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, [math]\displaystyle{ 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, }[/math] is given by single integration [math]\displaystyle{ f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t. }[/math]
Proof
A proof is given by induction. The base case with n=1 is trivial, since it is equivalent to: [math]\displaystyle{ f^{(-1)}(x) = \frac1{0!}\int_a^x {(x-t)^0}f(t)\,\mathrm{d}t = \int_a^x f(t)\,\mathrm{d}t }[/math]Now, suppose this is true for n, and let us prove it for n+1. Firstly, using the Leibniz integral rule, note that [math]\displaystyle{ \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 . }[/math]
Then, applying the induction hypothesis, [math]\displaystyle{ \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 \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} }[/math]
- It has been shown that this statement holds true for the base case [math]\displaystyle{ n=1 }[/math].
- If the statement is true for [math]\displaystyle{ n=k }[/math], then it has been shown that the statement holds true for [math]\displaystyle{ n=k+1 }[/math].
- 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 [math]\displaystyle{ n \in \Z_{\geq 0} }[/math] is replaced by [math]\displaystyle{ \alpha \in \Complex,\ \Re(\alpha)\gt 0 }[/math], and the factorial is replaced by the gamma function. The two formulas agree when [math]\displaystyle{ \alpha \in \Z_{\geq 0} }[/math].
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.
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