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What is to differentiate under the integral

  1. Jul 24, 2014 #1
    Definition/Summary

    The trick is to change a difficult integral into an easier integral by treating a constant as a variable, and integrating with respect to that variable first.

    For example, in normalising Gaussians, the difficult integrand [itex]x^2 \exp ( - \lambda x^2)[/itex] where [itex]\lambda[/itex] is a constant can be changed to the easier [itex]\exp ( - \lambda x^2)[/itex]

    Equations

    [tex]\frac{d}{d\lambda}\int_a^b F(x,\lambda) dx\ =\ \int_a^b \frac{\partial}{\partial\lambda} F(x,\lambda) dx[/tex]

    enables us to change a (partial) differential "under" (or inside) the integral sign into an (ordinary) differential outside the integral sign:

    So if [itex]f(x,\lambda)[/itex] is difficult to integrate with respect to [itex]x[/itex] but [itex]F(x,\lambda)[/itex] is easier, where [tex]f(x,\lambda)\ =\ \frac{\partial}{\partial\lambda}F(x,\lambda),[/tex]then:

    [tex]\int_a^b f(x,\lambda) dx\ =\ \frac{d}{d\lambda}\int_a^b F(x,\lambda) dx[/tex]

    Extended explanation

    Sometimes we have to normalize Gaussians, and in this case we can use differentiation under the integral sign (as mentioned in "Feymnan's Surely You're Joking, Mr. Feynman!"). Take this example:

    [tex]
    \begin{align*}
    \int_{ - \infty }^\infty {x^2 \exp \left( { - \lambda x^2 } \right)} dx = - \int_{ - \infty }^\infty {\frac{\partial }{{\partial \lambda }}\exp \left( { - \lambda x^2 } \right)} dx = - \frac{\partial}{{\partial\lambda }}\int_{ - \infty }^\infty {\exp \left( { - \lambda x^2 } \right)} dx = - \frac{d}{{d\lambda }}\sqrt {\frac{\pi }{\lambda }}
    \end{align*}
    [/tex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
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