Cauchy formula for repeated integration

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress n antidifferentiations of a function into a single integral (cf. Cauchy's formula).

Scalar case

Let ƒ be a continuous function on the real line. Then the nth repeated integral of ƒ based at a,

$f^{(-n)}(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1$,

is given by single integration

$f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t$.

A proof is given by induction. Since ƒ is continuous, the base case follows from the Fundamental theorem of calculus:

$\frac{\mathrm{d}}{\mathrm{d}x} f^{(-1)}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\,\mathrm{d}t = f(x)$;

where

$f^{(-1)}(a) = \int_a^a f(t)\,\mathrm{d}t = 0$.

Now, suppose this is true for n, and let us prove it for n+1. Apply the induction hypothesis and switching the order of integration,

\begin{align} f^{-(n+1)}(x) &= \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n}} f(\sigma_{n+1}) \, \mathrm{d}\sigma_{n+1} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1 \\ &= \frac{1}{(n-1)!} \int_a^x \int_a^{\sigma_1}\left(\sigma_1-t\right)^{n-1} f(t)\,\mathrm{d}t\,\mathrm{d}\sigma_1 \\ &= \frac{1}{(n-1)!} \int_a^x \int_t^x\left(\sigma_1-t\right)^{n-1} f(t)\,\mathrm{d}\sigma_1\,\mathrm{d}t \\ &= \frac{1}{n!} \int_a^x \left(x-t\right)^n f(t)\,\mathrm{d}t \end{align}

The proof follows.

Applications

In fractional calculus, this formula can be used to construct a notion of differintegral, allowing one to differentiate or integrate a fractional number of times. Integrating a fractional number of times with this formula is straightforward; one can use fractional n by interpreting (n-1)! as Γ(n) (see Gamma function).