How to integrate $\int \frac{1}{y^s (1-y)} dy$? I want to integrate the above function where s can take any value (positive or negative). Also, I would like to integrate this without limits, so I won't be able to use Gamma function as per my understanding.
For this specific problem, $0 < y < 1$, hence I tried expanding $(1-y)$ as
$$
1 + y + y^2 + y^3 + \cdots = \frac{1}{1-y}
$$
Multiplying this with $\frac{1}{y^s}$ and integrating gave
$$
\int \frac{1}{y^s (1-y)} dy = \frac{-1}{y^s} \left(\frac{y}{-s+1} + \frac{y^2}{-s+2} + \cdots \right)
$$
For $s = 0$, the expression comes out to be $\ln(1-y)$ which is expected, however, I would like to obtain a closed-form solution for any general s. Any hints are greatly appreciated.

Not entirely sure whether this works but here's what I have come up with
$$\begin{align}\int y^{-s}(1-y)^{-1}dy &= \int y^{-s}\sum_{x=0}^{\infty}y^xdy \\ &=\sum_{x=0}^{\infty} \int y^{-s}y^xdy \\ &=\sum_{x=0}^{\infty} \frac{y^{x-s+1}}{x-s+1} \\ &= y^{1-s}\Phi(y, 1, s-1) \end{align}$$
Given that $0<y<1$, where $\Phi(y, 1, s-1)$ is the Lerch Transcendent
