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 $$f^{(-n)}(x)=\frac{1}{(n-1)!}{\int }_a^x{(x-t)}^{n-1}f(t)\mathrm{d}t.$$

Proof

A proof is given by induction. The base case with n = 1 is trivial, since it is equivalent to $$f^{(-1)}(x)=\frac{1}{0!}{\int }_a^x(x-t{)}^{0}f(t)\mathrm{d}t={\int }_a^{x}f(t)\mathrm{d}t.$$

Now, suppose this is true for n, and let us prove it for n + 1. Firstly, using the Leibniz integral rule, note that $$\frac{\mathrm{d}}{\mathrm{d}x}[\frac{1}{n!}{\int }_a^x(x-t{)}^{n}f(t)\mathrm{d}t]=\frac{1}{(n-1)!}{\int }_a^x(x-t{)}^{n-1}f(t)\mathrm{d}t.$$ Then, applying the induction hypothesis, $$\begin{array}{cc}{f}^{-(n+1)}(x)& ={\int }_a^x{\int }_a^{{\sigma }_1}\cdots {\int }_a^{{\sigma }_n}f({\sigma }_{n+1})\mathrm{d}{\sigma }_{n+1}\cdots \mathrm{d}{\sigma }_2\mathrm{d}{\sigma }_1\\ & ={\int }_a^x[{\int }_a^{{\sigma }_1}\cdots {\int }_a^{{\sigma }_n}f({\sigma }_{n+1})\mathrm{d}{\sigma }_{n+1}\cdots \mathrm{d}{\sigma }_2]\mathrm{d}{\sigma }_1.\end{array}$$ 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 $$\begin{array}{cc}& ={\int }_a^x\frac{1}{(n-1)!}{\int }_a^{{\sigma }_1}({\sigma }_1-t{)}^{n-1}f(t)\mathrm{d}t\mathrm{d}{\sigma }_1\\ & ={\int }_a^x\frac{\mathrm{d}}{\mathrm{d}{\sigma }_1}[\frac{1}{n!}{\int }_a^{{\sigma }_1}({\sigma }_1-t{)}^{n}f(t)\mathrm{d}t]\mathrm{d}{\sigma }_1\\ & =\frac{1}{n!}{\int }_a^x(x-t{)}^{n}f(t)\mathrm{d}t.\end{array}$$

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

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.

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). Template:ISBN

External links

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