# 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,

${\displaystyle f^{(-n)}(x)=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n-1}}f(\sigma _{n})\,\mathrm {d} \sigma _{n}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1}}$,

is given by single integration

${\displaystyle f^{(-n)}(x)={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t)\,\mathrm {d} t}$.

A proof is given by induction. Since ƒ is continuous, the base case follows from the Fundamental theorem of calculus:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}f^{(-1)}(x)={\frac {\mathrm {d} }{\mathrm {d} x}}\int _{a}^{x}f(t)\,\mathrm {d} t=f(x)}$;

where

${\displaystyle f^{(-1)}(a)=\int _{a}^{a}f(t)\,\mathrm {d} t=0}$.

Now, suppose this is true for n, and let us prove it for n+1. Firstly, using the Leibniz integral rule, note that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t)\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t)\,\mathrm {d} t}$.

Then, applying the induction hypothesis,

{\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}{\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}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n}\mathrm {d} \sigma _{1}\right]f(t)\,\mathrm {d} t\\&={\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