Cauchy formula for repeated integration

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

Scalar case
Let f be a continuous function on the real line. Then the nth repeated integral of f with base-point 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 $$f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.$$

Proof
A proof is given by induction. The base case with n=1 is trivial, since it is equivalent to:

$$f^{(-1)}(x) = \frac1{0!}\int_a^x {(x-t)^0}f(t)\,\mathrm{d}t = \int_a^x f(t)\,\mathrm{d}t$$ Now, suppose this is true for n, and let us prove it for n+1. Firstly, using the Leibniz integral rule, note that

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

Then, applying the induction hypothesis,

$$\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 \\

&= \int_a^x \left[\int_a^{\sigma_1} \cdots \int_a^{\sigma_{n}} f(\sigma_{n+1}) \, \mathrm{d}\sigma_{n+1} \cdots \, \mathrm{d}\sigma_2 \right] \, \mathrm{d}\sigma_1 \\ \end{align}$$

Note, the term within square bracket has n-times succesive integration, and upper limit of outermost integral inside the square bracket is $$ \sigma_1$$. Thus, comparing with the case for n=n, and replacing $$ x, \sigma_1, \cdots , \sigma_n $$ of the formula at induction step n=n with $$ \sigma_1, \sigma_2 , \cdots , \sigma_{n+1} $$ respectively to obtain

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

Putting this expression inside the square bracket results in

$$\begin{align} &= \int_a^x \frac{1}{(n-1)!} \int_a^{\sigma_1}\left(\sigma_1-t\right)^{n-1} f(t)\,\mathrm{d}t\,\mathrm{d}\sigma_1 \\

&= \int_a^x \frac{\mathrm{d}}{\mathrm{d}\sigma_1} \left[\frac{1}{n!} \int_a^{\sigma_1} \left(\sigma_1-t\right)^n f(t)\,\mathrm{d}t\right] \,\mathrm{d}\sigma_1 \\

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


 * It has been shown that this statement holds true for the base case $$n=1$$.
 * If the statement is true for $$n=k$$, then it has been shown that the statement holds true for $$n=k+1$$.
 * Thus this statement has been proven true for all positive integers.

This completes the proof.

Generalizations and applications
The Cauchy formula is generalized to non-integer parameters by the Riemann-Liouville integral, where $$n \in \Z_{\geq 0}$$ is replaced by $$\alpha \in \Complex,\ \Re(\alpha)>0$$, and the factorial is replaced by the gamma function. The two formulas agree when $$\alpha \in \Z_{\geq 0}$$.

Both the Cauchy formula and the Riemann-Liouville integral are generalized to arbitrary dimensions by the Riesz potential.

In fractional calculus, these formulae can be used to construct a differintegral, allowing one to differentiate or integrate a fractional number of times. Differentiating a fractional number of times can be accomplished by fractional integration, then differentiating the result.