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).
Best Answer
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.