Cauchy formula for repeated integration

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress n antidifferentiations 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 basepoint 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. Since f is continuous, the base case follows from the fundamental theorem of calculus:

{\frac {\mathrm {d} }{\mathrm {d} x}}f^{(-1)}(x)={\frac {\mathrm {d} }{\mathrm {d} x}}\int _{a}^{x}{\frac {(x-t)^{0}}{0!}}f(t)\,\mathrm {d} t={\frac {\mathrm {d} }{\mathrm {d} x}}\int _{a}^{x}f(t)\,\mathrm {d} t=f(x);

where

f^{(-1)}(a)=\int _{a}^{a}f(t)\,\mathrm {d} t=0.

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{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}{\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{aligned}}

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 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.