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The fractional calculus is partly about nested indefinite integrals. Is there any study or body of knowledge on nested DEFINITE integrals? For example, the fractional calculus helps with this integral: $$\int_0^{l_2}{\left(\int_0^{l_1}{f(l_1)dl_1}\right)dl_2}$$ It can be readily observed that these forms don't allow for any limits of integration other than a new variable to replace the previous variable of integration.

What I am looking for is nested integrals that have functions for the limits of integration: $$\int_{g_2(x_1,x_2,\dots,x_n)}^{f_2(x_1,x_2,\dots,x_n)}{\left(\int_{g_1(x_1,x_2,\dots,x_n)}^{f_1(x_1,x_2,\dots,x_m)}{f(x_1,x_2,\dots,x_m)dx_i}\right)dx_j}$$

Where or how can I find more about this second form of nesting?

REFINEMENT

The first expression above is combined into an operator, say $J$, to the second power in the fractional calculus. I'm looking for an extension of this so that the second expression can be combined into a similar operator, say $J_2$. I'm wondering where this has been done. It seems to be more than just iterated integrals.

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I don't quite follow how this connects back to fractional calculus. Fractional integrals can be defined as a type of integral transform. No "nesting" is necessary. As for "nested integrals" with functions as limits, look up information about iterated integrals in any multivariable calculus text. –  Bill Cook Jul 7 '12 at 1:14
    
@BillCook: I'm seeking to combine iterated integrals into a single integral or operator, if you will. The idea is to create a set of operators that perform various operations using the "power" of integrals to do the dirty work. I guess I'm looking to see what kind of work has been done that combines operator theory and iterated integrals. –  Matt Groff Jul 7 '12 at 1:39
    
You do realize that differintegrals in general necessarily require a lower limit as part of their complete representation? (The exception, of course, being nonnegative integer orders...) –  J. M. Jul 7 '12 at 3:55
    
@J.M.: Is this lower limit used in every step of the integration, or just at the end? I'm hoping to find a way to use limits of integration at every "iteration". –  Matt Groff Jul 7 '12 at 4:12

1 Answer 1

I do not have the answer you are waiting for. I will try to put my ideas about the subject. I work in fractional calculus (FC) over eighteen years and I sent a lot of time in thinking about the relations between the classic concepts and the FC. In all my texts on FC I tried to obtain formulations that recover the classic when the orders become integers. My vision is particular and is not the orthodox vision. In fact I defined a generalised Grunwald_letnikov derivative valid for any order (real or complex) and for functions defined in R (or C) - the generalization for R^n is immediate. When the order in negative we can use the term anti-derivative. When the order is -1 it is the usual primitive. This has nothing with the integral definition. The notion of integral can be done in several ways and we know that the integral between a and b is the difference between the anti-derivative values at a and b. However, there is no equivalent definition of integral for any order. Could we say that is the difference between that anti-derivative (anyn order) values at a and b? No, because the "integral" from a to b would be dfferent from the sum of the integrals frm a to c plus the integral from c to b. This means that, before going into fractional integral computation we should try to define it carefully. You may ask. And what about Riemann-Liouville and Caputo integrals aren't they fractional integrals? Not in the above sense. They are essentially system interpretations (based on convolution) of the fractional derivative. The imposition of two integral limits are nothing else than requiring that the integrand function is null outside the given interval. I did not think on the subject enough time, I want suggest to transport the integral limits into the integrand using the Heaviside function. For example, the integral between a and b of f(t) is the integral between -infinite and +infinte of [u(t-b) - u(t-a)]f(t) dt, where u(t) is the Heaviside function. I always uese this trick.

M. Ortigueira mdo@fct.unl.pt

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