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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?


marked as duplicate by Did, kingW3, Arnaldo, Jack D'Aurizio sequences-and-series Jun 16 '17 at 15:23

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  • $\begingroup$ Complex analysis seems the way! $\endgroup$ – vidyarthi Jun 16 '17 at 10:10
  • $\begingroup$ You can't prove it, obviously. Look for identities and special values of polylogarithm (trilogarithm, in this case), cf. math.stackexchange.com/questions/555961/… $\endgroup$ – Professor Vector Jun 16 '17 at 10:17
  • $\begingroup$ @vidyarthi I don't see how. $\endgroup$ – MereMortal47 Jun 16 '17 at 10:43
  • $\begingroup$ Why is it obvious, @professorvector? $\endgroup$ – MereMortal47 Jun 16 '17 at 10:43
  • $\begingroup$ I guess you asked how to prove that, because you don't know. If that's wrong, I'm sorry. $\endgroup$ – Professor Vector Jun 16 '17 at 10:46


Let $$f(x):=\sum_{n=1}^\infty\frac{x^n}{n^3}.$$




$$(x(xf'(x))')'=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}.$$

Now backward, integrating from $0$ to $x$,




  • $\begingroup$ but we require complex analysis to evaluate $\int_0^x\frac{\log(1-x)}{x}dx$ isnt it? $\endgroup$ – vidyarthi Jun 16 '17 at 11:00
  • $\begingroup$ 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? $\endgroup$ – MereMortal47 Jun 16 '17 at 11:42
  • $\begingroup$ @Asemismaiel: maybe Ramanujan... :) $\endgroup$ – Yves Daoust Jun 16 '17 at 11:58

$$S=\sum_{n=1}^{\infty}\frac{x^2}{n^3}=\text{Li}_3(x)$$ where appears the polylogarithm function. Have a look here for special values; in your case, $x=\frac 12$.

  • $\begingroup$ Your 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. $\endgroup$ – MereMortal47 Jun 16 '17 at 11:47

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