Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
546
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Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$
Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about
$$\approx 111....
- 44.4k
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6
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A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Let $H_q$ denote harmonic numbers (generalized to a non-integer index $q$):
$$H_q=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+q}\right)=\int_0^1\frac{1-x^q}{1-x}dx=\gamma+\psi(q+1),\tag1$$
where $\psi(z)=\...
- 47.3k
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4
answers
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Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
- 5,626
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2
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Closed form for $\sum_{n=0}^\infty\frac{\Gamma\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}$
I was experimenting with hypergeometric-like series and discovered the following conjecture (so far confirmed by more than $5000$ decimal digits):
$$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\...
- 47.3k
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3
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Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$
I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!...
- 47.3k
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How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
How to prove
$$\int_0^1x\ln^2(1+x)\ln\left(\frac{x^2}{1+x}\right)\frac{dx}{1+x^2}=-\frac{7}{32}\cdot\zeta{(3)}\ln2+\frac{3\pi^2}{128}\cdot\ln^22-\frac{1}{64}\cdot\ln^42-\frac{13\pi^4}{46080}$$
The ...
- 5,507
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A conjectured identity for tetralogarithms $\operatorname{Li}_4$
I experimentally discovered (using PSLQ) the following conjectured tetralogarithm identity:
$$720 \,\text{Li}_4\!\left(\tfrac{1}{2}\right)-2160 \,\text{Li}_4\!\left(\tfrac{1}{3}\right)+2160 \,\text{Li}...
- 47.3k
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Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$
It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are ...
- 44.4k
19
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3
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Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$
I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed ...
- 11.6k
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2
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Conjectured closed form for $\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)$
There are few known closed form for values of the dilogarithm at specific points. Sometimes only the real part or only the imaginary part of the value is known, or a relation between several different ...
- 47.3k
18
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3
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Closed form for $\int_0^e\mathrm{Li}_2(\ln{x})\,dx$?
Inspired by this question and this answer, I decided to investigate the family of integrals
$$I(k)=\int_0^e\mathrm{Li}_k(\ln{x})\,dx,\tag{1}$$
where $\mathrm{Li}_k(z)$ represents the polylogarithm of ...
- 856
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8
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About the integral $\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx$
I came across the following Integral and have been completely stumped by it.
$$\large\int_{0}^{1}\dfrac{\log(x)\log^2(1+x)}{x}dx$$
I'm extremely sorry, but the only thing I noticed was that the ...
17
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1
answer
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A rare integral involving $\operatorname{Li}_2$
A rare but interesting integral problem:
$$\int_{0}^{1}
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} }
=\...
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3
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A conjectured value for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$
In evaluating the integral given here it would seem that:
$$\operatorname{Re} \operatorname{Li}_4 (1 + i) \stackrel{?}{=} -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{...
- 11.8k
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How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
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