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Is there a theorem that says

  1. Aug 25, 2005 #1
    [tex] \frac {d} {dt} \int_{a}^{b} f(x,t) dx = \int_{a}^{b} \frac {\delta} {\delta t} f(x,t) dx [/tex]

    From trial and error, it seems true but I can't find it in my textbook anywhere. Am I missing something obvious?
  2. jcsd
  3. Aug 25, 2005 #2


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    Yes, this is true provided f and [itex]\frac{\partial f}{\partial t}[/itex] are continuous. This is an example of interchanging limits. Be careful about doing that because in general you can't. This borders on advanced calculus, but my regular calc textbook gives a proof in the appendix. Maybe yours does too.
  4. Aug 25, 2005 #3
    There's a very thorough proof on page 245 of the Bartle if you have access to one of those. Also, note that as a notational thing
    \frac{\delta}{\delta g}
    usually denotes a functional derivative, while what you want is \partial, not \delta in LaTeX, like
    \frac{\partial}{\partial t}
  5. Aug 25, 2005 #4
    Ok, thanks. Just needed to confirm that the statement was true (my physics textbook used it in a proof and never justified the statement, so I wanted to make sure)
  6. Aug 28, 2005 #5
    This is only true though when the bounds (a & b) are held constant, correct? I mean, if they were a=u(x) and b=v(x), the limits couldn't be switched.
  7. Aug 28, 2005 #6


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    If the limits themselves are also functions of t, the theorem (called Leibniz integral rule, IIRC) is:

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

    I know a sufficient condition for this to hold is that f be uniformly continuous in the region of interest.
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