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Generalized Log-Trig series related to the Hurwitz Zeta

DreamWeaver

Well-known member
Sep 16, 2013
337
This thread is dedicated to the study of Log-Trig series of the form:



\(\displaystyle \mathscr{S}_{(m, n)} (z) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\, \sin 2\pi k z\)


\(\displaystyle \mathscr{C}_{(m, n)} (z) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\, \cos 2\pi k z\)


Where \(\displaystyle m, n \in \mathbb{Z} \ge 1\), and \(\displaystyle 0 < z < 1 \in \mathbb{Q}\).


This is NOT a tutorial, so by all means DO chime in, if it tickles yer fancy... :D
 

DreamWeaver

Well-known member
Sep 16, 2013
337
A few preliminaries... I'll finish off the rest tomorrow... (Headbang)



\(\displaystyle \zeta(x) = \sum_{k=1}^{\infty}\frac{1}{k^x} \, \Rightarrow \, \zeta^{(m)}(x) = \frac{d^m}{dx^m} \, \sum_{k=1}^{\infty}\frac{1}{k^x} = (-1)^m\, \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^x}\)



\(\displaystyle \eta (x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^x} = \left(1-2^{1-x}\right)\, \zeta(x) = \)


\(\displaystyle \sum_{k=0}^{\infty} \frac{1}{(2k+1)^x} - \sum_{k=0}^{\infty} \frac{1}{(2k+2)^x} \Rightarrow\)


\(\displaystyle \eta^{(m)}(x) = (-1)^m \, \sum_{k=1}^{\infty} \frac{(-1)^{k+1}(\log k)}{k^x} \equiv \)


\(\displaystyle (-1)^m \sum_{k=0}^{\infty}\Bigg\{ \frac{\log^m (2k+1)}{(2k+1)^x} - \frac{\log^m (2k+2)}{(2k+2)^x} \Bigg\}\)


Furthermore,


\(\displaystyle \eta^{(m)}(x) = \frac{d^m}{dx^m} \, \Bigg\{ \zeta(x)- 2^{\,1-x}\, \zeta(x) \Bigg\}=\)


\(\displaystyle \zeta^{(m)}(x) - \sum_{j=0}^{m} 2^{\,1-x} \binom{m}{j} (-\log 2)^{m-j}\, \zeta^{(j)}(x) =\)



\(\displaystyle \left(1-2^{1-x}\right)\, \zeta^{(m)}(x) - \sum_{j=0}^{m-1} 2^{\,1-x} \binom{m}{j} (-\log
2)^{m-j}\, \zeta^{(j)}(x)\)




\(\displaystyle \beta(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^x} = \sum_{k=0}^{\infty} \Bigg\{
\frac{1}{(4k+1)^x} - \frac{1}{(4k+3)^x} \Bigg\} =\)



\(\displaystyle \sum_{k=0}^{\infty} \Bigg\{ \frac{1}{(8k+1)^x} - \frac{1}{(8k+3)^x} + \frac{1}{(8k+5)^x} -
\frac{1}{(8k+7)^x} \Bigg\}\)


\(\displaystyle \beta(x) = \frac{1}{4^x} \, \Bigg\{ \zeta\left(x,\, \tfrac{1}{4} \right) - \zeta\left(x,\,
\tfrac{3}{4} \right) \Bigg\} = \)



\(\displaystyle \frac{1}{8^x} \, \Bigg\{ \zeta\left(x,\, \tfrac{1}{8} \right) - \zeta\left(x,\, \tfrac{3}{8}
\right) + \zeta\left(x,\, \tfrac{5}{8} \right) - \zeta\left(x,\, \tfrac{7}{8} \right)
\Bigg\}\)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
A few more expressions and relations for the Dirichlet Beta function, \(\displaystyle \beta(x)\):



\(\displaystyle \beta^{(m)}(x) = (-1)^m\, \sum_{k=0}^{\infty} \frac{(-1)^k\log^m(2k+1)}{(2k+1)^x} = \)



