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,
,
is given by single integration
.
A proof is given by induction. Since ƒ is continuous, the base case follows from the Fundamental theorem of calculus:
;
where
.
Now, suppose this is true for n, and let us prove it for n+1. Apply the induction hypothesis and switching the order of integration,
![{\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}\\&={\frac {1}{(n-1)!}}\int _{a}^{x}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n-1}f(t)\,\mathrm {d} t\,\mathrm {d} \sigma _{1}\\&={\frac {1}{(n-1)!}}\int _{a}^{x}\int _{t}^{x}\left(\sigma _{1}-t\right)^{n-1}f(t)\,\mathrm {d} \sigma _{1}\,\mathrm {d} t\\&={\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t)\,\mathrm {d} t\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a2ca5157e52a5f9e8d69df4cd803c4ab3076e3b)
The proof follows.
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).
References
- Gerald B. Folland, Advanced Calculus, p. 193, Prentice Hall (2002). ISBN 0-13-065265-2
External links