I want to know how can i solve this function.
$\int{(1-y^d)^n}dy$
Is it possible to solve it? If you know the method, please teach me.
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Are $d$ and $n$ positive integers? – John Hughes Jul 19 '16 at 15:12
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yes. those are positive integer. – S. Jun Jul 19 '16 at 15:14
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Assuming that $d$ and $n$ are positive integers, you can just use the binomial theorem to expand the integrand, and integrate term by term, since it's a polynomial in $y$. $$ (1 - u)^n = \sum_{i = 0}^n (1)^{n-i}{n \choose i} (-u)^i = \sum_{i = 0}^n (-1)^i{n \choose i} u^i $$ In your case, $u = y^d$, so this becomes $$ (1 - y^d)^n = \sum_{i = 0}^n (-1)^i{n \choose i} (y^d)^i = \sum_{i = 0}^n (-1)^i{n \choose i} y^{di} $$ and \begin{align} \int (1 - y^d)^n ~dy &= \int \left( \sum_{i = 0}^n (-1)^i{n \choose i} y^{di} \right) ~dy\\ &= \sum_{i = 0}^n (-1)^i{n \choose i} \int y^{di} ~dy \\ &= \sum_{i = 0}^n (-1)^i {n \choose i} \frac{y^{di+1}}{di + 1}. \end{align}
John Hughes
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$(1-u)^n = \sum_{i=0}^n (-1)^i \dbinom{n}{i} u^n$ ?! Following that, $(1-u)^2 = 0$. You obviously made a typo in that $u^n$. (: Consequently, the result is wrong. – Matt Jul 19 '16 at 15:24
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