I was wondering if it would be possible to get an analytically continuation of the following function:
$$ J(x) = \sum_{r=1}^\infty \ln(r)x^r $$
My attempt
Consider the following:
(1) $$ J'(x) = \sum_{r=1}^\infty r \ln(r) x^{r-1}$$
We also know:
(2) $$ \frac{J}{1-x} = \sum_{r=1}^\infty \ln(r!) x^r$$
Elaborating on (2)
$$ J = \ln(2) x^2 + \ln(3) x^3 + \dots $$
Dividing by $1-x$ and expanding on the R.H.S:
$$ \implies \frac{J}{(1-x)} = \ln(2)(x^2 + x^3 + \dots) + \ln(3)(x^3 + x^4 + \dots) + \dots$$
Combining all the powers of $x$
$$ \frac{J}{1-x} = \sum_{r=1}^\infty \ln(r!) x^r$$
Adding (1) and (2) and multiplying by $x$
$$ x J'(x) + \frac{x J(x)}{1-x} = \sum_{r=1}^\infty (r \ln (r) + \ln(r-1)!) x^r$$
Using the identity: $ \sum_{r=1}^n (r \ln (r) + \ln(r-1)!) = n \ln n!$
$$ \implies \frac{x J'(x)}{(1-x_)} + \frac{x J(x)}{(1-x)^2} = \sum_{r=1}^\infty r \ln(r!) x^r$$
Differentiating (2) and multiplying $x$ and comparing:
$$ \implies \frac{x J'(x)}{(1-x_)} + \frac{x J(x)}{(1-x)^2} = \frac{J'(x)}{1-x} - \frac{J(x)}{(1-x)^2} $$
Solving:
$$ J(x) = c \exp(\frac{-2}{x-1})(x-1) $$
But this can't be correct! Can someone point out where I went wrong?