Cauchy formula for repeated integration

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The Cauchy formula for repeated integration allows one to compress n antidifferentiations of a function into a single integral.

[edit] Scalar case

Let f be a continuous function on the real line. Then the nth antidifferentiation of f,

f^{[n]}(x) = \int_{0}^x\int_0^{\sigma_1}\cdots\int_0^{\sigma_{n-1}}f(\sigma_{n})d\sigma_{n}\cdots d\sigma_2 d\sigma_1,

is given by single integration

f^{[n]}(x) = \frac{1}{(n-1)!}\int_0^x\left(x-y\right)^{n-1}f(y)dy.

A proof is given by induction. Since f is continuous, the base case is given by

\frac{d}{dx}f^{[1]}(x) = \frac{d}{dx}\int_0^xf(y)dy = f(x).

A little work shows that we also have

\frac{d}{dx}f^{[n]}(x) = \frac{d}{dx}\frac{1}{(n-1)!}\int_0^x\left(x-y\right)^{n-1}f(y)dy = f^{[n-1]}(x).

Hence, f[n](x) gives the nth antidifferentiation of f(x).

[edit] References

[edit] See also