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Interchanging partial derivatives and integrals

  1. May 13, 2010 #1
    In the midst of https://www.physicsforums.com/showthread.php?t=403002", I came upon a rather interesting, though probably elementary, question. Analagous to the fundamental theorem of calculus, is there a formula or theorem concerning the expression [tex] \frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau [/tex]. In other words, does partial differentiation with respect to one variable commute with partial integration with respect to the other variable? If they don't commute, what is the relation between the two operations?

    The fundamental theorem of calculus (well, the first part anyway) came from the non-rigorous intuition that [tex]\int^{x+dx}_{x}f(t)dt=f(x)dx[/tex]. Can a similar intuitive statement be made about [tex]\int^{t+dt}_{t}F(t, \tau) \partial \tau[/tex]? We know it has to be proportional to [tex]dt[/tex], but what is the constant of proportionality?
     
    Last edited by a moderator: Apr 25, 2017
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  3. May 13, 2010 #2

    gabbagabbahey

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    If you don't want to use the Heaviside step function, as I suggested in the other thread, you might prefer this
     
  4. May 13, 2010 #3
    Thanks. That's exactly the kind of thing I wanted. Using the formula in the end of the subsection entitled "General form with variable limits," I get
    [tex]\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau = F(t,t)+\int^{t}_{0}\frac{\partial F}{\partial t} \partial \tau[/tex]
     
  5. May 13, 2010 #4

    gabbagabbahey

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    That's correct; provided [itex]F[/itex] is piecewise smooth everywhere, exponentially bounded, and continuous at [itex]\tau=t[/itex] . You'd get the same thing using the Heaviside step function (really a distribution or generalized function, not an ordinary function) since its derivative is a delta function.
     
  6. May 13, 2010 #5
    Based on this formula, in the other thread I just tried to write an expression for the derivative of a convolution. I hope I'm right.
     
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