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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 ...
Jack's user avatar
  • 424
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 ...
Dr Potato's user avatar
  • 812
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}-\...
garondal's user avatar
  • 889
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 ...
Rudyard's user avatar
  • 305
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}...
Max's user avatar
  • 910
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 ...
user47363's user avatar
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} ...
Dens's user avatar
  • 303
1 vote
0 answers
98 views

Using Contour Integral to find the value of $\int_{-1}^{+1}\frac{\ln{(1+t)}}{t}dt$

$\newcommand{LogI}{\operatorname{Li}}$ We know that the value of $\LogI_{2}(-1)$ is -$\frac{\pi^2}{12}$ and $\LogI_{2}(1)$ is $\frac{\pi^2}{6}$. The value of the polylogarithms has already been ...
Sourangshu Ghosh's user avatar
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, ...
mr_e_man's user avatar
  • 5,726
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, ...
dan_fulea's user avatar
  • 34.2k
2 votes
1 answer
326 views

Evaluate $\Im(\operatorname{Li}_3(2i) + \operatorname{Li}_3(\frac i2))$

Applying the trilogarithm identity $$ \operatorname{Li}_{3}\left(z\right) - \operatorname{Li}_{3}\left(1 \over z\right) = -{1 \over 6}\ln^{3}\left(-z\right) - {\pi^{2} \over 6}\ln\left(-z\right)\tag{1}...
Quanto's user avatar
  • 99.7k
2 votes
2 answers
226 views

Evaluation of a log-trig integral in terms of the Clausen function (or other functions related to the dilogarithm)

Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the definite integral $$\mathcal{I}{\left(a,\theta\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2a\cos{\left(\...
David H's user avatar
  • 30.7k
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 ...
Integrand's user avatar
  • 8,369
2 votes
1 answer
111 views

Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.

I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
zalm's user avatar
  • 125
1 vote
0 answers
145 views

Dilogarithm of a negative real number outside unit circle

The dilogarithm is defined in $\mathbb{C}$ as $$ Li_2(z) = -\int_0^1 \frac{\ln(1 - zt)}{t} dt $$ Because $1-zt \in \mathbb{C}$, then you can write $\ln(1 - zt) = \ln|1 - zt| + i·\arg(1 - zt)$ As ...
Vicky's user avatar
  • 539
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 ...
Diger's user avatar
  • 6,277
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 $$\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}=\sum_{k\geq 1}\frac{\cos(kx)}{k^2}.$$ ...
46andpi's user avatar
  • 158
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 $...
Diger's user avatar
  • 6,277
3 votes
0 answers
75 views

Approaching a branch point along different paths

There's a very nice characterization of the three main types of isolated singularities of an analytic function $f(z)$ that takes oriented curves $\gamma$ that terminate at the singularity and ...
tparker's user avatar
  • 6,280
3 votes
2 answers
304 views

Jump of dilogarithm

I am reading about the dilogarithm function $$ \mathrm{Li}_2(z):= - \int_0^z \frac{\log(1-u)}{u}du, \quad z \in \mathbb{C} \backslash [1, \infty).$$ I found it stated that the "jump" of the ...
57Jimmy's user avatar
  • 6,326
3 votes
1 answer
500 views

Bose-Einstein function as real part of polylogarithm: $\overline{G}_{s}(x)= \Re \mathrm{Li}_{s+1}(e^x)$

For real $x<0$ the Bose-Einstein integral of order $s$ is given at https://dlmf.nist.gov/25.12.E15 as $$G_{s}(x)=\frac{1}{\Gamma\left(s+1\right)}\int_{0}^{\infty}\frac{t^{s}}{e^{t-x} -1}\mathrm{d}...
gammatester's user avatar
  • 18.9k
2 votes
1 answer
686 views

Calculate a $\operatorname{Li}_{2}(-1)$ using Integral Representation

$\newcommand{LogI}{\operatorname{Li}}$ I know that $\LogI_{2}(-1)=-\frac{\pi^2}{12}$, but I have never seen a proof of this result without using a functional identity of the Dilogarithm or a series ...
Biggs's user avatar
  • 436
0 votes
0 answers
135 views

Validity of argument in dilogarithm identities on Wolfram

I've come across a series of identities existing between dilogarithms and powers of logarithms but I am not sure about when such equations are valid in terms of the restriction of the domain of the ...
CAF's user avatar
  • 2,850
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 ...
dfnu's user avatar
  • 7,613
1 vote
1 answer
1k views

on the (double) discontinuity of dilogarithm along a branch cut

Define the function $$Li_s(z)=\sum_{k=1}^\infty \frac{z^k}{k^s}$$ for |z|<1. Let's focus on $s=2$. It can be extended to a holomorphic function on $\mathbb C \setminus [1,\infty)$ $$Li_2(z)= -\...
jj_p's user avatar
  • 2,390
1 vote
2 answers
640 views

Sum of Dilogarithm and its Complex Conjugate

$$\newcommand{\dl}[1]{\operatorname{Li}_2\left( #1\right)}$$ I have the following expression involving Dilogarithims, where $z\in\mathbb{R}$ and $b\in\mathbb{C}\setminus\lbrace 0\rbrace$: $$f(z,b) = ...
Zyrax's user avatar
  • 13
3 votes
1 answer
709 views

What's about $\int_0^\infty x^{-z}Li_{z+2}(e^{-xz})dx$ as $-\zeta(3)z^2\Gamma(-z)$?

Because $$-\int_0^\infty \frac{e^{-zt}}{t^z}dt=z^2\Gamma(-z),$$ holds for $0<\Re z<1$ then using the change of variable $t=nx$ one has that $$-\frac{1}{n^{z-1}}\int_0^\infty \frac{e^{-znx}dx}{x^...
user avatar
1 vote
0 answers
98 views

What's the worst sequence that still leads to a converging series?

As a background, I'm initially interested in sequences $a_n$ giving rise to functions $\sum_{n=0}^\infty a_nx^n$ for $x\in(0,1)$ and their diverging behavior for $x$ to $1$. E.g. the geometric series $...
Nikolaj-K's user avatar
  • 12.3k
0 votes
1 answer
406 views

Evaluating Fermi Dirac integrals of order j<0

The complete Fermi Dirac integral (I'm purposely leaving off the gamma prefactor)... $$F_j(x) =\int\limits_{0}^{\infty} \frac{t^j}{e^{t-x}+1} \: dt$$ is generally defined for j>-1. Is there a way to ...
Johnny's user avatar
  • 1
5 votes
2 answers
438 views

Extract imaginary part of $\text{Li}_3\left(\frac{2}{3}-i \frac{2\sqrt{2}}{3}\right)$ in closed form

We know that polylogarithms of complex argument sometimes have simple real and imaginary parts, e.g. $\mathrm{Re}[\text{Li}_2(i)]=-\frac{\pi^2}{48}$ Is there a closed form (free of polylogs and ...
user12588's user avatar
  • 369

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