Cauchy formula for repeated integration

The Cauchy formula for repeated integration allows one to compress $n$ antidifferentiations of a function into a single integral.

[edit] Scalar case

Let $f$ be a continuous function on the real line. Then the $n t h$ antidifferentiation of $f$ ,

$f^{[n]}(x) = \int_0^x\int_0^{\sigma_1}\cdots\int_0^{\sigma_{n-1}}f(\sigma_{n})d\sigma_{n}\cdots d\sigma_2 d\sigma_1$ ,

is given by single integration

$f^{[n]}(x) = \frac{1}{(n-1)!}\int_0^x\left(x-y\right)^{n-1}f(y)dy$ .

A proof is given by induction. Since $f$ is continuous, the base case is given by

$\frac{d}{dx}f^{[1]}(x) = \frac{d}{dx}\int_0^xf(y)dy = f(x)$ .

A little work shows that we also have

$\frac{d}{dx}f^{[n]}(x) = \frac{d}{dx}\frac{1}{(n-1)!}\int_0^x\left(x-y\right)^{n-1}f(y)dy = f^{[n-1]}(x)$ .

Hence, $f [n] (x)$ gives the $n t h$ antidifferentiation of $f (x)$ .

Gerald B. Folland, Advanced Calculus, p. 193, Prentice Hall (2002). ISBN 0-13-065265-2