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1 vote
1 answer
111 views

asymptotic behaviour of polylogarithmic function

I would like to understand the asymptotic behaviour as $a \rightarrow 0$ of the function $$ f(a) := \sum\limits_{k=2}^{\infty} e^{ - a^2 k}{k^{-3/2}} $$ More precisely, I would like to obtain an ...
QuantumLogarithm's user avatar
6 votes
1 answer
282 views

Calculate $\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$

this integral got posted on a mathematics group by a friend $$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$$ I tried seeing what I'd get from ...
logandetner's user avatar
0 votes
1 answer
93 views

$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$ as polylogarithm

$$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$$ It is very clear for me that it has to be polylogarithm function but as it is partial sum I tried to split the series as $$\sum_{i=1}^{\infty} \frac {x^{...
Adolf L.'s user avatar
3 votes
0 answers
316 views

Two tough integrals with logarithms and polylogarithms

The following two integrals are given in (Almost) Impossible Integrals, Sums, and Series (see Sect. $\textbf{1.55}$, page $35$), $$i) \int_0^{\pi/2} \cot (x) \log (\cos (x)) \log ^2(\sin (x)) \...
user97357329's user avatar
  • 5,495
3 votes
1 answer
314 views

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

How to show that $$\int_0^1\frac{\ln^2(1+x)+2\ln(x)\ln(1+x^2)}{1+x^2}dx=\frac{5\pi^3}{64}+\frac{\pi}{16}\ln^2(2)-4\,\text{G}\ln(2)$$ without breaking up the integrand since we already know: $$\int_0^1\...
Ali Shadhar's user avatar
  • 25.8k
0 votes
1 answer
148 views

Evaluate: ${{\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx}}$

Evaluate: $${{I=\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx.}}$$ The answer is given below: $$ I=-\frac{7}{12}\pi^4\ln^2(2)-\...
Setness Ramesory's user avatar
1 vote
1 answer
163 views

Integral involving product of dilogarithm and an exponential

I am interested in the integral \begin{equation} \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast) \end{equation} where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to ...
o0BlueBeast0o's user avatar
12 votes
2 answers
718 views

General expressions for $\mathcal{L}(n)=\int_{0}^{\infty}\operatorname{Ci}(x)^n\text{d}x$

Define $$\operatorname{Ci}(x)=-\int_{x}^{ \infty} \frac{\cos(y)}{y}\text{d}y.$$ It is easy to show $$ \mathcal{L}(1)=\int_{0}^{\infty}\operatorname{Ci}(x)\text{d}x=0 $$ and $$\mathcal{L}(2)=\int_{0}^{\...
Setness Ramesory's user avatar
10 votes
1 answer
784 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
6 votes
1 answer
493 views

Is the closed form of $\int_0^1 \frac{x\ln^a(1+x)}{1+x^2}dx$ known in the literature?

We know how hard these integrals $$\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx; \int_0^1 \frac{x\ln^2(1+x)}{1+x^2}dx; \int_0^1 \frac{x\ln^3(1+x)}{1+x^2}dx; ...$$ can be. So I decided to come up with a ...
Ali Shadhar's user avatar
  • 25.8k
3 votes
1 answer
145 views

Integral representation of Dilogarithm

I have a question regarding how to obtain a certain integral representation of the dilogarithm, namely: $$\DeclareMathOperator{\li}{Li_2}\li(x) = -\int_{0}^{1}\frac{x\ln t}{1-tx}\mathrm dt\quad ,|x|&...
Brian Constantinescu's user avatar
17 votes
2 answers
892 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
  • 34.1k
23 votes
3 answers
2k views

Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$

The following problem is proposed by a friend: $$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$ $$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
Ali Shadhar's user avatar
  • 25.8k
8 votes
1 answer
495 views

How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.

How can i evaluate $$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$ $$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }...
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