\(\displaystyle (-1)^m\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log^m(4k+1)}{(4k+1)^x} - \frac{\log^m(4k+3)}{(4k+3)^x} \Bigg\} =\)



\(\displaystyle (-1)^m\, \sum_{k=0}^{\infty} \Bigg\{ \frac{\log^m(8k+1)}{(8k+1)^x} - \frac{\log^m(8k+3)}{(8k+3)^x} - \frac{\log^m(8k+5)}{(8k+5)^x} -
\frac{\log^m(8k+7)}{(8k+7)^x} \Bigg\}\)



\(\displaystyle \beta^{(m)}(x) = \frac{(-1)^m}{4^x}\, \sum_{j=0}^m(-1)^j\binom{m}{j}\, (2\log 2)^{m-j} \, \Bigg\{ \zeta^{(j)} \left(x,\, \tfrac{1}{4} \right) - \zeta^{(j)} \left(x,\,
\tfrac{3}{4} \right) \Bigg\} = \)



\(\displaystyle \frac{(-1)^m}{8^x} \, \sum_{j=0}^m(-1)^j\binom{m}{j}\, (3\log 2)^{m-j}\, \Bigg\{ \zeta^{(j)}\left(x,\, \tfrac{1}{8} \right) - \zeta^{(j)}\left(x,\, \tfrac{3}{8}
\right) + \zeta^{(j)}\left(x,\, \tfrac{5}{8} \right) - \zeta^{(j)}\left(x,\, \tfrac{7}{8} \right)
\Bigg\}\)




The Polygamma functions:


\(\displaystyle \psi_0(z) = -\gamma +\sum_{k=0}^{\infty} \Bigg\{ \frac{1}{k+1}-\frac{1}{k+z} \Bigg\}\)


\(\displaystyle \psi_{m \ge 1}(z) = (-1)^{m+1}m! \, \sum_{k=0}^{\infty}\frac{1}{(k+z)^{m+1}}\)


\(\displaystyle \psi_{n \ge 0}(z) + (-1)^{n+1}\, \psi_{n \ge 0}(1-z) = \frac{d^n}{dz^n}\, \pi\cot \pi z\)


\(\displaystyle \psi_{m \ge 1}(1) = (-1)^{m+1}m! \, \zeta(m+1) \)



For \(\displaystyle m \in \mathbb{Z} \ge 1\), we can write the Dirichlet Beta function as:


\(\displaystyle \beta(m) = \frac{(-1)^m}{4^m(m-1)! }\, \Bigg\{ \psi_{m-1}\left( \tfrac{1}{4} \right) - \psi_{m-1}\left( \tfrac{3}{4} \right) \Bigg\}=\)


\(\displaystyle \frac{(-1)^m}{8^m(m-1)! }\, \Bigg\{ \psi_{m-1}\left( \tfrac{1}{8} \right) - \psi_{m-1}\left( \tfrac{3}{8} \right) + \psi_{m-1}\left( \tfrac{5}{8} \right) - \psi_{m-1}\left( \tfrac{7}{8}\right)\Bigg\} \)




The Legendre Chi function:


\(\displaystyle \chi(x) = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^x} = \Bigg( 1-2^{\, -x}\Bigg)\, \zeta(x) \Rightarrow\)


\(\displaystyle \chi^{(m)}(x) = (-1)^m\, \sum_{k=0}^{\infty}\frac{\log^m(2k+1)}{(2k+1)^x}=\)


\(\displaystyle \zeta^{(m)}(x) - 2^{\, -x}\, \sum_{j=0}^m\binom{m}{j} (-\log 2)^{m-j} \zeta^{(j)}(x)\)




Nearly done wiv teh prepwork... (Heidy)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
For the sake of brevity later on, I will occasionally make use of the following Dirichlet L-series, where \(\displaystyle \chi_1\) and \(\displaystyle \chi_2\) are characters on \(\displaystyle \mathbb{Z}/8\mathbb{Z}\) defined by:


