Let $x \in \mathbb{C}$. What is the Taylor series for $x^w/w$ around $w=1$?
I have tried by finding the $n$-th derivative of $\frac{x^w}{w}$ with respect to $w$, and then evaluating it at $w=1$. I only have a recurrence relation which doesn’t seem to help much. $$R_n := \left( \frac{x^w}{w} \right)^{(n)} = R_{n-1} \ln x -\left( \frac{x^w}{w^n} \right)^{(n-1)}$$.
If you want the Taylor series of $\;f(z)=\cfrac{a^z}z\;$ , with more usual symbols, around $\;z=1\;$ , we can do as follows:
$$\frac{a^z}z=\frac1{1+(z-1)}\cdot a\cdot a^{z-1}=a\sum_{n=0}^\infty(z-1)^n\cdot e^{(z-1)\log a}=a\sum_{n=0}^\infty(z-1)^n\cdot\sum_{n=0}^\infty\frac{\log^na}{n!}(z-1)^n$$
Continue...and the above is valid for $\;|z-1|<1\;$ (for the part of the geometric series)