The extension of this derivative notation $f^{(0)} = f, f^{(1)} = f', ...$

Question

The extension of this derivative notation $f^{(0)} = f, f^{(1)} = f', ...$

Just a quick question on this notation, is $f^{(-1)}$ used for antiderivatives?

Answer 1

Sometimes people use that notation, yes. But as you know antiderivatives are not unique, so there is some problem with using this notation naively (since we will tend to assume it means one function instead of a family of functions). This non-uniqueness problem goes away if you choose a "base point" for your integration (meaning that you define $f^{(-1)}(x)=\int_a^x f(y) dy$ for some particular $a$). For example, it is sometimes natural to pick the left endpoint of your domain. But this can sometimes be inconvenient. For example, the "natural" antiderivative of $x^k$ is $x^{k+1}/(k+1)$, but there is no fixed $a$ such that $\int_a^x y^k dy = x^{k+1}/(k+1)$ for all $k \in \mathbb{R} \setminus \{ -1 \}$.

For more information, you can look up "fractional calculus", which gives a notion of $f^{(s)}$ for arbitrary complex $s$.