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What's $\displaystyle f(x)=\sum_{n=1}^\infty{\frac{x^n}{n^3}}$?

Note its derivative:

$$\displaystyle f'(x)=\sum_{n=1}^\infty{\frac{x^{n-1}}{n^2}}$$

and the next derivative:

$$\displaystyle f''(x)=\sum_{n=2}^\infty{\frac{x^{n-2}(n-1)}{n^2}}$$

I'm asking because $f(1)$ is Apéry's constant $=\zeta(3)$, $$f'(1)=\frac{\pi^2}{6}=\zeta(2) \text{ and } f''(1)=\infty.$$

Asaf Karagila
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Mark Hurd
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    So, you know that $f(1)$ doesn't have a closed form, but you ask if $f(x)$ does? – Karolis Juodelė Aug 06 '14 at 11:50
  • Note that if you multiply by $x$ before the second derivative, then that derivative brings down $n$ again rather than $n-1$. That makes computations more straightforward. – Semiclassical Aug 06 '14 at 11:55

1 Answers1

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It is the definition of a special function called trilogarithm $\operatorname{Li}_3(x)$. This is a special case of the polylogarithm $$\operatorname{Li}_n(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^n},$$ which has the properties $x\operatorname{Li}'_n(x)=\operatorname{Li}_{n-1}(x)$ and $\operatorname{Li}_n(1)=\zeta(n)$. These identities naturally generalize those mentioned in the question.

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