Is that will work as analytic continuation for mandelbrot function, if not why?
Mandelbrot function can be defined with Catalan numbers as $\sum_{n=1}^{\infty}\frac {(2n-2)!}{(n-1)!(n)!}x^n$
From (Infinite series as integral representation) we know that $\sum_{k=x}^{\infty} f(k)=-2\pi \int_{-\infty}^{\infty}\frac {F (x-1/2+it)}{(e^{\pi t}+e^{-\pi t})^2}dt$, which works for divergent series (and this is antiderivative without +C). So it should technically work.
I'm not saying that this is possible to give exact value of this integral, but at least we could count it by numerical methods
PS: If that will work, pls post image of this function in complex plane
The following will work as an analytical continuation. Take $$\sum_{n=1}^\infty (-1)^n \frac{x^n(2n-2)!}{(n-1)!n!} $$ Then, rewriting as a contour integral it becomes $$-\int_C \frac{\csc(\pi n)}{2i} x^n \frac{(2n-2)!}{(n-1)!n!} dn$$ Where $n!$ is of course the gamma function, but I've written it as the factorial to emphasize the correspondence between this integral and the original series. A linear contour from $1/2 < c < 1$ will work, in particular, we can compute the series as $$-\int_{3/4 - i \infty}^{3/4 + i \infty} \frac{\csc(\pi n)}{2i} x^n \frac{(2n-2)!}{(n-1)!n!} dn$$ For instance, here is what the function looks like along the real line, with the original function in blue and the continuation in orange. I suspect this integral doesn't converge for $x$ negative, so some more work might be required to get a continuation there (this may suggest there is a branch cut there by the way).
Looks like I was correct about that branch cut, the series evaluates to simply $-\frac{1}{2}\sqrt{1-4x}+\frac{1}{2}$. Which is of course the solution to $f(x) = f(x)^2 + x$ (i.e. the mandelbrot iteration).