# Ngô Quốc Anh

## April 25, 2013

### The Cauchy formula for repeated integration

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 23:41

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress $n$ antidifferentiations of a function into a single integral.

Let $f$ be a continuous function on the real line. Then the $n$-th repeated integral of $f$ 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 $f$ 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 $n+1$. Apply the induction hypothesis and switching the order of integration,

$\begin{array}{lcl} {f^{ - (n + 1)}}(x) &=& \displaystyle\int_a^x {\int_a^{{\sigma _1}} \cdots } \int_a^{{\sigma _n}} f ({\sigma _{n + 1}}){\mkern 1mu} {\text{d}}{\sigma _{n + 1}} \cdots {\mkern 1mu} {\text{d}}{\sigma _2}{\mkern 1mu} {\text{d}}{\sigma _1} \hfill \\ &=& \displaystyle\frac{1}{{(n - 1)!}}\int_a^x {\int_a^{{\sigma _1}} {{{\left( {{\sigma _1} - t} \right)}^{n - 1}}} } f(t){\mkern 1mu} {\text{d}}t{\mkern 1mu} {\text{d}}{\sigma _1} \hfill \\ &=& \displaystyle\frac{1}{{(n - 1)!}}\int_a^x {\int_t^x {{{\left( {{\sigma _1} - t} \right)}^{n - 1}}} } f(t){\mkern 1mu} {\text{d}}{\sigma _1}{\mkern 1mu} {\text{d}}t \hfill \\ &=& \displaystyle\frac{1}{{n!}}\int_a^x {{{\left( {x - t} \right)}^n}} f(t){\mkern 1mu} {\text{d}}t. \hfill \\ \end{array}$

The proof follows.

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 $\Gamma (n)$, that is the Riemann–Liouville integral which is defined by

$\displaystyle I^\alpha f(x) = \frac{1}{\Gamma(\alpha)}\int_a^xf(t)(x-t)^{\alpha-1}\,dt .$

Another notation, which emphasizes the basepoint, is

$\displaystyle {}_aD_x^{-\alpha}f(x) = \frac{1}{\Gamma(\alpha)}\int_a^x f(t)(x-t)^{\alpha-1}\,dt.$

This also makes sense if $a =-\infty$, with suitable restrictions on $f$. The fundamental relations hold

$\displaystyle \frac{d}{dx}I^{\alpha+1} f(x) = I^\alpha f(x),\quad I^\alpha(I^\beta f) = I^{\alpha+\beta}f,$

the latter of which is a semigroup property. These properties make possible not only the definition of fractional integration, but also of fractional differentiation, by taking enough derivatives of $I^\alpha f$.

One can define fractional-order derivatives of $f$ as well by

$\displaystyle \frac{d^\alpha}{dx^\alpha} f=\frac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}} I^{\lceil\alpha\rceil-\alpha}f$

where $\lceil\cdot\rceil$ denotes the ceiling function. One also obtains a differintegral interpolating between differentiation and integration by defining

$\displaystyle D^\alpha_x f(x) = \begin{cases} \dfrac{d^{\lceil\alpha\rceil}}{dx^{\lceil\alpha\rceil}} I^{\lceil\alpha\rceil-\alpha}f(x)& \alpha>0\\ f(x) & \alpha=0\\ I^{-\alpha}f(x) & \alpha<0. \end{cases}$

Let take a look at the following picture.

The above picture illustrate the half derivative (purple curve) of the function $f(x)=x$ (blue curve) together with the first derivative (red curve).

An alternative fractional derivative was introduced by Caputo in 1967, and produces a derivative that has different properties: it produces zero from constant functions and, more importantly, the initial value terms of the Laplace Transform are expressed by means of the values of that function and of its derivative of integer order rather than the derivatives of fractional order as in the Riemann-Liouville derivative. The Caputo fractional derivative with base point $x$, is then

$\displaystyle D_x^{\alpha}f(y)=\frac{1}{\Gamma(1-\alpha)}\int_x^y f'(y-u)(u-x)^{-\alpha}du.$

Another representation is

$\displaystyle {}_a\tilde{D}^\alpha_x f(x)=I^{\lceil \alpha\rceil-\alpha}\left(\frac{d^{\lceil \alpha\rceil}f}{dx^{\lceil \alpha\rceil}}\right).$

Source: Wikipedia.