 Cauchy Formula For Repeated Integration
Get Cauchy Formula For Repeated Integration essential facts below. View Videos or join the Cauchy Formula For Repeated Integration discussion. Add Cauchy Formula For Repeated Integration to your PopFlock.com topic list for future reference or share this resource on social media.
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 f be a continuous function on the real line. Then the nth repeated integral of f 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 f 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.

## Generalizations and Applications

The Cauchy formula is generalized to non-integer parameters by the Riemann-Liouville integral, where $n\in \mathbb {Z} _{\geq 0}$ is replaced by $\alpha \in \mathbb {C} ,\ {\mathfrak {R}}(\alpha )>0$ , and the factorial is replaced by the gamma function. The two formulas agree when $\alpha \in \mathbb {Z} _{\geq 0}$ .

Both the Cauchy formula and the Riemann-Liouville integral are generalized to arbitrary dimension 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.