I'm looking to find a series for the inverse of Euler's analytic continuation of the harmonic numbers
$$H_x=\int_0^1\frac{1-t^x}{1-t}dt$$
From what I've seen, this integral is equal to the power series
$$-\sum_{k=1}^\infty {x\choose k}\frac{(-1)^k}{k}$$
Using these facts, is it possible to find a series for $g(x)$, acting as the inverse of $H_x$, such that $g(H_x)=x$?
I'm unsure if there is a general process to get the inverse of a power series of known coefficients. This website seems to lead the general idea but I'm not certain if this is applicable here.