# Derive a recursive formula using integration by parts

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 $$\log(y)$$).

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