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Can I evaluate the series $$ \sum_{k=1}^{\infty} \frac{\ln(k)\sin(2\pi kx)}{k}$$ in closed form, for any specific non integer values of $x$ , without using derivatives of the polylogarithm?

It looks similar to the fourier series for the fractional part of x, which is, $$\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^{\infty} \frac{\sin(2\pi kx)}{k}$$

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For $x=\pi/4$ this is like the derivative of riemann zeta, which rarely has a closed form. –  Peter Sheldrick Dec 5 '12 at 6:21
    
Id consider derivatives of the reiman zeta closed form expressions, if you could find more, id appreciate it –  Ethan Dec 5 '12 at 7:14

1 Answer 1

up vote 2 down vote accepted

Consider the polylogarithm function $\operatorname{Li}$ defined by : $$\tag{1}\operatorname{Li}_s(z)=\sum_{k=1}^\infty \frac {z^k}{k^s}\quad\text{for}\quad |z|<1$$ Since $\ k^{-s}=e^{-s\log(k)}\ $ we have : $$\tag{2}\frac d{ds}\operatorname{Li}_s(z)=-\sum_{k=1}^\infty \log(k)\frac {z^k}{k^s}\quad\text{for}\quad |z|<1$$ Let's use the (abusive) notation $\ \operatorname{Li}_{1^{'}}(z)\,$ for the limit as $s\to 1$ of $(2)$ : $$\tag{3}\operatorname{Li}_{1^{'}}(z)=\lim_{s\to 1} \frac d{ds}\operatorname{Li}_s(z)=-\sum_{k=1}^\infty \log(k)\frac {z^k}k$$ then we want $$\tag{4}\ f(x)=-\Im\bigl(\operatorname{Li}_{1^{'}}\bigl(e^{2\pi ix}\bigr)\bigr)$$ Our $\,z=e^{2\pi ix}\,$ verifies $|z|=1$ for $x\in\mathbb{R}\ $ so that the series $(1)$ and following may not converge but we may use its analytic extension (i.e. the polylogarithm function itself) to get expressions for the cases $|z|>=1$.

I am not sure that a closed formula may be found for $f(x)$ but some alternative expressions may be obtained (we will suppose that $|x|<\frac 12$ since the imaginary part of $\operatorname{Li}_{1^{'}}(z)$ seems to become zero out of this interval) :

Starting with the classical integral expression of $\operatorname{Li}_s(z)$ $$\operatorname{Li}_s(z)=\frac 1{\Gamma(s)}\int_0^\infty \frac {t^{s-1}}{e^t/z-1}dt$$ we get (from the definition of the digamma function $\psi(s)$) : $$\frac d{ds}\operatorname{Li}_s(z)=-\frac{\psi(s)}{\Gamma(s)}\int_0^\infty \frac {t^{s-1}}{e^t/z-1}dt+\frac 1{\Gamma(s)}\int_0^\infty \frac {\log(t)t^{s-1}}{e^t/z-1}dt$$ using $\,\psi(1)=-\gamma$ (the Euler constant), $\ \Gamma(1)=1\,$ and the $\operatorname{Li}_1(z)$ expression we get : $$\tag{5}\operatorname{Li}_{1^{'}}(z)=-\gamma\,\log(1-z)+\int_0^\infty \frac {\log(t)}{e^t/z-1}dt$$

You may obtain faster convergence with the expression (52) (page 36) of Crandall's recent paper 'Unified algorithms for polylogarithm, L-series, and zeta variants' ($\gamma_1$ is a Stieltjes constant as well as $\gamma$ and $\zeta$ is the zeta function of course) : $$\tag{6}\operatorname{Li}_{1^{'}}(z)=\sum_{n=1}^\infty \zeta'(1-n)\frac{\log(z)^n}{n!}-\gamma_1-\frac {\gamma^2+\zeta(2)}2-\gamma\log(-\log(z))-\frac{\log(-\log(z))^2}2$$

This simplifies a little in our $\,z=e^{2\pi ix}\,$ case. Minus the imaginary part (see $(4)$) will give the wished result.

For $\ 0<\epsilon \ll 1\ $ you should get $\ f\bigl(\frac 12-\epsilon\bigr)\approx -\epsilon\,\pi\log\bigl(\frac {\pi}2\bigr)\ $.

Concluding with a picture of the odd function $f$ :

f

Hoping this helped,

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Nice work, Raymond! –  000 Dec 6 '12 at 1:21
    
Thanks @Limitless ! –  Raymond Manzoni Dec 6 '12 at 1:22
    
Amazing job, I rarely get long detailed answers like this one, thanks alot –  Ethan Dec 6 '12 at 9:07
    
Glad you liked this @Ethan ! There are many other ways to write this and the polylogarithms, zeta sums and so on have many undisclosed 'veins'... –  Raymond Manzoni Dec 6 '12 at 17:59

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