Articles of polylogarithm

Find the derivative of a polylogarithm function

I was trying to find to which function the next series converges. $$ \sum_{n=1}^{\infty} \ln(n)z^n $$ If we take the polylogarithm function $Li_s(z)$ defined as $$ Li_s(s)=\sum_{n=1}^{\infty} \frac{z^n}{n^s} $$ Then it is easily seen that $$ \sum_{n=1}^{\infty} \ln(n)z^n = – \left( \frac{\partial}{\partial s}Li_s(z)\right)_{s=0} $$ Now, my question is how to calculate $ \frac{\partial}{\partial s}Li_s(z)$, using […]

Generating function for $\sum_{k\geq 1} H^{(k)}_n x^ k $

Is there a generating function for $$\tag{1}\sum_{k\geq 1} H^{(k)}_n x^ k $$ I know that $$\tag{2}\sum_{n\geq 1} H^{(k)}_n x^n= \frac{\operatorname{Li}_k(x)}{1-x} $$ But notice in (1) the fixed $n$.

Evaluating a logarithmic integral in terms of trilogarithms

For $a,c\in\mathbb{R}\land-1\le a\land-1<c$, define the function $J{\left(a,c\right)}$ to be the value of the dilogarithmic integral $$J{\left(a,c\right)}:=\int_{0}^{1}\mathrm{d}y\,\frac{\operatorname{Li}_{2}{\left(\frac{c}{1+c}\right)}-\operatorname{Li}_{2}{\left(\frac{ay}{1+ay}\right)}}{c-ay}.$$ In principle, $J{\left(a,c\right)}$ may be evaluated in terms of trilogs, dilogs, and elementary functions. In the process of trying to develop my own solution, I managed to obtain partial solutions valid over various subsets of the parameter space $(a,c)\in[-1,\infty)\times(-1,\infty)$, […]

A special value polylogarithm identity involving $\text{Li}_3(-1/2),\,\text{Li}_3(-1/3),\,\text{Li}_3(2/3),\,\text{Li}_2(-1/3),\,\text{Li}_2(2/3)$

I’ve found that \begin{align} \mathcal{L} = 2\operatorname{Li}_3\left(-\frac{1}{2}\right)+\operatorname{Li}_3\left(-\frac{1}{3}\right)+2\operatorname{Li}_3\left(\frac{2}{3}\right)+\operatorname{Li}_2\left(-\frac{1}{3}\right) \ln(3)+2\operatorname{Li}_2\left(\frac{2}{3}\right) \ln(3) \end{align} equals to $$ \mathcal{L} = \frac{\pi^2}{3}\ln(2)+\frac{1}{3}\ln^3(2)-\frac{1}{3} \ln^2(3) \ln\left(\frac{27}{8}\right)-\frac{\zeta(3)}{6}. $$ How could we prove this identity? A numerical approximation: $$ \mathcal{L} \approx 1.701652530545172752791574942340971991312113932043\dots $$

Tough quadrilogarithm integral

Solve the follolwing definite integral $$\int \frac{\operatorname{Li}_4(z)}{1-z}\, dz$$ It is easy for lower powers!

Strategies for evaluating sums $\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n}$

I’m looking for strategies for evaluating the following sums for given $z$ and $m$: $$ \mathcal{S}_m(z):=\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n}, $$ where $H_n^{(m)}$ is the generalized harmonic number, and $|z|<1$, $m \in \mathbb{R}$. Using the generating function of the generalized harmonic numbers, an equivalent problem is to evaluate the following integral: $$ \mathcal{S}_m(z) = \int_0^z \frac{\operatorname{Li}_m(t)}{t(1-t)}\,dt, $$ where […]

A possible dilogarithm identity?

I’m curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before? $$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{Li}_2\left(\frac{1}{2+\sqrt{2}}\right)\right)+\text{Li}_2\left(3-2 \sqrt{2}\right)$$

Is the following Harmonic Number Identity true?

Is the following identity true? $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right] + \operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}$$ In this accepted answer, @Tunk Fey proved the above. (See $(4)$). However, I have the following $3$ queries : Why can we add the integrals after the substitution $x \mapsto 1-x$ in the following step? I doubt it since […]

Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$

I stumbled upon the interesting definite integral \begin{equation} \int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2 \end{equation} Here is my proof of this result. Let $u=\sin^{-1}(x)$ then integrate by parts, \begin{align} \int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\ &= u \ln\sin(u) – \int \ln\sin(u) du \tag{1} \label{eq:20161030-1} \end{align} \begin{align} \int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} – […]

Closed-forms of real parts of special value dilogarithm identities from inverse tangent integral function

The inverse tangent integral is defined as $$\operatorname{Ti}_2(x)=\Im\operatorname{Li}_2\left(ix\right)$$ Because this we have some special value identitiy. Let $c_1 = \operatorname{Li}_2(i)$, then $\Im c_1 = G$, where $G$ is Catalan’s constant. Let $c_2 = 4\operatorname{Li}_2\left(\frac{i}{2}\right) + 2\operatorname{Li}_2\left(\frac{i}{3}\right)+\operatorname{Li}_2\left(\frac{3i}{4} \right)$, then $\Im c_2=6G – \pi \ln 2$. Let $c_3 = \operatorname{Li}_2\left(2i-\sqrt 3\,i\right)$, then $\Im c_3 = \frac{2G}{3}-\frac{\pi}{12}\ln(2+\sqrt 3)$. […]