When |z| < 1, from geometric series,
$$\sum_{k=0}^{\infty} z^k = \frac{1}{1-z}$$
By integration,
$$\sum_{k=1}^{\infty} \frac{z^k}{k} = -\log{(1-z)}$$
This formula also can be verified by Taylor series of $-\log{(1-z)}$.
Then, what about $$\sum_{k=1}^{\infty} \frac{z^{k+x}}{k+x}$$
for $0 \le x < 1$ ?
In the same way,
$$\sum_{k=0}^{\infty} z^{k+x} = \frac{z^x}{1-z}$$
By integration,
$$ \begin{align} \sum_{k=1}^{\infty} \frac{z^{k+x}}{k+x} =& \int_0^z \frac{w^x}{1-w} dw \\ =& \left[ \frac{w^x}{(1-w)\log(w)} \right]_0^z dw \\ =& \frac{z^x}{(1-z)\log(z)} \end{align} $$
This formula should be same with $-\log(1-z)$ when $x=0$.
Howevery, it is not.
What mistake I did?
What is the correct formula?
I stuck here.
You are in the domain of special functions $$\sum_{k=1}^{\infty} \frac{z^{k+x}}{k+x}=z^{x+1} \Phi (z,1,x+1)$$ where appears the Hurwitz-Lerch transcendent function.
If you want to compute the antiderivative $$I=\int \frac{w^x}{1-w}\, dw =\frac{w^{x+1} }{x+1} \, _2F_1(1,x+1;x+2;w)$$ where appears the gaussian hypergeometric function.
Now, for $$J=\int_0^z \frac{w^x}{1-w}\, dw =B_z(x+1,0) \qquad \text{if} \qquad (\Re(z)<1\lor z\notin \mathbb{R})\land \Re(x)>-1$$ where appears the incomplete beta function.