Answer by Ali Shadhar for Closed Form for the Imaginary Part of...
Start with the trilog identity$$\text{Li}_3(x)+\text{Li}_3(1-x)+\text{Li}_3\left(\frac{x-1}{x}\right)$$$$=\zeta(3)+\frac16\ln^3(x)+\zeta(2)\ln(x)-\frac12\ln^2(x)\ln(1-x)\tag1$$set $x=i$ and consider...
View ArticleAnswer by Paul Enta for Closed Form for the Imaginary Part of...
Using the relation between 3 trilogarithms given here:\begin{equation}...
View ArticleAnswer by user97357329 for Closed Form for the Imaginary Part of...
I'll sketch a simple solution to the fact that $$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3),$$ as the OP stated.First step: We...
View ArticleAnswer by Tito Piezas III for Closed Form for the Imaginary Part of...
I'm seriously late for this party, but to complement Cleo's answer, if we consider a hypergeometric as a closed-form, then using a different hypergeometric, the family...
View ArticleAnswer by user153012 for Closed Form for the Imaginary Part of...
Form here we know that$$\operatorname{Li}_3(z)-\operatorname{Li}_3\left(\frac{1}{z}\right)=-\frac{1}{6} \ln^3(z)-\frac{\pi\sqrt{-(z-1)^2}}{2(z-1)}\ln^2(z)+\frac{\pi^2}{3}\ln(z)\tag{1}.$$If we put...
View ArticleAnswer by Cleo for Closed Form for the Imaginary Part of...
If you consider a hypergeometric function to be a closed form, you can have the following...
View ArticleAnswer by Gahawar for Closed Form for the Imaginary Part of...
This is equivalent to finding a closed form for the following series$$I :=\mathrm{Im} \left[\mathrm{Li}_3\left(\frac{1+i}{2}\right)\right] = \sum_{k=1}^{\infty} \frac{\sin \pi k/4}{2^{k/2}k^3}.$$I was...
View ArticleClosed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
$\qquad\qquad$Is there any closed form expression for the imaginary part of$~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?Motivation: We alreadyknow...
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