log 2 + \left(1-2^{\, 1-x}\right)\, \zeta ' (x)$$$$ \therefore \, \zeta(2) = \frac{\pi^2}{6} \, \Rightarrow \, \eta ' (2) = \frac{\zeta(2)}{2}\, \log 2 + \frac{\zeta ' (2)}{2} = $$$$\frac{\pi^2}{12}\, \log 2 + \frac{\zeta ' (2)}{2} \, \Rightarrow $$$$-\frac{\eta(2)}{8}\, \log 2 + \frac{\eta ' (2) }{16}= $$$$-\frac{\pi^2}{96}\, \log 2 + \frac{1}{16} \Bigg( \frac{\pi^2}{12}\, \log 2 + \frac{\zeta ' (2)}{2} \Bigg) = \frac{\pi^2}{96}\, \log 2 +\frac{\zeta ' (2)}{32} $$

Hence$$\mathscr{S}_{\infty}\left( \frac{1}{8} \right) \equiv$$ $$\frac{\pi^2}{96}\, \log 2 + \frac{\zeta ' (2)}{32} \, -\frac{\beta ' (2)}{ \sqrt{2} } \, +\, \sqrt{2} \, \sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^2} $$ $$
-\, \sqrt{2} \, \sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^2} $$Nearly done! Part 3 to follow shortly... (Drunk)
 
The 8-case - PART 3:
Those last two, final series are evaluated as before - in the first example, for $$z=1/5$$ - in terms of Trigamma functions and derivatives of the Hurwitz Zeta function:$$\sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^2} = $$$$\frac{1}{8^2} \Bigg\{
\sum_{k=0}^{\infty} \frac{\log 8}{[k+(5/8)]^2} +
\sum_{k=0}^{\infty} \frac{\log [k+(5/8)]}{[k+(5/8)]^2}
\Bigg\} = $$$$\frac{1}{64} \Bigg\{ \psi_1\left( \tfrac{5}{8}\right)\, \log 8 - \zeta ' \left( 2, \tfrac{5}{8} \right) \Bigg\} = $$$$\frac{3\, \psi_1\left( \tfrac{5}{8}\right) }{64}\, \log 2 - \frac{ \zeta ' \left( 2, \tfrac{5}{8} \right) }{64}$$
Applying the same process to the other series, we have:$$\sum_{k=0}^{\infty} \frac{\log (8k+5)}{(8k+5)^2} =
\frac{3\, \psi_1\left( \tfrac{5}{8}\right) }{64}\, \log 2 - \frac{ \zeta ' \left( 2, \tfrac{5}{8} \right) }{64}$$$$\sum_{k=0}^{\infty} \frac{\log (8k+7)}{(8k+7)^2} =
\frac{3\, \psi_1\left( \tfrac{7}{8}\right) }{64}\, \log 2 - \frac{ \zeta ' \left( 2, \tfrac{7}{8} \right) }{64}$$
Our evaluation is now complete...$$\mathscr{S}_{\infty}\left( \frac{1}{8} \right) \equiv$$$$\frac{1}{32\, \sqrt{2}} \Bigg\{
3\, \psi_1\left( \tfrac{5}{8}\right)\, \log 2
-3\, \psi_1\left( \tfrac{7}{8}\right)\, \log 2
-\zeta ' \left( 2, \tfrac{5}{8} \right)
+\zeta ' \left( 2, \tfrac{7}{8} \right)
\Bigg\}$$$$+\, \frac{\pi^2}{96}\, \log 2 + \frac{\zeta ' (2)}{32} \, -\frac{\beta ' (2)}{ \sqrt{2} } $$
 
Last edited:
I got

$$
\sum_{n=1}^\infty \frac{\log(n)\cos(2\pi n x)}{n^2}=-\zeta'(2)+\pi^2 \left(x-x^2\right)\left( \log(2\pi)+\gamma -1\right) -2\pi^2 x \log \Gamma(x)+2\pi^2 \log \text{G}(x+1)+\frac{\pi}{2}\text{Cl}_2(2\pi x)
$$

where $G(z)$ denotes the Barnes G Function and $\text{Cl}_2(z)$ is the Clausen Function.

This can be derived using Kummer's Fourier Expansion of the log gamma function.
 
Last edited:
The case $x=\frac{1}{4}$ gives

$$\sum_{n=1}^\infty \frac{\log(n)}{n^2}\cos \left( \frac{\pi}{2}n\right)=\frac{\pi^2}{48}\log \left(\frac{2\pi e^{\gamma}}{A^{12}} \right)$$
 
Hey S nice generalization. I found this paper which seems interesting.
 
