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B Why can't a chain rule exist for integration?

  1. Nov 17, 2016 #1
    I was thinking if the known methods of integration are enough to integrate any given function. In differentiation, we've evaluated the derivatives of all the basic functions by first principles and then we have the chain rule and product rule to differentiate any possible combination (product or composition) of those basic functions.
    But, in integration, if I need to integrate something like sin(x^3)*log(sin(e^(x2)) or something more complicated then all of the methods ,like substitution or integration by parts,will be of no use.
    Isn't a more direct method like something similar to the chain rule required for integration?
     
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  3. Nov 17, 2016 #2

    BvU

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  4. Nov 17, 2016 #3

    fresh_42

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  5. Nov 17, 2016 #4
    I don't think we will ever be able to integrate the function I've written #1 using partial integration. And, there are even more complicated ones.
     
  6. Nov 17, 2016 #5
    Integration by parts wouldn't be of much use in more complicated product functions because we have to integrate another product function after using it. It isn't in terms of the anti-derivatives of the original function.
    I looked up implicit function theorem. It was about describing the graph of a relation, which is not a function, by two or more functions. How does it help in integrating something like e^(sqrt(tan(log(arcsin(sqrt(x))))? Differentiating it is kid's stuff but integrating it is quite stressful.
     
  7. Nov 17, 2016 #6

    fresh_42

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    Well in the end, my math teacher at school has been right, as he said:
    "Differentiation can be done by everyone, integration is an art."

    However, the statement is a bit unfair, since for differentiation we rely on polynomials and a list of functions which we already calculated. But integrating polynomials or looking up lists can be done for integration, too. When sea becomes rough, we have to fall back on the definition by limits. In both cases. And in both cases it can be done this way.
     
  8. Nov 17, 2016 #7

    BvU

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    Since we're in a physics forum: I have seen very few of such pathological integrals (or ones like $$e^{\sqrt { \tan\left (\log\left (\arcsin\left (\sqrt x\right ) \right )\right ) } } $$ pass by in the textbooks I've used in my career.

    Nor in practice: one resorts to numerical integration rather quickly (admittedly sometimes too quickly :smile:).
     
  9. Nov 17, 2016 #8

    fresh_42

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    I have found this link (university of TX) which might be helpful to see the connection between integration and the implicit function theorem:
    https://www.ma.utexas.edu/users/sbutt/m427l-summer07/Imp-Func-Thm-Integration.pdf [Broken]
     
    Last edited by a moderator: May 8, 2017
  10. Nov 17, 2016 #9

    fresh_42

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    I dare to claim that it is equally pathological from a mathematical point of view. :smile:
     
  11. Nov 17, 2016 #10

    jedishrfu

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    but not so much from abnormally pathological psychological view :smile::smile:
     
  12. Nov 17, 2016 #11

    Erland

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  13. Nov 17, 2016 #12

    fresh_42

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    You're right, substitution is the direct counterpart of the chain rule. With the connection to implicit functions I meant more "how to deal with nested functions" as the chain rule also deals with nested functions and I therefore only referred to it as "in a way" to tone it down from a direct correspondence. Perhaps even this remark has been a little bit too optimistic, and I certainly wouldn't insist on it. It just automatically came to my mind when it's about to get a hand on nested functions (and I admit to have abused the question a little bit to emphasizes on the importance of this theorem :sorry:).
     
  14. Nov 17, 2016 #13
    When there is some transformation of a function that tells you its derivative (like xn → nxn-1), then since indefinite integration is almost the inverse of differentiation, that always tells you a corresponding rule for integration: Just go backwards. (So an antiderivative of nxn-1 is xn.) In general, of course, you have to add a constant when integrating, unless it's for a definite integral.

    Therefore if the chain rule says that f(g(x)) → f'(g(x)⋅g'(x) under differentiation, then we know the integral of any expression of the form f'(g(x)⋅g'(x) is f(g(x)).

    (Only slightly related puzzle: Given the function f(x) = 1/x defined for all x ≠ 0, find its indefinite integral.)
     
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