5
$\begingroup$

Referring to trivial form, sometimes called Cavalieri's formula: $$ \int x^p \mathrm{d}x=\frac{x^{p+1}}{p+1} + C, \qquad p\neq-1 $$ I was wondering what other restrictions on $p$ exist besides $p \neq -1$.

Initially I thought $p \in \mathbb{Z}\backslash\{-1\}$, but the formula is also used to obtain forms for radicals and root functions. Then I thought perhaps $p \in \mathbb{R}\backslash\{-1\}$, but then I doubted: would it work on complex numbers? I carried out some integrations with $p\in \mathbb{C}\backslash\{-1\}$, and the formula seems to work fine.

I could not find a proof with any restrictions on $p$. Any ideas on how general can $p$ become without breaking the equality?

Thank you for the insight.

$\endgroup$
4
  • $\begingroup$ I get the feeling that it does not hold for quaternions in general, owing to the fact that multiplication in $\Bbb H$ doesn't commute, and per Wikipedia, for instance, $$\frac{d}{dx} x^2 = x \otimes 1 + 1 \otimes x$$ I think multiplying by $3$ on both sides would get you to where the antiderivative on the left-hand side would be $x^3$, but I can't imagine the right-hand side would look anything like that. $\endgroup$ Commented Mar 15, 2020 at 7:49
  • 1
    $\begingroup$ But I do not at all know enough about even the basics of quaternion analysis to be able to justify that claim beyond a mere feeling. But considering you've gone through the real and complex numbers, quaternions seems like a logical next place to look. (And I'm sorry if the earlier comments were just dead wrong.) $\endgroup$ Commented Mar 15, 2020 at 7:50
  • 1
    $\begingroup$ Fun fact, it can work for $p=-1$ too if you write $C+\lim\limits_{q\to p}\frac{x^{q+1}-1}{q+1}$. $\endgroup$
    – anon
    Commented Oct 4, 2020 at 4:26
  • $\begingroup$ I used wolfram alpha and it worked for $i$ and $3+2i$. $\endgroup$
    – user
    Commented Oct 4, 2020 at 4:47

0

You must log in to answer this question.