All Questions
4
questions
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votes
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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
$$\sum_{i=1}^{\infty} \frac {x^{...
- 3
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0
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Difference of polylogarithms of complex conjugate arguments
I have the expression
$$\tag{1}
\operatorname{Li}_{1/2}(z)-\operatorname{Li}_{1/2}(z^*)
$$
Where $\operatorname{Li}$ is the polylogarithm and $^*$ denotes complex conjugation. The expression is ...
- 4,817
3
votes
1
answer
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On $\int_0^{2\pi }\frac{\prod_{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod_{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx$
OP of this post evaluated a lot of remarkable polylog integrals (without proof), from which I conjecture a generalized one ($a_k\in \mathbb N$):
$$\int_0^{2 \pi } \frac{\prod _{k=1}^m \text{Li}_{a_k}(...
- 8,654
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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}...
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