\(\displaystyle \chi_1(2k) = \chi_2(2k) \equiv 0\)


\(\displaystyle \chi_1(k) =
\begin{cases}
\, 1, & \text{if }k = 1, 3, 9, 11, \, \cdots \, \\
-1, & \text{if }k = 5, 7, 13, 15, \, \cdots \,
\end{cases}\)


\(\displaystyle \chi_2(k) =
\begin{cases}
\, 1, & \text{if }k = 1, 7, 9, 15, \, \cdots \, \\
-1, & \text{if }k = 3, 5, 11, 13, \, \cdots \,
\end{cases}\)




In terms of the characters \(\displaystyle \chi_1\) and \(\displaystyle \chi_2\) we define the following Dirichlet L-series:



\(\displaystyle L(s, \chi_1) = \sum_{k=0}^{\infty}\frac{\chi_1(k)}{k^s} = 1+\frac{1}{3^s}-\frac{1}{5^s}-\frac{1}{7^s}+ \, \cdots \)



\(\displaystyle L(s, \chi_2) = \sum_{k=0}^{\infty}\frac{\chi_2(k)}{k^s} = 1-\frac{1}{3^s}-\frac{1}{5^s}+\frac{1}{7^s}+ \, \cdots \)




\(\displaystyle L^{(m)}(s, \chi_1) = (-1)^m\,\sum_{k=0}^{\infty}\frac{\chi_1(k)\, (\log k)^m}{k^s} = \)


\(\displaystyle (-1)^m\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log^m(8k+1)}{(8k+1)^s}+
\frac{\log^m(8k+3)}{(8k+3)^s}-\frac{\log^m(8k+5)}{(8k+5)^s}- \frac{\log^m(8k+7)}{(8k+7)^s}+ \, \cdots \Bigg\}\)



\(\displaystyle L^{(m)}(s, \chi_2) = (-1)^m\,\sum_{k=0}^{\infty}\frac{\chi_2(k)\, (\log k)^m}{k^s} = \)


\(\displaystyle (-1)^m\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log^m(8k+1)}{(8k+1)^s}-\frac{\log^m(8k+3)}{(8k+3)^s}-\frac{\log^m(8k+5)}{(8k+5)^s}+ \frac{\log^m(8k+7)}{(8k+7)^s}+ \, \cdots \Bigg\}\)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
Relations between the Dirichlet L-series, the Dirichlet Beta function, and Hurwitz Zeta function:



\(\displaystyle \beta(x) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x} = 1-\frac{1}{3^x}+\frac{1}{5^x}-\frac{1}{7^x}+ \, \cdots \)


\(\displaystyle L(x, \chi_1) = \sum_{k=0}^{\infty}\frac{\chi_1(k)}{k^x} = 1+\frac{1}{3^x}-\frac{1}{5^x}-\frac{1}{7^x}+ \, \cdots \)


\(\displaystyle L(x, \chi_1) =\beta(x) - 2\, \sum_{k=0}^{\infty}\frac{1}{(8k+3)^x}+ 2\, \sum_{k=0}^{\infty}\frac{1}{(8k+5)^x}=\)


\(\displaystyle \beta(x)- \frac{2}{8^x}\, \zeta\left(x, \tfrac{3}{8}\right) +
\frac{2}{8^x}\, \zeta\left(x, \tfrac{5}{8}\right)\)




\(\displaystyle L(x, \chi_2) = \sum_{k=0}^{\infty}\frac{\chi_2(k)}{k^x} = 1-\frac{1}{3^x}-\frac{1}{5^x}+\frac{1}{7^x}+ \, \cdots \)


\(\displaystyle L(x, \chi_2) =\beta(x) + 2\, \sum_{k=0}^{\infty}\frac{1}{(8k+5)^x}- 2\, \sum_{k=0}^{\infty}\frac{1}{(8k+7)^x}=\)


