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Changing derivative inside an integral

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello.

    My teacher did something on the blackboard today that I really didn't figure out how was done.

    He took this equation:
    [tex]{{\mathcal{L}}^{\,\left( \alpha \right)}}=\frac{2{{e}^{2}}\tau }{m}\int_{-\infty }^{\infty }{g\left( \varepsilon \right)\varepsilon {{\left( \varepsilon -\mu \right)}^{\alpha }}\left( -\frac{\partial f}{\partial \varepsilon } \right)d\varepsilon }[/tex]
    And then made a change into:
    [tex]{{\mathcal{L}}^{\,\left( \alpha \right)}}=\frac{2{{e}^{2}}\tau }{m}\int_{-\infty }^{\infty }{\overbrace{\frac{\partial }{\partial \varepsilon }\left( g\left( \varepsilon \right)\varepsilon {{\left( \varepsilon -\mu \right)}^{\alpha }} \right)}^{H\left( \varepsilon \right)}f\left( \varepsilon \right)d\varepsilon }[/tex]
    So he was able to use it in the Sommerfeld expansion.

    But I don't know how he was able to change from the derivative of f (The fermi function) to that not being derived but the other expression.

    So I was kinda hoping someone could help me figure this out :)

    Thanks in advance.


    Regards
     
  2. jcsd
  3. Apr 30, 2013 #2

    Office_Shredder

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    He just did integration by parts, and the boundary terms were zero
     
  4. Apr 30, 2013 #3
    That seems fair :)

    Thank you very much.
     
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