Integral | Logic, Astronomy, Science, and Ideas Too

Suppose that $f(s)$ is a real or complex valued function defined for almost all $s>0$ , and that $n$ is a positive real number. Let $\Gamma(n)$ denote the gamma function. The fractional integral $I_n\,f$ is the function defined by $I_n\,f(t)={{1}\over{\Gamma(n)}}\,\int_0^t\,{(t-s)^{n-1}\,f(s)\,ds}$ for all $t>0$ for which the integral exists (as a Lebsegue integral).

The fractional integral is a handy generalization of the ordinary operation of integration. The concept goes back to Liebnitz, who discussed the idea of a 1/2-th integral in a letter. The fractional integral is usually called the Riemann-Liouville integral.

If $n$ is a positive integer, the fractional integral is simply repeated integration, as can be proved using integration by parts, or by using an inversion of the order of integration, eg via Fubini’s theorem. For instance, if $n=1$ , then $I_1\,f(t)={\int_0^t{\,f(s)\,ds}}$ since $\Gamma(1)=1$ . If $n=2$ , then $I_2\,f(t)={\int_0^t{(t-s)\,f(s)\,ds}}$ . If $n=3$ , then $I_3\,f(t)={{{1}\over{2}}\,\int_0^t {(t-s)^2\,f(s)\,ds}}$ . And so on.

The fractional integration operator $I_n$ has a neat property: if $m$ and $n$ are positive real numbers, then the operator $I_m$ composed with the operator $I_n$ equals the operator $I_{m+n}$ . That is, the $m$ -th integral of the $n$ -th integral of a function $f$ , equals the $(m+n)$ -th integral of $f$ . The equality is only almost everywhere, if $m+n<1$ , since a fractional integral of order less than 1 of an arbitrary Lebesgue-integrable function is not necessarily defined for all argument values.

It is well known that, if an ordinary (order 1) integral of a function is equal to a constant, then the constant must be zero, and the function being integrated is almost everywhere equal to zero. That fact is used in many contexts, usually in some proof which concludes that a function must be zero because its integral is constant.

What is not widely known is that it is possible for a fractional integral, of order $n$ where $n$ lies between 0 and 1, to equal a nonzero constant. In fact, if the function $f$ is defined by $f(s)=s^{-n}$ , then the fractional integral $I_n\,f(t)$ will be constant for all $t>0$ . Up to multiplication by a fixed number, and alteration of the function $f$ on a set of measure zero, the function $f(s)=s^{-n}$ is the unique function whose $n$ -th fractional integral is equal to a constant. These facts are detailed in a paper which I wrote quite a while ago; see “On fractional integrals equivalent to a constant”, in Canadian Mathematical Bulletin, vol 25, no 3, September 1982, pp 335-338.

Very recently, I discovered a simple insight into why the solution $f(s)$ of the equation $I_n\,f(t)=$ constant must be of the form $f(s)=s^{-n}$ . It uses Euler’s concept of a homogenous function. See David Widder’s book “Advanced Calculus”, 2nd edition, 1961, pp 19-22 for a quick intro to the concept, and a proof of Euler’s theorem on homogenous function, and also a proof of the converse. Widder’s book is marvelously direct and clear on this and other topics; it is a real pleasure to read, and I recommend it for your enjoyment. However, one does not need to consult Widder for the following. One does not even need Euler’s theorem or its converse for solving the “constant fractional integral” problem. Just the concept of a homogenous function is sufficient.

A homogenous function of order $k$ is a function of one or more real or complex variables, let’s say $h(u,v)$ for illustration, such that for all real multipliers $p$ , we have $h(pu,pv)=p^k\,h(u,v)$ . We can think of multiplying by $p$ as a scale change, eg a change of units in a physical formula. Working with two variables, the expressions $u^k$ , $(v-u)^k$ , and $(v-u)^{k-j}\,u^j$ are each homogenous of order $k$ . Linear combinations of such forms are also homogenous of order $k$ . The closure of the space of such forms, equipped with a reasonable metric, gets essentially all the homogenous functions of interest.

Now consider the constant fractional integral problem. Dropping the gamma function multiplier, let $g(t)={\int_0^t\,{h(t,s)\,ds}}={\int_0^t{(t-s)^{n-1}f(s)ds}}$
and suppose that $g(t)$ is constant. A constant is a homogenous function of order 0. Integration of $h(t,s)$ over the interval $(0,t)$ raises the homogenous index of $h(t,s)$ by 1, while reducing the number of independent variables by 1. Hence the function $f(s)$ must be homogenous of order $k$ , where $k$ satisfies $1+(n-1)+k=0$ . That is, $k = -n$ and $f(s)$ is $s^{-n}$ multiplied by some constant.

I was really happy to discover the above simple solution of the constant fractional integral problem. I hope that you find something in the concept of fractional integral to enjoy or use in your own work.

Regards,

Ken Roberts

November 30, 2013

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Constant Fractional Integral