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 ƒ be a continuous function on the real line. Then the nth repeated integral of ƒ based at 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

A proof is given by induction. Since ƒ 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 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.

Applications

In fractional calculus, this formula can be used to construct a notion of differintegral, allowing one to differentiate or integrate a fractional number of times. Integrating a fractional number of times with this formula is straightforward; one can use fractional n by interpreting (n-1)! as Γ(n) (see Gamma function). Differentiating a fractional number of times can be accomplished by fractional integration, then differentiating the result.

References

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.

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