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, $$f^{(-n)}(x)=\underset{a}{\overset{x}{\int }}\underset{a}{\overset{{\sigma }_1}{\int }}\cdots \underset{a}{\overset{{\sigma }_{n-1}}{\int }}f({\sigma }_n)\mathrm{d}{\sigma }_n\cdots \mathrm{d}{\sigma }_2\mathrm{d}{\sigma }_1,$$ is given by single integration $$f^{(-n)}(x)=\frac{1}{(n-1)!}\underset{a}{\overset{x}{\int }}{(x-t)}^{n-1}f(t)\mathrm{d}t.$$

Proof

A proof is given by induction. The base case with n=1 is trivial, since it is equivalent to: $$f^{(-1)}(x)=\frac{1}{0!}\underset{a}{\overset{x}{\int }}(x-t{)}^{0}f(t)\mathrm{d}t=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t$$Now, suppose this is true for n, and let us prove it for n+1. Firstly, using the Leibniz integral rule, note that $$\frac{\mathrm{d}}{\mathrm{d}x}[\frac{1}{n!}\underset{a}{\overset{x}{\int }}{(x-t)}^{n}f(t)\mathrm{d}t]=\frac{1}{(n-1)!}\underset{a}{\overset{x}{\int }}{(x-t)}^{n-1}f(t)\mathrm{d}t.$$

Then, applying the induction hypothesis, $$\begin{array}{rl}{f}^{-(n+1)}(x)& =\underset{a}{\overset{x}{\int }}\underset{a}{\overset{{\sigma }_1}{\int }}\cdots \underset{a}{\overset{{\sigma }_n}{\int }}f({\sigma }_{n+1})\mathrm{d}{\sigma }_{n+1}\cdots \mathrm{d}{\sigma }_2\mathrm{d}{\sigma }_1\\ & =\underset{a}{\overset{x}{\int }}\frac{1}{(n-1)!}\underset{a}{\overset{{\sigma }_1}{\int }}{({\sigma }_1-t)}^{n-1}f(t)\mathrm{d}t\mathrm{d}{\sigma }_1\\ & =\underset{a}{\overset{x}{\int }}\frac{\mathrm{d}}{\mathrm{d}{\sigma }_1}[\frac{1}{n!}\underset{a}{\overset{{\sigma }_1}{\int }}{({\sigma }_1-t)}^{n}f(t)\mathrm{d}t]\mathrm{d}{\sigma }_1\\ & =\frac{1}{n!}\underset{a}{\overset{x}{\int }}{(x-t)}^{n}f(t)\mathrm{d}t.\end{array}$$

• 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 $n\in {\mathbb{\mathbb{Z}}}_{\ge 0}$ is replaced by $\alpha \in \mathbb{ℂ},\mathrm{\Re }(\alpha )>0$, and the factorial is replaced by the gamma function. The two formulas agree when $\alpha \in {\mathbb{\mathbb{Z}}}_{\ge 0}$.

Both the Cauchy formula and the Riemann-Liouville integral are generalized to arbitrary dimension 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.

References

• Augustin-Louis Cauchy: Trente-Cinquième Leçon. In: Résumé des leçons données à l’Ecole royale polytechnique sur le calcul infinitésimal. Imprimerie Royale, Paris 1823. Reprint: Œuvres complètes II(4), Gauthier-Villars, Paris, pp. 5–261.
• Gerald B. Folland, Advanced Calculus, p. 193, Prentice Hall (2002). ISBN:0-13-065265-2

External links

• Alan Beardon (2000). "Fractional calculus II". University of Cambridge. http://nrich.maths.org/public/viewer.php?obj_id=1369. 

• Maurice Mischler (2023). "About some repeated integrals and associated polynomials". https://sites.google.com/site/mathmontmus/accueil/pages-math%C3%A9matiques-dures/int%C3%A9grales-ni%C3%A8mes-et-polyn%C3%B4mes-sympas.