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Derivative of an Integral

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data
    F is an antiderivative of f, so F’=f.
    [tex]\begin{array}{l}
    \int_{g(x)}^{h(x)} {f(t)\,dt} = F\left( {h\left( x \right)} \right) - F\left( {g\left( x \right)} \right) \\
    \frac{d}{{dx}}\int_{g(x)}^{h(x)} {f(t)\,dt} = F'\left( {h\left( x \right)} \right)h'\left( x \right) - F'\left( {g\left( x \right)} \right)g'\left( x \right) \\
    \end{array}[/tex]




    2. Relevant equations
    Can somebody show how to find the derivative of the following integral?

    [tex]\frac{d}{{ds}}\int_0^\infty {f\left( t \right){e^{ - st}}dt} [/tex]



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 8, 2012 #2

    vela

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    Just reverse the order of integration and differentiation. The fundamental theorem doesn't really apply here.
     
  4. Jun 8, 2012 #3
    Use integration by parts. Then d/ds every term. Then write the solution as a sum.
     
    Last edited: Jun 8, 2012
  5. Jun 8, 2012 #4
    I would think the answer is 0, since the definite integral is just a number, and the derivitive of a number if 0.
     
  6. Jun 8, 2012 #5
    This isn't true. The definite integral for multivariable functions can be a function of a different variable, and if you actually tried the problem, you'd see that the definite integral becomes a function of s.
     
  7. Jun 9, 2012 #6

    Dick

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  8. Jun 9, 2012 #7
    Leibniz's Integral rule as shown by Dick does the trick:

    [tex]\frac{d}{{ds}}\int_0^\infty {f\left( t \right){e^{ - st}}dt} = \int_0^\infty {\frac{\delta }{{\delta s}}f\left( t \right){e^{ - st}}dt} = \int_0^\infty {f\left( t \right)\left( { - t{e^{ - st}}} \right)dt} = \int_0^\infty { - tf\left( t \right){e^{ - st}}dt} [/tex]
     
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