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3 votes
1 answer
62 views

Euler Sums of Weight 6

For the past couple of days I have been looking at Euler Sums, and I happened upon this particular one: $$ \sum_{n=1}^{\infty}\left(-1\right)^{n}\, \frac{H_{n}}{n^{5}} $$ I think most people realize ...
Jessie Christian's user avatar
1 vote
0 answers
67 views

How to integrate $\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x$ [duplicate]

Question; how to integrate $$\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x$$ here is my attempt to solve the integral \begin{align} I&=\int_0^\frac{1}{2}\frac{\ln(1+...
Mods And Staff Are Not Fair's user avatar
1 vote
0 answers
51 views

Polylogarithmically solving $\int\frac{\log(a_1x+b_1)\cdots\log(a_nx+b_n)}{px+q}\,dx$

I am now trying a direct approach to solving my question about $$\int_0^\infty\frac{\arctan a_1x\arctan a_2x\dots\arctan a_nx}{1+x^2}\,dx$$ where the $a_i$ are all positive. Note that the $\arctan$s ...
Parcly Taxel's user avatar
8 votes
1 answer
285 views

Evaluate $\int_0^\infty\frac{dx}{1+x^2}\prod_i\arctan a_ix$ (product of arctangents and Lorentzian)

Define $$I(a_1,\dots,a_n)=\int_0^\infty\frac{dx}{1+x^2}\prod_{i=1}^n\arctan a_ix$$ with $a_i>0$. By this answer $\newcommand{Li}{\operatorname{Li}_2}$ $$I(a,b)= \frac\pi4\left(\frac{\pi^2}6 -\Li\...
Parcly Taxel's user avatar
12 votes
2 answers
504 views

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$? No context, this is just a curiosity o'mine. Yes, I am aware there is no reason to believe a random power ...
Alma Arjuna's user avatar
  • 3,881
3 votes
0 answers
186 views

how to find closed form for $\int_0^1 \frac{x}{x^2+1} \left(\ln(1-x) \right)^{n-1}dx$?

here in my answer I got real part for polylogarithm function at $1+i$ for natural $n$ $$ \Re\left(\text{Li}_n(1+i)\right)=\left(\frac{-1}{4}\right)^{n+1}A_n-B_n $$ where $$ B_n=\sum_{k=0}^{\lfloor\...
Faoler's user avatar
  • 1,637
4 votes
0 answers
112 views

Calculate an integral involving polylog functions

Im my recent answer https://math.stackexchange.com/a/4777055/198592 I found numerically that the following integral has a very simple result $$i = \int_0^1 \frac{\text{Li}_2\left(\frac{i\; t}{\sqrt{1-...
Dr. Wolfgang Hintze's user avatar
4 votes
0 answers
83 views

Closed form of dilogarithm fucntion involving many arctangents

I am trying to find closed form for this expression: $$ - 2{\text{L}}{{\text{i}}_2}\left( {\frac{1}{3}} \right) - {\text{L}}{{\text{i}}_2}\left( {\frac{1}{6}\left( {1 + i\sqrt 2 } \right)} \right) - {\...
OnTheWay's user avatar
  • 2,702
5 votes
1 answer
193 views

Evaluating $\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{(ax)}\operatorname{arsinh}{(bx)}}{x}$ in terms of polylogarithms

Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by the definite integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{\left(ax\right)}\...
David H's user avatar
  • 30.7k
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
5 votes
1 answer
288 views

Closed forms of the integral $ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x $

(This is related to this question). How would one find the closed forms the integral $$ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x? $$ I tried using Nielsen Generalized Polylogarithm as mentioned ...
Anomaly's user avatar
  • 107
2 votes
2 answers
93 views

Finding a recurrence relation to evaluate $\int_{a}^{1}\mathrm{d}x\,\frac{x^{n}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}$

For each $n\in\mathbb{Z}_{\ge0}$, define the function $\mathcal{J}_{n}:(0,1)\rightarrow\mathbb{R}$ via the doubly improper integral $$\mathcal{J}_{n}{\left(a\right)}:=\int_{a}^{1}\mathrm{d}x\,\frac{x^{...
David H's user avatar
  • 30.7k
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
4 votes
1 answer
257 views

Find closed-form of: $\int_{0}^{1}\frac{x\log^{3}{(x+1)}}{x^2+1}dx$

I found this integral: $$\int_{0}^{1}\frac{x\log^{3}{(x+1)}}{x^2+1}dx$$ And it seems look like this problem but i don't know how to process with this one. First, i tried to use series of $\frac{x}{x^...
OnTheWay's user avatar
  • 2,702
3 votes
1 answer
215 views

Is there an analytic solution to $\int_a^b \frac{\arctan(A+Bt)}{C^2 + (t-Z)^2}dt$?

