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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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    actually$k1$and$k2$are positive integers(including 0).. Also, does the integral become more tractable if alpha is restricted to rational numbers2012-01-27

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

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

${\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.