  • #10
Yes, that's a good paper. By the way, I worked out the case $x=\frac{1}{3}$:

$$\sum_{n=1}^\infty \frac{\log(n)}{n^2}\cos\left(\frac{2\pi}{3}n \right)=\frac{\pi^2}{36}\log\left(\frac{12\pi^2 e^{2\gamma}}{A^{24}} \right)$$
 
Last edited:
  • #11
Shobhit said:
I got

$$
\sum_{n=1}^\infty \frac{\log(n)\cos(2\pi n x)}{n^2}=-\zeta'(2)+\pi^2 \left(x-x^2\right)\left( \log(2\pi)+\gamma -1\right) -2\pi^2 x \log \Gamma(x)+2\pi^2 \log \text{G}(x+1)+\frac{\pi}{2}\text{Cl}_2(2\pi x)
$$

where $G(z)$ denotes the Barnes G Function and $\text{Cl}_2(z)$ is the Clausen Function.

This can be derived using Kummer's Fourier Expansion of the log gamma function.
Quite right there, Shobhit! That's actually why I've been studying this series for small rational arguments; as an alternative way of evaluating lesser-known arguments of the Barnes' G-function... ;)
 
  • #12
Setting $$z=1/12$$ gives:
$$\mathscr{S}_{\infty}\left(\frac{1}{12}\right)=\sum_{k=1}^{\infty}\frac{\log k}{k^2}\cos\left(\frac{\pi k}{6}\right)=$$$$\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+1)}{(12k+1)^2}+
\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+2)}{(12k+2)^2}+$$

$$\cos\left(\frac{\pi}{2}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+3)}{(12k+3)^2}+
\cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+4)}{(12k+2)^4}+$$

$$\cos\left(\frac{5\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+5)}{(12k+5)^2}+
\cos\left(\pi \right)\, \sum_{k=0}^{\infty}\frac{\log(12k+6)}{(12k+6)^2}+$$

$$\cos\left(\frac{7\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+7)}{(12k+7)^2}+
\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+8)}{(12k+8)^2}+$$

$$\cos\left(\frac{3\pi}{2}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+9)}{(12k+9)^2}+
\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+10)}{(12k+10)^2}+$$

$$\cos\left(\frac{11\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+11)}{(12k+11)^2}+
\cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+12)}{(12k+12)^2}$$Using $$\cos(\pi \pm \theta) = -\cos \theta$$ and $$\cos(2\pi - \theta) = \cos \theta$$, we express the second 6 cosines in terms of the first 6:
$$\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+1)}{(12k+1)^2}+
\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+2)}{(12k+2)^2}+$$

$$
-\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+4)}{(12k+4)^2}+$$

$$-\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+5)}{(12k+5)^2}- \sum_{k=0}^{\infty}\frac{\log(12k+6)}{(12k+6)^2}+$$

$$-\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+7)}{(12k+7)^2}-
\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+8)}{(12k+8)^2}+$$

$$
\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+10)}{(12k+10)^2}+$$

$$\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty}\frac{\log(12k+11)}{(12k+11)^2}+
\sum_{k=0}^{\infty}\frac{\log(12k+12)}{(12k+12)^2}$$Combining series with equivalent Cosine coefficients, this becomes:$$\cos\left(\frac{\pi}{6}\right)\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log(12k+1)}{(12k+1)^2} -
\frac{\log(12k+5)}{(12k+5)^2} -
\frac{\log(12k+7)}{(12k+7)^2} +
\frac{\log(12k+11)}{(12k+11)^2}
\Bigg\} +$$$$\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log(12k+2)}{(12k+2)^2} -
\frac{\log(12k+4)}{(12k+4)^2} -
\frac{\log(12k+8)}{(12k+8)^2} +
\frac{\log(12k+10)}{(12k+10)^2}
\Bigg\} +$$$$- \sum_{k=0}^{\infty}\frac{\log(12k+6)}{(12k+6)^2}+ \sum_{k=0}^{\infty} \frac{\log(12k+12)}{(12k+12)^2}=$$
$$\frac{ \cos\left(\frac{\pi}{6}\right)}{144} \, \log12 \, \Bigg\{
\psi_1\left(\tfrac{1}{12}\right)
- \psi_1\left(\tfrac{5}{12}\right)
- \psi_1\left(\tfrac{7}{12}\right)
+\psi_1\left(\tfrac{11}{12}\right)
\Bigg\}$$$$-\frac{ \cos\left(\frac{\pi}{6}\right)}{144} \Bigg\{
\zeta ' \left(2, \tfrac{1}{12} \right)
- \zeta ' \left(2, \tfrac{5}{12} \right)
- \zeta ' \left(2, \tfrac{7}{12} \right)
+ \zeta ' \left(2, \tfrac{11}{12} \right)

