Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Advanced calculus

  1. Dec 8, 2006 #1
    I'm looking for a book/paper/online source where the following formula is shown to be true:

    [tex]\frac d{dt} \int_{a(t)}^{b(t)} f(t,\tau)d\tau =\int_{a(t)}^{b(t)} \frac {\partial f(t,\tau)}{\partial t} d\tau\,+\, \frac {da(t)}{dt}f(t,a(t))\,-\, \frac {db(t)}{dt}f(t,b(t))[/tex]

    I know one should post attempts at solving the problem but to be honest I wouldn't know where to start. That's one of the reasons why I asked for a reference that would explain/elucidate the proof rather than the proof itself (as well as such a source probably being very useful to me in general). I will say that I can show the Fundamental Theorem of Calculus to be true where the integrand is a function of only one variable if that helps..

    Thanks for taking the time to read. Any help you offer will be very much appreciated.
  2. jcsd
  3. Dec 8, 2006 #2


    User Avatar
    Homework Helper

    When the bounds are independent of t, you have:

    [tex]\frac{d}{dt} \int_a^b f(t,\tau)dt = \lim_{h \rightarrow 0} \frac{1}{h} \left( \int_a^b f(t+h,\tau)d\tau-\int_a^b f(t,\tau)d\tau \right)[/tex]

    [tex]=\lim_{h \rightarrow 0} \int_a^b \frac{f(t+h,\tau)-f(t,\tau)}{h} d\tau[/tex]

    If you could justify exchanging the limit and integral, you'd get that this is equal to:

    [tex]=\int_a^b \lim_{h \rightarrow 0} \frac{f(t+h,\tau)-f(t,\tau)}{h} d\tau = \int_a^b \frac{\partial f(t,\tau)}{\partial t}d \tau[/tex]

    Then you can get the formula you mentioned by treating the integral as a function F(a,b,t) and using the chain rule and the fundamental theorm of calculus.

    One time you can justify switching the limit and the integral is when the derivative is bounded on the integration range, which would be true, eg, if the function was continuously differentiable and the integral was finite.
    Last edited: Dec 8, 2006
  4. Dec 8, 2006 #3


    User Avatar
    Science Advisor
    Homework Helper

    Perhaps it will help if you can visualize where the terms in the derivative come from in terms of the area under the curve.

    Attached Files:

    Last edited: Dec 8, 2006
  5. Dec 8, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    "Perhaps it will help if you can visualize where the terms in the derivative come from in terms of the area under the curve."

    Indeed it will. It's more useful to learn how to do that than try to remember "the formula". Learning that skill will also tell you that your equation is wrong, just by looking at it - the da/dt term should be minus and the db/dt term should be plus.

    I don't have a book recommendation (I learned this stuff a very long time ago) but try and find a book that explains the principles with pictures, rather than just proving things formally.

    Learning to think these things out for yourself is like learning to ride a bike, once you have "got it" you will never forget how to do it. You can then get by without remembering formulas, or having to look them up in books.
  6. Dec 11, 2006 #5
    thankyou so much OlderDan for your awesome and pedagogical rendition of the problem and AlephZero for that inspirational and, in my humble opinion, profound message :smile:

    I'll be sure to apply the methods you described in the future and hopefuly, as in this case, with much success.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook