Compute the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$ or $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$ I need to calculate the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$, where $a$, $b$ are REAL NUMBERS and $b>0$.
(my goal is to determine the definite integral $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$, where $d<1$).
Thanks !
 A: \begin{align}
\int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy}  = \left\{ \begin{array}{l}
 \int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a < 0,\,\,\text{setting}\,\,a = 1 - r,r > 1 \\ 
 \frac{1}{b}\left( {1 - d} \right)^b ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 0 \\ 
 \int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a > 0,\,\,\text{setting}\,\,a = 1 - r,r < 1 \\ 
 \end{array} \right.
\end{align}
So that, you can continue with incomplete beta function.
Another method:
Case 1:If $a=0$ and $b>0$, then $$I=
\int_d^1 {\left( {1 - y} \right)^{b - 1} dy}  = \frac{1}{b}\left( {1 - d} \right)^b 
$$
Case 2:If $b=0$ and $a>0$ and $a\ne1$, then $$
\int_d^1 {y^{ - a} dy}  = \frac{{1 - d^{1 - a} }}{{1 - a}}
$$ 
Case 3:If $b>0$ with $b\ne1$ and $a>0$, $-1<d<1$, then the binomial series 
\begin{align}
\left( {1 + x} \right)^\alpha   = \sum\limits_{k = 0}^\infty  {\left( \begin{array}{l}
 \alpha  \\ 
 k \\ 
 \end{array} \right)x^k } , \alpha \in \mathbb{R}-\{0\},\,\,\, \forall x \in (-1,1),
\end{align}
where $$
\left( \begin{array}{l}
 \alpha  \\ 
 k \\ 
 \end{array} \right) = \frac{{\alpha \left( {\alpha  - 1} \right)\left( {\alpha  - 2} \right) \cdots \left( {\alpha  - \left( {k - 1} \right)} \right)}}{{k!}}$$
may be used by setting $\alpha:=b-1$, $(b\ne1)$, the interval $(d,1)\subseteq (-1,1)$, we have
\begin{align}
\left( {1 - y} \right)^{b - 1}  = \sum\limits_{k = 0}^\infty  {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
   {b - 1}  \\
   k  \\
\end{array}} \right)y^k } 
\end{align}
Multiplying both sides by $y^{ - a} $ and then integrating both sides with respect to $y$, we get
\begin{align}
&\int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy}  
\\
&= \int_d^1 {\left( {\sum\limits_{k = 0}^\infty  {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
   {b - 1}  \\
   k  \\
\end{array}} \right)y^{k - a} } } \right)dy} 
\\
&= \sum\limits_{k = 0}^\infty  {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
   {b - 1}  \\
   k  \\
\end{array}} \right)\left( {\int_d^1 {y^{k - a} dy} } \right)} 
\\ 
&= \sum\limits_{k = 0}^\infty  {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
   {b - 1}  \\
   k  \\
\end{array}} \right)\frac{{1 - d^{k - a + 1} }}{{k - a + 1}}}, \,\,\,\, \text{with} \,\,
a \ne k + 1 
\end{align}
where, we used the fact that the series converges uniformly for all $y \in (d,1)\subseteq (-1,1)$.
We note that the condition $a\ne k+1$ is equivalent to say that $a \notin \mathbb{N}$ i.e., $a\in\mathbb{R}-\mathbb{N}$. One more point, If $d<-1$ the series diverges so that we cannot integrate. All these additional conditions should be added.
A: HINT use integration by parts where function to be differentiated is (1-y)^b-1 and integrable function is y^-a 
YOu have reduction formula setup .
then when you get values of m and n .