\Bigg\}$$$$+ \frac{ \cos\left(\frac{\pi}{3}\right)}{144} \, \log12 \, \Bigg\{
\psi_1\left(\tfrac{1}{6}\right)
- \psi_1\left(\tfrac{1}{3}\right)
- \psi_1\left(\tfrac{2}{3}\right)
+\psi_1\left(\tfrac{5}{6}\right)
\Bigg\}$$$$-\frac{ \cos\left(\frac{\pi}{3}\right)}{144} \Bigg\{
\zeta ' \left(2, \tfrac{1}{6} \right)
- \zeta ' \left(2, \tfrac{1}{3} \right)
- \zeta ' \left(2, \tfrac{2}{3} \right)
+ \zeta ' \left(2, \tfrac{5}{6} \right)

\Bigg\}$$$$-\frac{(\eta(2)\, \log 6-\eta ' (2))}{36} $$By the reflection formula for the Trigamma function, all of those Trigammas can be reduced to squared cosecants:$$\frac{ \cos\left(\frac{\pi}{6}\right)}{144} \, \log12 \, \Bigg\{
\pi^2 \csc^2 \left(\frac{\pi}{12}\right)
- \pi^2 \csc^2 \left(\frac{5\pi}{12}\right)
\Bigg\} +$$$$\frac{ \cos\left(\frac{\pi}{3}\right)}{144} \, \log12 \, \Bigg\{
\pi^2 \csc^2 \left(\frac{\pi}{6}\right)
- \pi^2 \csc^2 \left(\frac{\pi}{3}\right)
\Bigg\}$$$$-\frac{ \cos\left(\frac{\pi}{6}\right)}{144} \Bigg\{
\zeta ' \left(2, \tfrac{1}{12} \right)
- \zeta ' \left(2, \tfrac{5}{12} \right)
- \zeta ' \left(2, \tfrac{7}{12} \right)
+ \zeta ' \left(2, \tfrac{11}{12} \right)

\Bigg\}$$$$-\frac{ \cos\left(\frac{\pi}{3}\right)}{144} \Bigg\{
\zeta ' \left(2, \tfrac{1}{6} \right)
- \zeta ' \left(2, \tfrac{1}{3} \right)
- \zeta ' \left(2, \tfrac{2}{3} \right)
+ \zeta ' \left(2, \tfrac{5}{6} \right)

\Bigg\}$$$$-\frac{(\eta(2)\, \log 6-\eta ' (2))}{36} $$Expressing all of the trig terms in surd form (see Sine and Cosine here: Elementary Functions ), we get the following:
$$\frac{\pi^2}{48}\left(4+3\sqrt{6}\right) -\frac{(\eta(2)\, \log 6-\eta ' (2))}{36} $$$$-\frac{ 1}{ 96\sqrt{3} } \Bigg\{
\zeta ' \left(2, \tfrac{1}{12} \right)
- \zeta ' \left(2, \tfrac{5}{12} \right)
- \zeta ' \left(2, \tfrac{7}{12} \right)
+ \zeta ' \left(2, \tfrac{11}{12} \right)

\Bigg\}$$$$-\frac{ 1}{288} \Bigg\{
\zeta ' \left(2, \tfrac{1}{6} \right)
- \zeta ' \left(2, \tfrac{1}{3} \right)
- \zeta ' \left(2, \tfrac{2}{3} \right)
+ \zeta ' \left(2, \tfrac{5}{6} \right)

\Bigg\}$$
 

Similar threads

I Geometry of series terms of the Riemann Zeta Function
Replies
3
Views
3K
Insights The Analytic Continuation of the Lerch and the Zeta Functions
Replies
0
Views
2K
I Fractional Integral of which function is equal to Riemann's Zeta-Function?
Replies
2
Views
2K
A Laurent series for algebraic functions
Replies
9
Views
3K
Insights Computing the Riemann Zeta Function Using Fourier Series
Replies
5
Views
3K
B How does the Riemann Hypothesis/Riemann Zeta function even work?
Replies
17
Views
843
MHB A curiosity about the Riemann Zeta Function....
Replies
10
Views
4K
MHB A derivation of the duplication formula for the Barnes-G function
Replies
2
Views
3K
MHB Generalized Log-Trig series related to the Hurwitz Zeta
Replies
11
Views
3K
MHB A complex exp-series and related trig-integral
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top