# Stuck with an integration

How do you integrate $$\int_0^1 y^{k_1} (1 - (1-y)^{\alpha})^{k_2}\,\mathrm{d}y$$ where $k_1$ and $k_2$ are non-negative integers and $\alpha>1$?

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Expand with binomial theorem and use beta functions. – anon Jan 27 '12 at 6:57
@anon $k1$ and $k2$ arent necessarily positive integers, no? – Ravi Jan 27 '12 at 7:06
actually k1 and k2 are positive integers(including 0).. Also, does the integral become more tractable if alpha is restricted to rational numbers – Nischal Jan 27 '12 at 7:16

$$I=\int\limits_0^1 {{y^a}{{\left( {1 - {{\left( {1 - y} \right)}^b }} \right)}^c}} dy$$

$$1-y=u$$ $$I=\int\limits_0^1 {{{\left( {1 - u} \right)}^a}{{\left( {1 - {u^b}} \right)}^c}} du$$

$${\left( {1 - {u^b }} \right)^c} = \sum\limits_{k = 0}^c {{{\left( { - 1} \right)}^{bk}}{c \choose k}{u^{bk}}}$$

$$I=\sum\limits_{k = 0}^c {{{\left( { - 1} \right)}^{bk}}{c \choose k}\int\limits_0^1 {{{\left( {1 - u} \right)}^a}{u^{bk}}} du}$$

$$\int\limits_0^1 {{{\left( {1 - u} \right)}^a}{u^{bk}}} du = B\left( {a+1,bk+1} \right)$$

$$\sum\limits_{k = 0}^c {{{\left( { - 1} \right)}^{bk}}{c \choose k}B\left( {a+1,bk+1} \right)}$$

Maybe there is some multiple argument formula you can use to simplify things.

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