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.

Let f be a continuous function on the real line. Then the nth repeated integral of f with base-point a, $${\displaystyle f^{(-n)}(x)=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n-1}}f(\sigma _{n}),円\mathrm {d} \sigma _{n}\cdots ,円\mathrm {d} \sigma _{2},円\mathrm {d} \sigma _{1},}$${\displaystyle f^{(-n)}(x)=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n-1}}f(\sigma _{n}),円\mathrm {d} \sigma _{n}\cdots ,円\mathrm {d} \sigma _{2},円\mathrm {d} \sigma _{1},} is given by single integration $${\displaystyle f^{(-n)}(x)={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t),円\mathrm {d} t.}$${\displaystyle f^{(-n)}(x)={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t),円\mathrm {d} t.}

A proof is given by induction. The base case with n = 1 is trivial, since it is equivalent to $${\displaystyle f^{(-1)}(x)={\frac {1}{0!}}\int _{a}^{x}{(x-t)^{0}}f(t),円\mathrm {d} t=\int _{a}^{x}f(t),円\mathrm {d} t.}$${\displaystyle f^{(-1)}(x)={\frac {1}{0!}}\int _{a}^{x}{(x-t)^{0}}f(t),円\mathrm {d} t=\int _{a}^{x}f(t),円\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 $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t),円\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}(x-t)^{n-1}f(t),円\mathrm {d} t.}$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t),円\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}(x-t)^{n-1}f(t),円\mathrm {d} t.} Then, applying the induction hypothesis, $${\displaystyle {\begin{aligned}f^{-(n+1)}(x)&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2},円\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2}\right],円\mathrm {d} \sigma _{1}.\end{aligned}}}$${\displaystyle {\begin{aligned}f^{-(n+1)}(x)&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2},円\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2}\right],円\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 ${\displaystyle \sigma _{1}}${\displaystyle \sigma _{1}}. Thus, comparing with the case for n = n and replacing ${\displaystyle x,\sigma _{1},\cdots ,\sigma _{n}}${\displaystyle x,\sigma _{1},\cdots ,\sigma _{n}} of the formula at induction step n = n with ${\displaystyle \sigma _{1},\sigma _{2},\cdots ,\sigma _{n+1}}${\displaystyle \sigma _{1},\sigma _{2},\cdots ,\sigma _{n+1}} respectively leads to $${\displaystyle \int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2}={\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t),円\mathrm {d} t.}$${\displaystyle \int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1}),円\mathrm {d} \sigma _{n+1}\cdots ,円\mathrm {d} \sigma _{2}={\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t),円\mathrm {d} t.} Putting this expression inside the square bracket results in $${\displaystyle {\begin{aligned}&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t),円\mathrm {d} t,円\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n}f(t),円\mathrm {d} t\right],円\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t),円\mathrm {d} t.\end{aligned}}}$${\displaystyle {\begin{aligned}&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t),円\mathrm {d} t,円\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),円\mathrm {d} t\right],円\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t),円\mathrm {d} t.\end{aligned}}}

• It has been shown that this statement holds true for the base case ${\displaystyle n=1}${\displaystyle n=1}.
• If the statement is true for ${\displaystyle n=k}${\displaystyle n=k}, then it has been shown that the statement holds true for ${\displaystyle n=k+1}${\displaystyle n=k+1}.
• Thus this statement has been proven true for all positive integers.

This completes the proof.

The Cauchy formula is generalized to non-integer parameters by the Riemann–Liouville integral, where ${\displaystyle n\in \mathbb {Z} _{\geq 0}}${\displaystyle n\in \mathbb {Z} _{\geq 0}} is replaced by ${\displaystyle \alpha \in \mathbb {C} ,\ \Re (\alpha )>0}${\displaystyle \alpha \in \mathbb {C} ,\ \Re (\alpha )>0}, and the factorial is replaced by the gamma function. The two formulas agree when ${\displaystyle \alpha \in \mathbb {Z} _{\geq 0}}${\displaystyle \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.

• 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

• Alan Beardon (2000). "Fractional calculus II". University of Cambridge.

• Maurice Mischler (2023). "About some repeated integrals and associated polynomials".

• Augustin-Louis Cauchy
• Integral calculus
• Theorems in mathematical analysis

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