# Closed form of $\displaystyle\sum_{n=1}^{\infty} \frac{x^\frac{1}{n}}{n^s}$

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).

• I doubt it has a closed form. Dec 1, 2022 at 17:14
• 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"? Dec 1, 2022 at 17:33
• Maybe this integral representation of $(1+z)^a$ helps? Then interchange the sum and integral if possible Dec 1, 2022 at 17:38

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.

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