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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 ) = ∫ a x ∫ a σ 1 ⋯ ∫ a σ n − 1 f ( σ n ) d σ n ⋯ d σ 2 d σ 1 , {\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},} is given by single integration f ( − n ) ( x ) = 1 ( n − 1 ) ! ∫ a x ( x − t ) n − 1 f ( t ) d t . {\displaystyle 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. The base case with n = 1 is trivial, since it is equivalent to f ( − 1 ) ( x ) = 1 0 ! ∫ a x ( x − t ) 0 f ( t ) d t = ∫ a x f ( t ) d t . {\displaystyle 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 d d x [ 1 n ! ∫ a x ( x − t ) n f ( t ) d t ] = 1 ( n − 1 ) ! ∫ a x ( x − t ) n − 1 f ( t ) d t . {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}(x-t)^{n-1}f(t)\,\mathrm {d} t.} Then, applying the induction hypothesis, f − ( n + 1 ) ( x ) = ∫ a x ∫ a σ 1 ⋯ ∫ a σ n f ( σ n + 1 ) d σ n + 1 ⋯ d σ 2 d σ 1 = ∫ a x [ ∫ a σ 1 ⋯ ∫ a σ n f ( σ n + 1 ) d σ n + 1 ⋯ d σ 2 ] d σ 1 . {\displaystyle {\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}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\right]\,\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 σ 1 {\displaystyle \sigma _{1}} . Thus, comparing with the case for n = n and replacing x , σ 1 , ⋯ , σ n {\displaystyle x,\sigma _{1},\cdots ,\sigma _{n}} of the formula at induction step n = n with σ 1 , σ 2 , ⋯ , σ n + 1 {\displaystyle \sigma _{1},\sigma _{2},\cdots ,\sigma _{n+1}} respectively leads to ∫ a σ 1 ⋯ ∫ a σ n f ( σ n + 1 ) d σ n + 1 ⋯ d σ 2 = 1 ( n − 1 ) ! ∫ a σ 1 ( σ 1 − t ) n − 1 f ( t ) d t . {\displaystyle \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 = ∫ a x 1 ( n − 1 ) ! ∫ a σ 1 ( σ 1 − t ) n − 1 f ( t ) d t d σ 1 = ∫ a x d d σ 1 [ 1 n ! ∫ a σ 1 ( σ 1 − t ) n f ( t ) d t ] d σ 1 = 1 n ! ∫ a x ( x − t ) n f ( t ) d t . {\displaystyle {\begin{aligned}&=\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}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n}f(t)\,\mathrm {d} t\right]\,\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t.\end{aligned}}}

• It has been shown that this statement holds true for the base case n = 1 {\displaystyle n=1} .
• If the statement is true for n = k {\displaystyle n=k} , then it has been shown that the statement holds true for n = k + 1 {\displaystyle 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 ∈ Z ≥ 0 {\displaystyle n\in \mathbb {Z} _{\geq 0}} is replaced by α ∈ C ,   ℜ ( α ) > 0 {\displaystyle \alpha \in \mathbb {C} ,\ \Re (\alpha )>0} , and the factorial is replaced by the gamma function. The two formulas agree when α ∈ Z ≥ 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

External links

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

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