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Differentiating under the integral sign

  1. Aug 25, 2006 #1
    Can someone point me to a reliable source (textbook, website) that teaches this technique? It seems like no course at my university covers this technique even though it is quite useful for solving particular integrals!
  2. jcsd
  3. Aug 25, 2006 #2
    Not quite sure what you mean, but it sounds like the fundamental theorem of calculus (and related Gauss's, Green's, Divergence, Stoke's Theorems, etc) Google and wikipedia.org should give plenty of good info.
  4. Aug 25, 2006 #3
    No it isn't really the classical theorems of vector calculus, it has something to do more with a method of solving integrals. For example integration by parts is a method and so is trig substitutions and so on.
  5. Aug 25, 2006 #4
  6. Aug 25, 2006 #5


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    It's called Leibnitz Rule:

    [tex]\frac{d}{dt}\int_{A(t)}^{B(t)} G(x,t)dx=G(B(t),t)\frac{dB}{dt}-G(A(t),t)\frac{dA}{dt}+
    \int_{A(t)}^{B(t)}\frac{\partial G}{\partial t}dx[/tex]

    Now look at that, see what's going on and then do this one.


    Note it's a littler easier than the general expression above since the limits are constants. What does that do to the final answer. Calculate it two ways:

    1. Do the integration first then just take the derivative of the answer with respect to t.

    2. Use Leibnitz Rule directly.

    Compare the answers.

    Now try one with variable limits of integration:

  7. Aug 26, 2006 #6
    Feynman mentions it in some of his literature, is this where you heard it? He read about the idea in Frederick Wood's Advanced Calculus, a now hard-to-obtain book, but a reasonably-sized library may have a copy. It does have a very good explanation of differentiation under the integral sign. If you can't find it though, pm me, and I can send you a copy of that part of the text. (I took pretty extensive notes at that part.)
  8. Aug 26, 2006 #7
    Everyone keeps asking me if Feynman is my motivation to learn this technique, unfortunately i didn't even know he used the technique til i googled it. My main reason is to find neat ways to solve those damn integrals on the putnam competition...
  9. Aug 31, 2006 #8


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    ?? I just checked 6 different Calculus books on my bookshelves. Every one had Leibniz' formula.
  10. Aug 31, 2006 #9
    If you're looking for a recepie, it is dealt with in Boas (2nd Ed.), at the end of the chapter on partial differntiation.
  11. Sep 11, 2006 #10
    Or in Apostol's Mathematical Analysis, theorem 7.40 (and again, but deeper, in the chapter about lebesgue integration).

    By the way, Neutrino, thanks for that video you post some months ago, about a metal version of the pachelbel's canon. It is really awesome !!!
  12. Sep 12, 2006 #11
    To take the derivative of an integral, you'd do like this:

    [tex]\frac{d} {dx} \int_{5x}^{x^{2}} f(x) dx = 2x * f(x^{2}) - 5 * f(5x)[/tex]

    If I recall correctly. the f(x) with the boundaries plugged in and times the derivatives of boundaries.

    What I usually do is pick a simple example like

    [tex]\frac{d} {dx} \int_{0}^{2x} x dx [/tex] and do it two ways: Integrate and then differentiate it back, and then try to apply the Polish Guy's rule...and see what I'm missing and add it in.

    EDIT: If it's of any use I learned this in the section dealing with the Fundamental Theorem of Calculus I think. It was chapter 5 I think of Calculus AB.
    Last edited: Sep 12, 2006
  13. Oct 6, 2006 #12
    hi i wanna know how to apply leibnitz thm for integration in partial differntaiation????????
  14. Oct 6, 2006 #13
  15. Sep 14, 2008 #14
    Pardon me for reviving this thread, but I've just read Feynman's book myself, and wanted to try this kind of differential work.

    Saltydog: I was very happy to see that you provided an example for us to try. However, I'm finding that solving your second example was easier the regular way (first integrating, then differentiating), rather than using Leibniz' rule. This is because of having to solve the integral of Sin(x*t)*x, in the third "part" of the rule. (Which becomes sin(t*x)/(t^2) - xcos(t*x)/t. )

    Did I miss out on something? Because I thought the whole idea was to make it simpler. Though it is a very way of doing it, apart from the last part.
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