# 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). For non-integer n it yields the definition of fractional integrals and (with n < 0) fractional derivatives.

## 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 $f^{(-1)}(x)={\frac {1}{0!}}\int _{a}^{x}{(x-t)^{0}}f(t)\,\mathrm {d} t=\int _{a}^{x}f(t)\,\mathrm {d} t.$
Now, suppose this is true for n, and let us prove it for n + 1. Firstly, using the Leibniz integral rule, note that ${\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}(x-t)^{n-1}f(t)\,\mathrm {d} t.$ Then, applying the induction hypothesis, ${\begin{aligned}f^{-(n+1)}(x)&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\right]\,\mathrm {d} \sigma _{1}.\end{aligned}}$ Note that the term within square bracket has n-times successive 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 leads to $\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}={\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t)\,\mathrm {d} t.$ Putting this expression inside the square bracket results in ${\begin{aligned}&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t)\,\mathrm {d} t\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n}f(t)\,\mathrm {d} t\right]\,\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t.\end{aligned}}$
* 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 $n\in \mathbb {Z} _{\geq 0}$ is replaced by $\alpha \in \mathbb {C} ,\ \Re (\alpha )>0$, and the factorial is replaced by the gamma function. The two formulas agree when $\alpha \in \mathbb {Z} _{\geq 0}$.
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.