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.

While this can be expressed in terms of the incomplete beta function, this runs into a problem if you want a real valued function. The beta function has a branch point at $y = 1$, so you get complex numbers for $y > 1$. However, the imaginary part is a constant $-i\pi$, and since the derivative of a constant is $0$, just ignoring that gives a real-valued antiderivative on $\mathbb R \setminus \{1\}$. This domain is the best you can do, as the integral diverges for $y = 1$. Thus
$$
\mathrm{Re}[B_y(1-s, 0)] + C(y)
$$
is the general form, where $C(y)$ is constant on $y < 1$ and $y > 1$, though not necessarily the same in both regions.
