Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
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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+...
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Evaluate $\int_{0}^{1} \operatorname{Li}_3\left [ \left ( \frac{x(1-x)}{1+x} \right ) ^2 \right ] \text{d}x$
Possibly evaluate the integral?
$$
\int_{0}^{1} \operatorname{Li}_3\left [
\left ( \frac{x(1-x)}{1+x} \right ) ^2 \right ]
\text{d}x.
$$
I came across this when playing with Legendre polynomials, ...
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Simplify the Laplace Transform for $E_{i}(-y)^{2}$
I want to simplify the Laplace transform expression of $E_{i}(-y)^{2}$, where $E_{i}(y)$ is the exponential integral defined by $E_{i}(y) = -\int\limits_{-y}^{\infty} \frac{e^{-t}}{t} dt$.
Question: ...
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How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way?
The following problem
\begin{align}
&\int_{0}^{\pi/2}
x\ln^{2}\left(\sin\left(x\right)\right)\,{\rm d}x \\[5mm] = & \
\frac{1}{2}\ln^{2}\left(2\right)\zeta\left(2\right)
- \frac{19}{32}\,\zeta\...
- 91k
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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 ...
- 104k
33
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Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$
This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:
Prove:
$$\int_0^1\ln(1-x)\ln(1+x)\ln^...
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$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$ as polylogarithm
$$\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^{...
- 33.2k
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Compute in closed form $\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$
I am trying to find closed form for this integral:
$$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Where $a>0$.
My try: Let: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Then:
$$\...
- 99.7k
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Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$
How would you solve the following
$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$
I might be able to relate the integral to Euler sums .
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How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?
How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$? No context, this is just a curiosity o'mine.
Yes, I am aware there is no reason to believe a random power ...
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Are these generalizations known in the literature?
By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(...
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Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$
How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$
where $\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$ , $|x|\leq1$
I came across this integral ...
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Behaviour of polylogarithm at $|z|=1$
I have the sum
$$
\sum_{n=1}^\infty \dfrac{\cos (n \theta)}{n^5} = \dfrac{\text{Li}_5 (e^{i\theta}) + \text{Li}_5 (e^{-i\theta})}{2},
$$
where $0\leq\theta < 2 \pi$ is an angle and $\text{Li}_5(z)$ ...
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Justify $\zeta(3)=2\int_0^1 \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx$
I don't know if this approach to get a formula involving the Apéry constant was in the literature. This idea was a simple idea few minutes ago, when I was studying the answers in this site Math Stack ...
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Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$
Please help me to evaluate this integral in a closed form:
$$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$
Using integration by parts I found that it could be expressed through ...