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Differentiating an integral

  1. Oct 21, 2013 #1
    something I often see without justification in my physics books. What is the justification for the following convolution property (pulling the derivative inside the integral)
    [tex] (f*g)^\prime = \frac{d}{dx} \int f(x) g(x-u) dx = \int f(x) \frac{d}{dx} g(x-u) dx = f(x) * g(x)^\prime [/tex]
     
  2. jcsd
  3. Oct 21, 2013 #2

    jbunniii

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    I think you mean
    $$(f*g)^\prime(x) = \frac{d}{dx} \int f(u) g(x-u) du = \int f(u) \frac{d}{dx} g(x-u) du = (f * g^\prime)(x)$$
    This is a special case of a more general situation. Suppose
    $$F(x) = \int_{a}^{b} f(x,y) dy$$
    where ##a## and ##b## are constants not dependent on ##x##. Then if ##f## is sufficiently nice, we have
    $$F'(x) = \int_{a}^{b} \frac{\partial}{\partial x} f(x,y) dy$$
    Sufficiently nice means, for example, that ##f## is continuous and that ##\partial f /\partial x## exists and is continuous. No point writing out a proof here; see this Wikipedia entry:

    http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Proof_of_Theorem
     
  4. Oct 21, 2013 #3
    Thanks for the correction. I guess I understand the justification.
     
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