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2 votes
0 answers
40 views

How is the dilogarithm defined?

I am pretty happy with the definition of the "maximal analytic contiuation of the logarithm" as $$ \operatorname{Log}_{\gamma}\left(z\right) = \int_\gamma ...
1 vote
1 answer
59 views

Connection between the polylogarithm and the Bernoulli polynomials.

I have been studying the polylogarithm function and came across its relation with Bernoulli polynomials, as Wikipedia site asserts: For positive integer polylogarithm orders $s$, the Hurwitz zeta ...
3 votes
0 answers
121 views

Show that $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$ (polylogarithm)

I am working with the polylogarithm function and want to find closed expressions for $\textrm{Li}_2(e^{ix})$. If I plot the function $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))$ I get $y=\dfrac{x^2}{4}-\...
1 vote
1 answer
60 views

Imaginary part of the dilogarithm of an imaginary number

I am wondering if I can simplify $${\rm Im} \left[ {\rm Li}_2(i x)\right] \ , $$ in terms of more elementary functions, when $x$ is real (in particular, I am interested in $0<x<1$). I checked ...
3 votes
0 answers
142 views

Prove that $-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$

Prove that $$-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$$ where ${\rm Li}_2(x)$ is the Poly Logarithm function and $\zeta(s)$ is the Riemann zeta function Let $$I=-\int_{0}^{1}...
3 votes
2 answers
416 views

Imaginary part of dilogarithm

I have evaluated a certain real-valued, finite integral with no general elementary solution, but which I have been able to prove equals the imaginary part of some dilogarithms and can write in the ...
2 votes
1 answer
311 views

Real Part of the Dilogarithm

It is well known that $$\frac{x-\pi}{2}=-\sum_{k\geq 1}\frac{\sin{kx}}{k}\forall x\in(0,\tau),$$ which gives $$\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}=\sum_{k\geq 1}\frac{\cos(kx)}{k^2}.$$ ...
2 votes
2 answers
797 views

Branch Points of the Polylog function

The polylogarithm $$ {\rm Li}_s(z) = \sum_{n=1}^\infty \frac{z^n}{n^s} $$ has obvious branch points at $z=1$. For integers $s\leq 0$ it is a rational function with a pole of order $1-s$ at $z=1$. If $...
17 votes
2 answers
894 views

A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...

Question: How to show the relation $$ J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\frac 1{64}\pi^4 $$ (using a "minimal industry" of relations, ...
2 votes
2 answers
288 views

Does the dilogarithm function (which is multi-valued) have a single-valued inverse?

The $p$-logarithm is defined for $|z|<1$ by $$\text{Li}_p(z)=\sum_{n=1}^\infty\frac{z^n}{n^p}$$ and defined elsewhere in $\mathbb C$ by analytic continuation, though it may be multi-valued, ...
1 vote
0 answers
128 views

Conjectured closed form for ${\it {Li_2}} \left( 1-{\frac {\sqrt {2}}{2}}-i \left( 1-{\frac {\sqrt { 2}}{2}} \right) \right)$

With Maple i find this closed form: ${\it {Li_2}} \left( 1-{\frac {\sqrt {2}}{2}}-i \left( 1-{\frac {\sqrt { 2}}{2}} \right) \right)$=$-{\frac {{\pi}^{2}}{64}}-{\frac { \left( \ln \left( 1+\sqrt {2} ...
4 votes
2 answers
260 views

Logarithmic integral $ \int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x $

At various places e.g. Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$ and How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$ logarithmic integrals are connected ...
8 votes
2 answers
2k views

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^...
5 votes
3 answers
320 views

Is there a closed-form for $\sum_{n=0}^{\infty}\frac{n}{n^3+1}$?

I'm reading a book on complex variables (The Theory of Functions of a Complex Variable, Thorn 1953) and the following is shown: Let $f(z)$ be holomorphic and single valued in $\mathbb{C}$ except at a ...
3 votes
3 answers
449 views

Indefinite integral $\int \arctan^2 x dx$ in terms of the dilogarithm function

I read about the integral $$\int \arctan^2 x dx$$ in this old post: Evaluation of $\int (\arctan x)^2 dx$ By replacing $$\arctan x = -\frac{i}{2}\left[\log(1+ix) - \log(i-ix)\right],$$ as suggested ...

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