\(\displaystyle \beta(x)+ \frac{2}{8^x}\, \zeta\left(x, \tfrac{5}{8}\right) -
\frac{2}{8^x}\, \zeta\left(x, \tfrac{7}{8}\right)\)




The Hurwitz Zeta function:


\(\displaystyle \zeta(x,a) = \sum_{k=0}^{\infty}\frac{1}{(k+a)^x} \Rightarrow\)


\(\displaystyle \zeta^{(m)} (x,a)= \frac{d^m}{dx^m}\, \zeta(x,a)=(-1)^m\, \sum_{k=0}^{\infty}\frac{\log^m(k+a)}{(k+a)^x}\)



Let \(\displaystyle p, q \in \mathbb{Z} \ge 1\), and \(\displaystyle q \le p\), then



\(\displaystyle \sum_{k=0}^{\infty}\frac{\log^m(kp+q)}{(kp+q)^x}= \frac{1}{p^x}\, \sum_{k=0}^{\infty} \frac{[\log p+\log (k+q/p)]^m}{(k+q/p)^x} =\)


\(\displaystyle \frac{1}{p^x}\, \sum_{j=0}^m\binom{m}{j}(\log p)^{m-j}\, \sum_{k=0}^{\infty} \frac{\log^j (k+q/p)}{(k+q/p)^x}=\)


\(\displaystyle \frac{1}{p^x}\, \sum_{j=0}^m(-1)^j\binom{m}{j}(\log p)^{m-j}\, \zeta^{(j)} \left(x, \tfrac{q}{p} \right)\)




That's all the groundwork out of the way. Phew...! (Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Special thanks to... Mathbalarka!


For letting me know that when defining the characters \(\displaystyle \chi_1\) and \(\displaystyle \chi_2\) above, they should have been for \(\displaystyle \mathbb{Z}/8\mathbb{Z}\) - since edited - rather than \(\displaystyle \mathbb{Z}/8\)...

I was using the notation from this 'ere paper (p.9) -->

http://arxiv.org/pdf/math/0411087v1.pdf


Many thanks! :D
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Just to recap - after all that blather about characters and zeta gubbins, we define...



\(\displaystyle \mathscr{S}_{(m, n)} (z) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\, \sin 2\pi k z\)


\(\displaystyle \mathscr{C}_{(m, n)} (z) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\, \cos 2\pi k z\)


Where \(\displaystyle m, n \in \mathbb{Z} \ge 1\), and \(\displaystyle 0 < z < 1 \in \mathbb{Q}\).




Since \(\displaystyle \sin \pi k\equiv 0\) for all integer \(\displaystyle k\), it's clear from the definition of \(\displaystyle \mathscr{S}_{(m, n)} (z)\) that



\(\displaystyle \mathscr{S}_{(m, n)} (1) = \mathscr{S}_{(m, n)} \left(\tfrac{1}{2}\right) \equiv 0\)


Similarly, since \(\displaystyle \cos 2\pi k = 1\) and \(\displaystyle \cos \pi k = (-1)^k\) for all integer \(\displaystyle k\), we have:


\(\displaystyle \mathscr{C}_{(m, n)} (1) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n} = (-1)^m\, \zeta^{(m)}(n)\)


and


\(\displaystyle \mathscr{C}_{(m, n)} \left(\frac{1}{2}\right) = \sum_{k=1}^{\infty}(-1)^k\frac{(\log k)^m}{k^n} = (-1)^{m+1}\, \eta^{(m)}(n)=\)


