Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $$\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^{\infty} \frac{\sin(2\pi kx)}{k}$$

share|cite|improve this question
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
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 $f$ :


Hoping this helped,

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.