0
$\begingroup$

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 $F(x,n)$.

$\endgroup$
4
$\begingroup$

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 $\log(y)$).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.