\(\displaystyle (-1)^{m+1} \Bigg\{ \Bigg(1-2^{\, 1-n}\Bigg)\, \zeta^{(m)}(n) - 2^{\, 1-n}\, \sum_{j=0}^{m-1}\binom{m}{j}\, (-\log 2)^{m-j}\zeta^{(j)}(n) \Bigg\}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Since \(\displaystyle \cos \left( \frac{4\pi}{3}\right)= \cos \left( 2\pi-\frac{2\pi}{3} \right) \equiv \cos \left( \frac{2\pi}{3} \right)=-1/2\), then when \(\displaystyle z=1/3\) we have



\(\displaystyle \mathscr{C}_{(m, n)} \left(\frac{1}{3}\right) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\, \cos \left(\frac{2\pi}{3}\right)=\)


\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{\log^m(3k+1)}{(3k+1)^n}
-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{\log^m(3k+2)}{(3k+2)^n}\)


\(\displaystyle + \sum_{k=0}^{\infty}\frac{\log^m(3k+3)}{(3k+3)^n} \equiv\)


\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty}\Bigg\{ \frac{\log^m(3k+1)}{(3k+1)^n}+ \frac{\log^m(3k+2)}{(3k+2)^n}+ \frac{\log^m(3k+3)}{(3k+3)^n}
\Bigg\}\)


\(\displaystyle + \frac{3}{2}\, \sum_{k=0}^{\infty}\frac{\log^m(3k+3)}{(3k+3)^n} =\)


\(\displaystyle -\frac{1}{2}\, \sum_{k=1}^m \frac{(\log k)^m}{k^n} + \frac{3^{\,1-n}}{2}\, \sum_{k=0}^{\infty}\frac{[\log 3+ \log(k+1)]^m}{(k+1)^n}=\)


\(\displaystyle \frac{(-1)^{m+1}}{2}\,\zeta^{(m)}(n) + \frac{3^{\,1-n}}{2}\, \sum_{j=0}^{m}\binom{m}{j}\,(\log 3)^{m-j}\, \zeta^{(j)}(n)=\)


\(\displaystyle \frac{ \left[(-1)^{m+1}+ 3^{\,1-n} \right] }{2}\,\zeta^{(m)}(n) + \frac{3^{\,1-n}}{2}\, \sum_{j=0}^{m-1}\binom{m}{j}\,(\log 3)^{m-j}\, \zeta^{(j)}(n)\)




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




\(\displaystyle \therefore \, \mathscr{C}_{(m, n)} \left(\frac{1}{3}\right)=\)


\(\displaystyle \frac{ \left[(-1)^{m+1}+ 3^{\,1-n} \right] }{2}\,\zeta^{(m)}(n) + \frac{3^{\,1-n}}{2}\, \sum_{j=0}^{m-1}\binom{m}{j}\,(\log 3)^{m-j}\, \zeta^{(j)}(n)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Due to the trigonometric nature of the functions \(\displaystyle \mathscr{C}_{(m,n)}(z)\) and \(\displaystyle \mathscr{S}_{(m,n)}(z)\), it should be possible to obtain a number of reflection and transformation formulae. This is indeed the case.

For positive integer \(\displaystyle k\), and \(\displaystyle 0 <z \le 1\), we have:


\(\displaystyle \cos 2\pi k(1-z) = \cos 2\pi kz\)

\(\displaystyle \sin 2\pi k(1-z) = -\sin 2\pi kz\)



Hence


\(\displaystyle \mathscr{C}_{(m,n)}(1-z)=\mathscr{C}_{(m,n)}(z)\)


\(\displaystyle \mathscr{S}_{(m,n)}(1-z)=-\mathscr{S}_{(m,n)}(z)\)



Applying the first reflection formula to the previous result for \(\displaystyle z=1/3\) gives the case for \(\displaystyle z=2/3\).



