Mathematica gives: $$S= -\frac{1}{12}\pi^2\log(2)+\frac{\log(2)^3}{6}+ \frac{7}{8}\zeta(3)$$ How can I prove it?
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Complex analysis seems the way! – vidyarthi Jun 16 '17 at 10:10
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You can't prove it, obviously. Look for identities and special values of polylogarithm (trilogarithm, in this case), cf. https://math.stackexchange.com/questions/555961/the-most-complete-reference-for-identities-and-special-values-for-polylogarithm – Jun 16 '17 at 10:17
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@vidyarthi I don't see how. – Burrrrb Jun 16 '17 at 10:43
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Why is it obvious, @professorvector? – Burrrrb Jun 16 '17 at 10:43
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I guess you asked how to prove that, because you don't know. If that's wrong, I'm sorry. – Jun 16 '17 at 10:46
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Why the downvote? – Burrrrb Jun 16 '17 at 11:48
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Hint:
Let $$f(x):=\sum_{n=1}^\infty\frac{x^n}{n^3}.$$
Then
$$f'(x):=\sum_{n=1}^\infty\frac{x^{n-1}}{n^2},$$
$$(xf'(x))'=\sum_{n=1}^\infty\frac{x^{n-1}}{n},$$
$$(x(xf'(x))')'=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}.$$
Now backward, integrating from $0$ to $x$,
$$(xf'(x))'=\frac{\log(1-x)}x,$$
$$f'(x)=\frac1x\int_0^x\frac{\log(1-x)}xdx,$$
$$f(x)=\int_0^x\left[\frac1x\int_0^x\frac{\log(1-x)}xdx\right]dx.$$
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but we require complex analysis to evaluate $\int_0^x\frac{\log(1-x)}{x}dx$ isnt it? – vidyarthi Jun 16 '17 at 11:00
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I don't think the integrals here can be easily evaluated because if they could be, then $lim_{x\rightarrow1^{+}}f(x)$ would give $\zeta(3)$, and this solves an open problem. Right? – Burrrrb Jun 16 '17 at 11:42
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$$S=\sum_{n=1}^{\infty}\frac{x^2}{n^3}=\text{Li}_3(x)$$ where appears the polylogarithm function. Have a look here for special values; in your case, $x=\frac 12$.
Claude Leibovici
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1Your link gives $\text{Li}_3(x)= -\frac{1}{12}\pi^2\log(2)+\frac{\log(2)^3}{6}+ \frac{7}{8}\zeta(3)$ and it says that it can be evaluated analytically but it doesn't tell me how. I'm asking for a derivation. – Burrrrb Jun 16 '17 at 11:47