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Is there a way to integrate over the intersection of a finite collection of convex regions, using only the definition of the regions (i.e. without actually calculating the intersections)?

The concrete problem I have, is to integrate over the intersection of two ellipses and a parallelogram, all centered at the origin (the ellipses and the parallelogram also depend on additional parameters).
While it would be possible to calculate the intersection, it lacks elegance and mathematical beauty and I wonder whether there is something analogous to lagrange multipliers.

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In view of the answers and comments, I see the need of stressing, that calculating the intersection of regions refers to determining the intersection of the respective point sets of the given convex regions and not to determining the measure (i.e. area or volume) of the intersection.
The motivation is to avoid tedious calculations of the intersection of region boundaries in order to determine the region-intersection via a description of its boundary.

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    $\begingroup$ I don't know if this is what you are looking for, but characteristic functions make a neat expression: $\int_{A_1\cap\cdots\cap A_n}f=\int\chi_{A_1}\cdots\chi_{A_n}f$. $\endgroup$
    – Joonas Ilmavirta
    Commented Aug 20, 2014 at 7:24
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    $\begingroup$ Are you talking about numerical integration or symbolic integration? $\endgroup$
    – Robert Israel
    Commented Aug 20, 2014 at 16:30
  • $\begingroup$ @RobertIsrael I'm lookiing for symbolic integration, because I need the exact value of the integral. $\endgroup$
    – Manfred Weis
    Commented Aug 21, 2014 at 3:32
  • $\begingroup$ @JoonasIlmavirta I think, I know now, how to turn your suggestion into a solution of my posed problem. $\endgroup$
    – Manfred Weis
    Commented Nov 5, 2016 at 10:07
  • $\begingroup$ @ManfredWeis, I'm glad if it helped. (+1) $\endgroup$
    – Joonas Ilmavirta
    Commented Nov 5, 2016 at 13:21

2 Answers 2

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rethinking the suggestion of Joonas Ilmavirta, to use characteristic functions of the convex sets, the following recipe seems to yield a method to calculate the integral of a function over the outcome of some set theoretic combination of a (finite) collection of compact regions of the respective euclideans space:

  • chose an orthonormal basis of at least all functions defined over an $n$-dimensional interval (which could e.g. mean a restriction to periodic functions in case of representing functions via their fourier series), which contains the union of the regions.

  • represent the characteristic function of each region in that basis.

  • model intersections of regions as products of the respective characteristic functions ,i.e. $\chi_{A\cap B} = \chi_A\ .\chi_B$ and unions according to the laws for combining probabilities, i.e. $\chi_{A\cup B} = \chi_A+\chi_B-\chi_A\ .\chi_B$, which in the special case of two regions means subtracting the product of the respective characteristic functions from their sum.

  • express the set theoretic combination of the regions via multiplications, additions and subtractions of the characteristic functions in the same manner as is done for calculating combined probabilities and multiply the outcome with the function to be integrated.

That method would also yield the area of arbitrary set theoretic combinations of arbitrary finite collections of compact regions of euclidean n-space.

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  • $\begingroup$ This will result in you calculating the area directly I think. That is what is required in order to express the final characteristic function in the final integral. Try it with the three disks maybe? $\endgroup$
    – apg
    Commented Nov 6, 2016 at 15:06
  • $\begingroup$ In case the function to be integrated is $(x,y)\mapsto 1$, this will certainly yield the area of their union or whatever set theoretic operation you want to perform on them. $\endgroup$
    – Manfred Weis
    Commented Nov 6, 2016 at 15:58
  • $\begingroup$ But the reason the area is difficult to calculate is because the necessary integral can't be done. $\endgroup$
    – apg
    Commented Nov 6, 2016 at 16:12
  • $\begingroup$ Why shouldn't it be possible to calculate the area via an integral? At least it can be approximated to arbitrary precision and in some cases it may also be possible to obtatain the exact value via symbolic integration. $\endgroup$
    – Manfred Weis
    Commented Nov 6, 2016 at 18:23
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One could work backwards from the answer to determine the area of intersection.

Any method would imply you had immediate knowledge of the area.

So it cannot be sidestepped.

The way to proceed is to approximate the area as well as possible. Even just the area of intersection of three random circles is practically impossible, see this question.

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  • $\begingroup$ Thank you for your answer, but I don't see, why it shouldn't be possible to determine the area of the intersection; do you have any proof for that, maybe a theorem I am not aware of? $\endgroup$
    – Manfred Weis
    Commented Nov 4, 2016 at 6:37
  • $\begingroup$ @ManfredWeis Integrate constant function 1. $\endgroup$
    – Boris Bukh
    Commented Nov 4, 2016 at 13:00
  • $\begingroup$ Imagine you integrate 1 over the area of intersection using some new technique, and so avoid calculating the area directly. Then you have calculated the area. So, already, even with the simplest integral, you are deducing the area in a way that you are trying to avoid. $\endgroup$
    – apg
    Commented Nov 4, 2016 at 13:45
  • $\begingroup$ @AlexanderGiles I am not trying to avoid calculating the area (as a measure) of the intersection of convex regions; what I try to avoid is to calculate the intersection of the boundary curves (try that for two or more ellipses in general position and you'll know, what I mean). $\endgroup$
    – Manfred Weis
    Commented Nov 4, 2016 at 18:14
  • $\begingroup$ @AlexanderGiles in your SE question you ask for the area of the union of disks and state, that is impossible to calculate without giving a proof for the impossibility. So the current state, as I see it, is that you don't know how to calculate the area of a union of disks. Using one's opinion on the answer to an open conjecture in subsequent proofs is moving on thin ice; there is a 50% chance of being right in such a yes/no decision, but also a 50% chance of being wrong. $\endgroup$
    – Manfred Weis
    Commented Nov 5, 2016 at 8:33

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