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




\(\displaystyle \therefore \, \mathscr{C}_{(m, n)} \left(\frac{2}{3}\right)=\)


\(\displaystyle \frac{ \left[(-1)^{m+1}+ 3^{\,1-n} \right] }{2}\,\zeta^{(m)}(n) + \frac{3^{\,1-n}}{2}\, \sum_{j=0}^{m-1}\binom{m}{j}\,(\log 3)^{m-j}\, \zeta^{(j)}(n)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
When considering the Sine case for \(\displaystyle z=1/3\), it's readily apparent that every third term vanishes, due to a coefficient congruous to \(\displaystyle \sin \pi k\):


\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{1}{3}\right)= \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\sin\left(\frac{2\pi k}{3}\right)=\)


\(\displaystyle \sin\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log^m(3k+1)}{(3k+1)^n}
+
\sin\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{\log^m(3k+2)}{(3k+2)^n}=\)


\(\displaystyle \frac{\sqrt{3}}{2}\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log^m(3k+1)}{(3k+1)^n}-\frac{\log^m(3k+2)}{(3k+2)^n}
\Bigg\}=\)


\(\displaystyle \frac{3^{1/2-n}}{2}\, \sum_{j=0}^m\binom{m}{j} (\log 3)^{m-j}\, \sum_{k=0}^{\infty} \Bigg\{
\frac{\log^j(k+1/3)}{(k+1/3)^n}-\frac{\log^j(k+2/3)}{(k+2/3)^n}
\Bigg\}=\)


\(\displaystyle \frac{3^{1/2-n}}{2}\, \sum_{j=0}^m(-1)^j\binom{m}{j} (\log 3)^{m-j}\, \Bigg\{
\zeta^{(j)}\left(n, \tfrac{1}{3}\right)-
\zeta^{(j)}\left(n, \tfrac{2}{3}\right)
\Bigg\}\)


The result for \(\displaystyle z=2/3\) is obtained from the above, and the reflection formula.



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



\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{1}{3}\right)=\)


\(\displaystyle \frac{3^{1/2-n}}{2}\, \sum_{j=0}^m(-1)^j\binom{m}{j} (\log 3)^{m-j}\, \Bigg\{
\zeta^{(j)}\left(n, \tfrac{1}{3}\right)-
\zeta^{(j)}\left(n, \tfrac{2}{3}\right)
\Bigg\}\)





\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{2}{3}\right)=\)


\(\displaystyle -\frac{3^{1/2-n}}{2}\, \sum_{j=0}^m(-1)^j\binom{m}{j} (\log 3)^{m-j}\, \Bigg\{
\zeta^{(j)}\left(n, \tfrac{1}{3}\right)-
\zeta^{(j)}\left(n, \tfrac{2}{3}\right)
\Bigg\}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
For the Sine series, the case \(\displaystyle z=1/4\) can be resolved in terms of derivatives of the Dirichlet Beta function; this is a natural consequence of the fact that every term with even index k vanishes:


\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{1}{4}\right)=\)


\(\displaystyle \sum_{k=0}^{\infty}\frac{(\log k)^m}{k^n}\sin\left(\frac{\pi k}{2}\right) \equiv\)


\(\displaystyle \sin\left(\frac{\pi}{2}\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+1)}{(4k+1)^n}+
\sin\left(\pi\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+2)}{(4k+2)^n}+ \)


\(\displaystyle \sin\left(\frac{3\pi}{2}\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+3)}{(4k+3)^n}+
\sin\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+4)}{(4k+4)^n} \equiv \)


\(\displaystyle \sum_{k=0}^{\infty}\frac{\log^m(4k+1)}{(4k+1)^n} - \sum_{k=0}^{\infty}\frac{\log^m(4k+3)}{(4k+3)^n}=\)


\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{\log^m(2k+1)}{(2k+1)^n}= \)


\(\displaystyle (-1)^m\, \lim_{x\to n} \, \frac{d^m}{dx^m} \left[\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}\right] = \)


\(\displaystyle (-1)^m\, \lim_{x\to n} \, \frac{d^m}{dx^m} \,\beta(x) = (-1)^m\beta^{(m)}(n)\)


The result for \(\displaystyle z=3/4\) is obtain from this one, via the reflection formula.