Is there a sensible analytic solution to the following integral: $$\int_a^b \frac{\arctan(A+Bt)}{C^2 + (t-Z)^2}dt$$ where all constants are real and $C>0$. This integral is part of the third term ...
Mathis's user avatar
  • 31
4 votes
0 answers
220 views

How can we prove a closed form for $\frac{1}{8} \text{Li}_2\left(\frac{2+\sqrt{3}}{4} \right)+\text{Li}_2\left(2+\sqrt{3}\right)$?

I have been working on a problem in number theory that I have reduced to the problem of showing that the two-term linear combination $$ \frac{1}{8} \text{Li}_2\left(\frac{2+\sqrt{3}}{4} \right) + \...
John M. Campbell's user avatar
0 votes
1 answer
130 views

Closed form for $\rm{Li }_2\left( -{\frac {i\sqrt {3}}{3}} \right)$

In my personal research with Maple i find this closed form : $$\operatorname{Li }_2\left( -{\frac {i\sqrt {3}}{3}} \right)={\frac {{\pi}^{2}}{24}}+{\frac {\ln \left( 2 \right) \ln \left( 3 \right) }...
Dens's user avatar
  • 303
10 votes
1 answer
790 views

A generalized "Rare" integral involving $\operatorname{Li}_3$

In my previous post, it can be shown that $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\...
Setness Ramesory's user avatar
17 votes
1 answer
1k views

A rare integral involving $\operatorname{Li}_2$

A rare but interesting integral problem: $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\...
Setness Ramesory's user avatar
8 votes
2 answers
494 views

Finding $\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{\pi^2+\ln^2\left(\frac{x-1}{2}\right)}\text{d}x$

Prove the integral $$\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{ \pi^2+\ln^2\left ( \frac{x-1}{2} \right ) }\text{d}x =\frac{96C\ln2+7\pi^3}{12(\pi^2+...
Setness Ramesory'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
0 votes
1 answer
126 views

Evaluate $\int_{{\frac {\pi}{8}}}^{{\frac {7\,\pi}{8}}}\!{\frac {\ln \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,{\rm d}t$

I'm interested in this integral: $\int_{{\frac {\pi}{8}}}^{{\frac {7\,\pi}{8}}}\!{\frac {\ln \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,{\rm d}t$ I found this particular ...
Dens's user avatar
  • 303
5 votes
1 answer
248 views

Closed form evaluation of a trigonometric integral in terms of polylogarithms

Define the function $\mathcal{K}:\mathbb{R}\times\mathbb{R}\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow\mathbb{R}$ via the definite ...
David H's user avatar
  • 30.7k
2 votes
1 answer
247 views

Finding a closed-form for the sum $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}$

Let $\mathcal{S}$ denote the sum of the following alternating series: $$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\approx-1.392562725547,$$ where $H_{n}$ denotes the $n$-...
David H's user avatar
  • 30.7k
3 votes
1 answer
502 views

Generating function of the polylogarithm.

Let $\operatorname{Li}_s(z)$ denote the polylogarithm function $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s}.$$ Does there exists a closed form or a known function which generates the ...
Dr Potato's user avatar
  • 812
2 votes
1 answer
126 views

Closed form evaluation of a class of inverse hyperbolic integrals

Define the function $\mathcal{I}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{\left(ax\...
David H's user avatar
  • 30.7k
7 votes
1 answer
196 views

Iterated integral involving polylogarithms

To establish notation the polylogarithm Li$_n(x)$ has the power series expansion $$ \text{Li}_n(x)= \sum_{k=1}^\infty \frac{x^k}{k^n} $$ and the Riemann zeta can be considered the special value $\zeta(...
user321120's user avatar
  • 6,760
3 votes
1 answer
368 views

Challenging integral $I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\cos x}dx$

My friend offered to solve this integral. $$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\cos x}dx=\frac{\pi^4}{32}-{4G^2} $$ Where G is the Catalan's constant. $$I=\int _0^{\infty }\frac{\arctan ^2\left(u\...
user178256's user avatar
  • 5,507
4 votes
1 answer
286 views

Evaluate $\int^1_0 x^a (1-x)^b \operatorname{Li}_2 (x)\, \mathrm dx$

For what $a,b$ the integral $$\int^1_0 x^a(1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx$$ has a closed form solution? I tried to solve it by expanding dilogarithm function, or by reducing it to linear ...
Machinato's user avatar
  • 2,903
6 votes
0 answers
306 views

Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?

