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Differentials of variable substitution

  1. Sep 22, 2010 #1
    Hi everyone,

    This question is a bit involved but it pertains to calculating the differential of a variable substitution used in the proof of the convolution theorem (http://en.wikipedia.org/wiki/Convolution_theorem)

    [tex] \int f(t) \int g(s - t) ds dt.[/tex]

    If we use the substitution

    [tex]r = s - t [/tex]

    we get the differential relation as

    [tex]dr = ds[/tex]

    so the above equation becomes

    [tex] = \int f(t) \int g(r) dr dt = \int f(t) dt \int g(r) dr [/tex]

    But why didn't we use the differential relation

    [tex]dr = ds - dt[/tex] ?
  2. jcsd
  3. Sep 22, 2010 #2


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    Staff Emeritus
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    You did, but the "d" of a constant function is =0. Note that you did the variable substitution r=s-t in the integral

    [tex]\int g(s-t)ds[/tex]

    where t is just a constant; it represents the constant function h defined by h(s)=t for all s.
  4. Sep 28, 2010 #3
    I see. Thanks for the explanation, Fredrik.
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