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Second derivative of an integral

  1. Dec 4, 2011 #1
    Good day,

    I don't understand the following:

    [tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)[/tex]

    All I know is:

    [tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon-\frac{d^{2}}{dt^{2}}\int_{0}^{t}\epsilon \cdot \phi (\epsilon)d\epsilon[/tex]

    Is it allowed to say:

    [tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\cdot t\int_{0}^{t}\phi (\epsilon)d\epsilon[/tex]?

    And if so, why is this correct? This is only correct when [tex]t\neq f(\epsilon )[/tex], right? But I am not sure whether this is the case...

    Thank you in advance!
     
  2. jcsd
  3. Dec 4, 2011 #2

    Mute

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    Homework Helper

    The result (if correct; I haven't checked) will follow from the Liebniz rule:

    [tex]\frac{d}{dt}\int_{\alpha(t)}^{\beta(t)} du~f(u,t) = \int_{\alpha(t)}^{\beta(t)}du~ \frac{\partial}{\partial t}f(u,t) + f(t,t)\frac{d\beta}{dt} - f(t,t)\frac{d\alpha}{dt}[/tex]

    Then you have to do the derivative again, so you would have to use the Liebniz rule on the integral term again.
     
  4. Dec 5, 2011 #3
    I think it's

    [tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi(t)[/tex]
     
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