I was evaluating an integral a while ago that appeared in a Youtube video, which I expressed in terms of the Riemann-zeta function (which seems to emerge from the argument of such sums, rather frequently - although this appearance of $\zeta(s)$ was alluded to by the thumbnail). However, my attention was soon centred on $\zeta(s)$ itself and the idea of plugging $x^\frac{1}{n}$ in the numerator each of the terms came to mind (the $s$ in the modified $\zeta(s)$ could be then be viewed as a parameter).

The sum would then look like this: $$\sum_{n=1}^{\infty} \frac{x^\frac{1}{n}}{n^s}.$$ This is similar to the polylogarithm, $Li_s(x) = \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n^s}$, except $\frac{1}{n}$ appears in the exponent of $x$ instead of $n$.

My question would then be: Does a closed form exist for such a sum? If so, what would it look like? I'm particularly interested in the closed form of this sum for $s=2$, which would look like this: $$\sum_{n=1}^{\infty} \frac{x^\frac{1}{n}}{n^2}$$

If you know the answer to this question, please could you tell me? It would be much appreciated.

Thank you in advance! (p.s. by closed form I really mean in terms of known functions that can be expressed together in a finite number of operations to construct the equivalent of this series, like log(x), sin(x), cos(x), or even things like $\Gamma(x)$, erf(x) etc.; it would be fascinating to see if the values of this function are algebraically independent of other rational, irrational, or transcendental constants - or if this is an entirely different function on its own, but I'm getting ahead of myself).

  • 4
    $\begingroup$ I doubt it has a closed form. $\endgroup$
    – xpaul
    Dec 1, 2022 at 17:14
  • $\begingroup$ Note that even the Polylogarithm function you mention does not have a "closed form". Usually in analysis when a function is used "often" some notation is invented for that function and then used. A very pedantic person might say that even the natural logarithm, trig functions and square roots do not have a "closed form" expression, because they cannot be calculated in a finite number of steps. What do you mean with "closed form"? $\endgroup$
    – student91
    Dec 1, 2022 at 17:33
  • $\begingroup$ Maybe this integral representation of $(1+z)^a$ helps? Then interchange the sum and integral if possible $\endgroup$ Dec 1, 2022 at 17:38

1 Answer 1


Using the series of $e^x$, you get a zeta series:

$$\sum_{n=1}^\infty\frac{\sqrt[n]x}{n^s}=\sum_{n=0}^\infty\frac{\zeta(n+s)\ln^n(x)}{n!}\iff \sum_{n=0}^\infty\frac{\zeta(n+s)x^n}{n!}=\sum_{n=1}^\infty\frac{e^\frac xn}{n^s} $$

shown here

Using $\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{e^t-1} dt$:

$$\sum_{n=1}^\infty\frac{\sqrt[n]x}{n^s}=\int_0^\infty \frac{t^{s-1}} {e^t-1}\,_0 \tilde{\text F}_1(s,t\ln(x))dt$$

Shown here where the regularized confluent hypergeometric function appears.

  • 1
    $\begingroup$ Also $\int_0^\infty y^{z-1}\sum_{n\ge 1} e^{-y/n} n^{-s}dy = \Gamma(z)\zeta(s-z)$ $\endgroup$
    – reuns
    Dec 1, 2022 at 18:18

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