I am not sure if there exists a closed form for $$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$ which seems non-trivial. I used the reflection and landen's identity, didn't help much. ...
Ali Shadhar's user avatar
  • 25.8k
8 votes
4 answers
688 views

How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$

Before you think I haven't tried anything, please read. I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$ But I can't find a way to simplify it. ...
user avatar
9 votes
2 answers
941 views

Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$

Here is a nice problem proposed by Cornel Valean $$ I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\, \ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x = -\frac{\pi}{16}\ln^{2}\left(2\right) - \frac{11}{...
Ali Shadhar's user avatar
  • 25.8k
44 votes
2 answers
3k views

Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$

We have the following result ($\text{Li}_{n}$ being the polylogarithm): $$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
pisco's user avatar
  • 19.1k
1 vote
2 answers
83 views

Closed-form expressions for the zeros of $\text{Li}_{-n}(x)$?

Consider the first few polylogarithm functions $\text{Li}_{-n}(x)$, where $-n$ is a negative integer and $x\in\mathbb R$ (plotted below). Observation suggests that $\text{Li}_{-1}(x)$ has one zero (at ...
WillG's user avatar
  • 6,692
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
1 answer
429 views

Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$

I encountered a hypergeometric integral while investigating harmonic sums $$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$ Based on my experience I suspect a nice ...
Infiniticism's user avatar
  • 8,654
2 votes
0 answers
142 views

Evaluating $\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$ without using $\sum_{n=1}^\infty\frac{H_n}{n^3}x^n$

I am trying to evaluate $$I=\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$$ Integration by parts yields $$I=\frac58\zeta(4)-\frac12\int_0^1\frac{\ln(1-x)\text{Li}_2(-x^2)}{x}dx$$ Another related ...
Ali Shadhar's user avatar
  • 25.8k
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
4 votes
3 answers
436 views

A closed form for the dilogarithm integral $\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx } $

$$\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx } $$ when I was solving an infinite series by using the beta function I encountered the above ...
Senna S's user avatar
  • 247
2 votes
2 answers
241 views

Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$

How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$ Any help will be appreciated.
Infiniticism's user avatar
  • 8,654
3 votes
4 answers
300 views

Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$

Any idea how ot approach $$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$ I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted ...
Ali Shadhar's user avatar
  • 25.8k
5 votes
1 answer
376 views

Evaluate $\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx$

$$\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx=-\frac{3\pi }{4} \Im(\operatorname{Li}_3(1+i))+\frac{189}{128} \zeta (3) \log (2)+\frac{C^2}{2}-\frac{1}{4} \pi C \log (2)+\frac{ \...
user178256's user avatar
  • 5,507
10 votes
3 answers
667 views

Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants

How can we evaluate $$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$ Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by ...
Infiniticism's user avatar
  • 8,654
7 votes
3 answers
282 views

The closed form for $\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n$

Is there a closed form for $$\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n\ ?$$ Where $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number. I managed to find the closed form but had hard ...
Ali Shadhar's user avatar
  • 25.8k
2 votes
1 answer
205 views

Verifying $\int_0^1 \ln^2x\ln(1+x)\operatorname{Li}_3\left(\frac1x\right)\ dx$

I managed to convert the integral $\large\int_0^1 \frac{\operatorname{Li}_3^2(-x)}{x^2}\ dx$ to $\int_0^1 \ln^2x\ln(1+x)\operatorname{Li}_3\left(\frac1x\right)\ dx$ in hope of evaluating it in a ...
Ali Shadhar's user avatar
  • 25.8k
6 votes
5 answers
315 views

How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$

$$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$ This number looks like $\zeta(5)$ value. We expand the terms $$\int_0^1\frac{\frac{\pi^2}{...
Sibawayh's user avatar
  • 1,343
30 votes
4 answers
2k views

Conjectural closed-form of $\int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$

Let $$I_n = \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$$ In a recently published article, $I_n$ are evaluated for $n\leq 6$: $$\begin{aligned}I_1 &= \frac{\log ^2(2)}{2}-\frac{\pi ^2}{...
pisco's user avatar
  • 19.1k
16 votes
3 answers
918 views

How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?

Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ? where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number. A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\...
Ali Shadhar's user avatar
  • 25.8k
14 votes
4 answers
2k views

Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$

Prove that I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean: The ...
Ali Shadhar's user avatar
  • 25.8k
8 votes
1 answer
173 views

Expressing $\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k\,2^k}$ as a finite sum involving $\zeta(\cdot)$, $Li_k(\cdot)$, $\pi$, and $\ln 2$

While working on the integral posted here, through a large amount of skulduggery, I managed to arrive at the following intriguing sum $$\begin{align}\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 ...
omegadot's user avatar
  • 11.8k

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