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




\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{1}{4}\right)=(-1)^m\beta^{(m)}(n)\)




\(\displaystyle \mathscr{S}_{(m,n)}\left(\frac{3}{4}\right)=(-1)^{m+1}\beta^{(m)}(n)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Setting \(\displaystyle z=1/4\) in the Cosine series gives:


\(\displaystyle \mathscr{C}_{(m,n)}\left(\frac{1}
{4}\right) = \sum_{k=1}^{\infty}\frac{(\log k)^m}{k^n}\cos\left(\frac{\pi k}{2}\right) \equiv\)


\(\displaystyle \cos\left(\frac{\pi}{2}\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+1)}{(4k+1)^n}+
\cos\left(\pi\right)\, \sum_{k=0}^{\infty}\frac{\log^m(4k+2)}{(4k+2)^n}+\)


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


\(\displaystyle -\, \sum_{k=0}^{\infty}\frac{\log^m(4k+2)}{(4k+2)^n}+
\, \sum_{k=0}^{\infty}\frac{\log^m(4k+4)}{(4k+4)^n}=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, \sum_{k=0}^{\infty} \Bigg\{ \frac{1}{(4k+2)^x}-\frac{1}{(4k+4)^x} \Bigg\}=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, \frac{1}{2^x}\, \Bigg\{ \sum_{k=0}^{\infty} \frac{1}{(2k+1)^x}-\sum_{k=0}^{\infty} \frac{1}{(2k+2)^x} \Bigg\}=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, \frac{1}{2^x}\, \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^x}=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, \frac{\eta(x)}{2^x}=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, \frac{1}{2^x}\left(1-\frac{1}{2^{\,x-1}}\right)\, \zeta(x)=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, 2^{-x}\, \zeta(x) - (-1)^{m+1}\, \lim_{x \to n}\, \frac{d^m}{dx^m}\, 2^{-(2x-1)}\, \zeta(x)=\)


\(\displaystyle (-1)^{m+1}\, \lim_{x \to n}\, 2^{-x} \, \sum_{j=0}^m(-1)^{m-j}\binom{m}{j}(\log 2)^{m-j}\, \zeta^{(j)}(x)\)


\(\displaystyle -(-1)^{m+1}\, \lim_{x \to n}\, 2^{-(2x-1)} \, \sum_{j=0}^m(-1)^{m-j}2^{m-j}\binom{m}{j}(\log 2)^{m-j}\, \zeta^{(j)}(x)=\)



\(\displaystyle (-1)^{m+1}\, 2^{-n} \, \sum_{j=0}^m(-1)^{m-j}\binom{m}{j}(\log 2)^{m-j}\, \zeta^{(j)}(n)\)


\(\displaystyle -(-1)^{m+1}\, 2^{-(2n-1)} \, \sum_{j=0}^m(-1)^{m-j}2^{m-j}\binom{m}{j}(\log 2)^{m-j}\, \zeta^{(j)}(n)=\)


\(\displaystyle \sum_{j=0}^m(-1)^{j+1}\binom{m}{j}\left(\frac{2^{n-1}-2^{m-j}}{2^{2n-1}}\right)\, (\log 2)^{m-j}\, \zeta^{(j)}(n)\)




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




\(\displaystyle \mathscr{C}_{(m,n)}\left(\frac{1}
{4}\right)= \sum_{j=0}^m(-1)^{j+1}\binom{m}{j}\left(\frac{2^{n-1}-2^{m-j}}{2^{2n-1}}\right)\, (\log 2)^{m-j}\, \zeta^{(j)}(n)\)





\(\displaystyle \mathscr{C}_{(m,n)}\left(\frac{3}
{4}\right)= \sum_{j=0}^m(-1)^{j+1}\binom{m}{j}\left(\frac{2^{n-1}-2^{m-j}}{2^{2n-1}}\right)\, (\log 2)^{m-j}\, \zeta^{(j)}(n)\)