Straightforward integration of $$F(x, n)=\int \frac{x^n}{(x-a)^\alpha}dx$$ yields quite a cryptic formula with a hypergeometric function which is not defined given my data (goes infinite), and since I'm using it as a part of some numeric computation it's not acceptable.

I have been advised to derive a recursive formula using integration by parts, but calculus classes are long gone and I struggle with what should be $u$ and $v$ correspondingly.

To be more precise, we're looking for derivation of $F(x, n)$ via $F(x,n-1)$. $F(x, 0)$ is trivial and therefore we will have nice looking algorithm to compute $F(x,n)$.


Change of variables $y=x-a$ gives you

$$ \eqalign{F(y+a,n) &= \int \frac{(y+a)^n}{y^\alpha} \; dy\cr &= \sum_{k=0}^n {n \choose k} a^{n-k}\int y^{k-\alpha}\; dy\cr &= \sum_{k=0}^n {n \choose k} a^{n-k} \frac{y^{k-\alpha+1}}{k-\alpha+1} + C }$$ (except, if $k-\alpha+1 = 0$, replace $y^0/0$ by $\log(y)$).

  • $\begingroup$ That's even better! $\endgroup$ – andrhua Mar 12 '20 at 17:47

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