Compute the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$ or $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$

Question

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 !

Answer 1

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 .

Answer 2

\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.