# Closed form of a series (dilogarithm)

We are all aware of the dilogarithm function (Spence's function):

$$\sum_{n=1}^{\infty} \frac{x^n}{n^2}, \;\; x \in (-\infty, 1]$$

Also it is known that:

$$\sum_{n=1}^{\infty} \frac{\cos n x}{n^2}= \frac{\pi^2}{6}-\frac{\pi x}{2} + \frac{x^2}{4}$$

This can be shown by using the known Fourier series $\displaystyle \sum_{n=1}^{\infty} \frac{\sin nx}{n} =\frac{\pi-x}{2}$ and integrating termwise.

My main goal here is to evaluate ${\rm Li}_2(e^{ix})$. I begin by using Euler's identity:

$$e^{ix}=\cos x + i \sin x$$

Hence:

$${\rm Li}_2\left ( e^{ix} \right )= \underbrace{\sum_{n=1}^{\infty }\frac{\cos n x}{n^2}}_{\frac{\pi^2}{6}- \frac{\pi x}{2}+ \frac{x^2}{4}}+ i \sum_{n=1}^{\infty}\frac{\sin nx}{n^2}$$

So what remains now is to evaluate the last series. The only thing I have come up with is to use the Fourier series:

$$\sum_{n=1}^{\infty} \frac{\cos nx}{n} = -\ln \left( 2 \sin \frac{x}{2} \right)$$

and integrate termwise. But I cannot integrate the RHS. I believe that there is no closed form, unless I am wrong ? I'd like your help for the last step.

• See here: en.wikipedia.org/wiki/… Aug 27, 2015 at 11:36
• Those series are known as Clausen functions, so far I have only managed to get the closed form of $\sum_{n>0} \frac{sin(nx)}{n^m}$ for odd values of $m$ Aug 27, 2015 at 11:37
• Indeed $\;\displaystyle \rm{Cl}_{2n}(\theta)=\sum_{k=1}^\infty \frac {\sin(k\theta)}{k^{2n}}\:$ and $\;\displaystyle \rm{Cl}_{2n+1}(\theta)=\sum_{k=1}^\infty \frac {\cos(k\theta)}{k^{2n+1}}\;$ define the Clausen functions. Note that we covered only half of the cases, the remaining ones give the Bernoulli polynomials. Aug 27, 2015 at 20:21

"What remains" is actually the only nontrivial part of $\operatorname{Li}_2\left(e^{ix}\right)$, given by Clausen function. The fact that one can evaluate cosine series is related to the identity $$\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(z^{-1}\right)=-\frac{\ln^2\left(-z\right)}{2}-\frac{\pi^2}{6}.$$ Slightly rephrasing, we can rewrite this as $\displaystyle 2\,\Re \operatorname{Li}_2\left(e^{ix}\right)=\frac{\left(x-\pi\right)^2}{2}-\frac{\pi^2}{6}$.

• He already evaluated the cosine series, I think what he wants is the sine series Aug 27, 2015 at 11:51
• @OussamaBoussif And what I'm saying is that there is no closed form for the sine series. Sad but true. Aug 27, 2015 at 11:54

Too long for a comment: Here's a little intuitive tip: What do $$~\dfrac{\sin t}t~$$ and $$~\dfrac{\cos t}{t^2}~$$ both

have in common? They are even functions. So, if you notice various series or integrals

whose summand or integrand belongs to this category having a nice closed form, that

should not surprise you. For instance, $$~\displaystyle\int_{-\infty}^\infty\frac{\sin x}x~dx=\pi,~$$ or $$~\displaystyle\int_{-\infty}^\infty\frac{\cos x}{1+x^2}~dx=\frac\pi e,~$$

or $$~\displaystyle\sum_{n=-\infty}^\infty'\frac1{n^{2k}}=a_k~\pi^{2k},~$$ where the apostrophe represents the omission of the divergent

term corresponding to $$n=0$$, and $$a_k\in\mathbb Q$$. Obviously, if one were to sum or integrate odd

functions over this entire interval, the result would be $$0$$ for integrals, and either $$0$$ or $$f(0)$$

for sums, since the values on $$(-\infty,0)$$ would cancel those on $$(0,\infty)$$. So, in this sense, if

one were to define odd $$\zeta$$ values as $$\zeta(2k+1)=\displaystyle\sum_{n=-\infty}^\infty'\frac1{n^{2k+1}},~$$ these would indeed possess

a very beautiful closed form, namely $$0$$. Indeed, $$~\displaystyle\int_0^\infty\frac{\sin x}{1+x^2}~dx~$$ also lacks a known closed

form, as does $$~\displaystyle\sum_{n=1}^\infty\frac{\sin nx}{n^2}.~$$ Please do not misunderstand me, there are exceptions to every

rule, and one might indeed find counter-examples of both kinds, but usually they are trivial

$$($$e.g., the odd integrand whose primitive can be expressed in closed form, and then evaluated

at the extremities of the integration interval, or, in the case of $$~\displaystyle\sum_{n=1}^\infty\frac{\cos nx}n,~$$ the famous

Mercator series for the natural logarithm; not to mention a whole infinity of even functions

whose summation or definite integral simply does not possess a closed form, for the trivial

reason that the overwhelming majority of functions simply do not have one, and those that

do are the exception rather than the rule$$)$$.

• Thank you Lucian. In another comment you had also presented closed forms for the integrals: $$\int_{-\infty}^\infty \frac{\sin x}{1+x^3}\, {\rm d}x$$ etc. Hope you remember. It was a question of mine asking for a closed form for the integral: $$\int_0^\infty \frac{\sin x}{1+x^3}\, {\rm d}x$$ The above was a very good explanation. (+1) Aug 28, 2015 at 8:28