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Finite-part integrals (Hadamard) integrals with Mathematica

  1. Dec 6, 2006 #1
    Finite-part integrals (Hadamard) with MATHEMATICA

    Hello guys,
    I have to calculate an integral defined in the Hadamard sense with MATHEMATICA, but I only found the option for the integrals defined in the CAUCHY sense.
    Just to be clear, suppose to have the integral with limits -1 and +1 of (1/x^2). This integral does not exist in the usual sense, but in the Hadamard sense its value is -2. Now, how do you calculate (numerically) the Hadamard integrals if the functions are complicated using MATHEMATICA?
    Last edited: Dec 6, 2006
  2. jcsd
  3. Dec 14, 2006 #2
    So??? Nobody knows Hadamard finite-part integrals? Is it so difficult for everybody?
  4. Dec 17, 2006 #3
    The concept of "Hadamard fnite integral" seems very curious to me could oyo provide a link where to find some info?.. perhaps you should use zeta regularization when calculating your integrals so:

    [tex] \sum_{n=1}^{\infty}n^{s}=\zeta (-s) [/tex]
  5. Dec 19, 2006 #4
    Hello Karlisbad, I know a very good article: "Numerical evaluation of ypersingular integrals" by Giovanni Monegato, Journal of Computational and Applied Mathematics 50 (1994) 9-31.

    I do not know how the zeta regularization works. Can you please explain? Thank you!
  6. Dec 25, 2006 #5
    A good intro for undergraduates can be found at ..


    Based on an Euler converntion about divergent series..

    resumming let be the divergent series..

    [tex] 1+2^{s}+3^{s}+...................... \rightarrow \zeta (-s) [/tex] (1)

    of course if s<0 then the series above makes sense (no pole at s=1) and is just the Riemann zeta function.. the algorithm of "Zeta regularization" makes some kind of analytic continuation (i'm not mathematician so i can't give a rigorous formulation :rolleyes: by the way really hate math rigour..:mad: ) and then the sum of the series (divergent) can be obtained (¿¿cheating??...if you believe this is cheating you should study renormalization in physics) as the negative values of the Riemann zeta function, for example if s=1 then..

    [tex] 1+2+3+......................=-1/12 [/tex]

    if s is an even number then the "sum" is taken to be 0.

    further info: "H.G Hardy divergent series" (you have it on E-mule but you need to download an .djvu viewer) :tongue2:
  7. Dec 26, 2006 #6

    Gib Z

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    [tex] 1+2+3+......................=-1/12 [/tex]

    That is not true, you need the Ramanujan Summation Operator. It is complete nonsense without that.
  8. Dec 27, 2006 #7
    I had a curious idea..let be a function f(x) with a singularity at x=1 then we define the function:

    [tex] f(x)=f*(x)+\delta(x-1) [/tex] and by definition [tex] f*(1)=0 [/tex]

    [tex] f(x)=f*(x) [/tex] for all x except x=1

    as you can see both function diverge at x=1 however using this definiton:

    [tex] \int_{0}^{2}dxf(x)=\int_{0}^{2}dx(f*(x)+\delta(x-1))=1+\int_{0}^{2}dxf*(x) [/tex] which is now finite...:wink:
  9. Dec 28, 2006 #8


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  10. Dec 28, 2006 #9


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    It's only finite if f and [itex]\delta[/itex] have a singularity of the same order, and whose coefficients are additive inverses.

    It's more typical to use analytically defined functions (like 1/x) to subtract off singularities of functions.
  11. Dec 28, 2006 #10

    matt grime

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    Sigh. Jose might hate the rigour of mathematics, but without it you just get apparent nonsense like this thread. (it should be s<-1, for instance, in post 5, not s<0), not to mention setting out in a confusing manner material which makes it appear that the sum 1+2+3+... actually does exist in the normal sense of summation. Still, never mind, eh? Is * supposed to be convolution? Those aren't functions, by the way. Why is that integral of that 'function' 'now finite'? What if I let f be some truly horrendous function with an essential singularity (infinitely many negative powers in the Laurent expansion)?

    If it helps, we don't object to 'non-rigorous' stuff at all; it just isn't maths in the rigorous formal proof sense. We do object to ill defined terms, confusing notation, indecipherable descriptions, and things that are so wrong it is impossible to know where to start to correct them. There is a difference between being rigorous and doing something clearly. No physicist would accept what you've written above as acceptable presentation either. It's nothing to do with mathematicians.
    Last edited: Dec 28, 2006
  12. Dec 29, 2006 #11
    - The problem with rigour is as perhaps user "eljose" has pointed before in his post :redface: is that you can put 'obstacles' to science, the same happened to infinitesimals [tex] dx [/tex] , the Dirac delta function [tex] \delta (x) [/tex] or "Feynmann (??) Path integrals" in QM or QFT (Field theory) to quantizy everthing..which can't even be defined (rigorously) in terms of Riemann sums¡¡ (sigh), i read a paper by A.Connes about "renormalization and Hopf Algebras" of course the paper were so much rigorous and math-defined that you coud hardly "extract" some info about it regarding renormalization, whereas there're lots and lots of math introduction without rigour in physics, conjectures or hypothesis that can be understood by almos everybody..that's why i critizy rigour..i don't hate mathematician i only say that sometimes they are a bit of "pedantry".
  13. Dec 29, 2006 #12

    matt grime

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    Who's putting obstacles there? Physicisist don't require the same level rigour that mathematicians do, so what? Point out one example where rigour held back the development of Feynmann path integrals, please.

    You aren't trained to read mathematics research papers; they aren't written with people with undergrad physics degrees in mind. They are written by professionals for professionals.

    Unrigorous work often helps, and mathematicians use it all the time. Your problem is not that mathematicians are pedantic about rigor. Your problems with getting your work read have been catalogued at length many times; you have had far more peer review than anyone else I can think of; you have yet to address more than one comment (to include proper references, occasionally).

    Pointing out that saying 'let f be a function with singularity at 1' is useless is not pedantry (that would be pointing out the grammatical errors, which is not helpful). There are infinitely many such functions. Which one? It is important because what kind of singularity you have affects what follows in your post, making it just plain wrong. And no, you can't get away with saying 'oh, choose one that makes it work'. You don't appear to make any effort to avoid mistakes, so we have no idea what is 'wrong but just a typo' and what is 'just plain wrong'.
    Last edited: Dec 29, 2006
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