The Basel problem consisted in finding the sum in closed form of the following numerical series, of which was already known its convergence and its numerical value with arbitrary precision (https://en.wikipedia.org/wiki/Basel_problem).
$$\sum_{n=1}^\infty\frac{1}{n^2}$$
The result was found by Euler about a hundred years later after the problem was posed for the first time.
$$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}=\zeta(2)$$
where $\zeta(n)$ is the Riemann zeta function.
Now, I would like to know what is the exact sum (in closed form) of the following functional series that, even if obviously different from the previous numerical series (even just for the fact that the previous is a numerical series while the following is a functional series), has in its structure the factor related to the Basel problem.
$$\qquad\qquad\qquad\ \sum_{n=1}^\infty \frac{x^\frac{1}{n}}{n^2}=\sum_{n=1}^\infty \frac{\sqrt[n]{x}}{n^2}\qquad\ \text{with}\quad x\in\Bbb N-\{0\}$$
I already know that the functional series is convergent, so I would kindly ask not to address that aspect: I just need the exact sum in closed form with the proof. Any other consideration or suggestion about how to proceed without the solution and the step-by-step proof would be useless; unless, of course, somebody can prove that there is something wrong in the question itself.
Thank you
-Update-
Here another way in which is possible to rewrite the functional series considered: $$\\$$
$$\frac{\sqrt[n]{x}}{n^2}=\frac{x^\frac{1}{n}}{n^2}= \frac{e^{\ln(x^\frac{1}{n})}}{n^2}=\frac{e^{\frac{1}{n}\ln(x)}}{n^2}=\frac{1+\frac{1}{n}\ln(x)+\frac{1}{n^2}\frac{\ln^2(x)}{2!}+\frac{1}{n^3}\frac{\ln^3(x)}{3!}+\frac{1}{n^4}\frac{\ln^4(x)}{4!}+...}{n^2}=\\ =\frac{1}{n^2}+\frac{1}{n^3}\ln(x)+\frac{1}{n^4}\frac{\ln^2(x)}{2!}+\frac{1}{n^5}\frac{\ln^3(x)}{3!}+\frac{1}{n^6}\frac{\ln^4(x)}{4!}+...$$
hence
$$\sum_{n=1}^\infty \frac{\sqrt[n]{x}}{n^2}=\sum_{n=1}^\infty \frac{x^\frac{1}{n}}{n^2}=\sum_{n=1}^\infty\Biggl(\frac{1}{n^2}+\frac{1}{n^3}\ln(x)+\frac{1}{n^4}\frac{\ln^2(x)}{2!}+\frac{1}{n^5}\frac{\ln^3(x)}{3!}+\frac{1}{n^6}\frac{\ln^4(x)}{4!}+...\Biggr)=\\ =\zeta(2)+\zeta(3)\ln(x)+\zeta(4)\frac{\ln^2(x)}{2!}+\zeta(5)\frac{\ln^3(x)}{3!}+\zeta(6)\frac{\ln^4(x)}{4!}+...= \sum_{n=1}^\infty \zeta(n+1)\frac{\ln^{^{n-1}}(x)}{(n-1)!}=\\=\sum_{n=0}^\infty \zeta(n+2)\frac{\ln^{^{n}}(x)}{n!} \qquad\ \qquad\ \text{with}\quad x\in\Bbb N-\{0\}$$