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B Does integration commute with substitution t=0?

  1. May 16, 2017 #1
    Let ##g(x,t)=\int f(k,x,t)\,dk##

    Under what conditions is the following true?

    ##g(x,0)=\int f(k,x,0)\,dk##

    That is, we can get the value of ##g(x,t)## when ##t=0##, by
    (1) either substituting ##t=0## into ##g(x,t)## or
    (2) by first substituting ##t=0## into ##f(k,x,t)## and then integrating wrt ##k##.

    Does it work for non-zero values of ##t##?

    EDIT: ##g(x,t)=\int f(x,t)\,dx## corrected to ##g(x,t)=\int f(k,x,t)\,dk##.
     
    Last edited: May 16, 2017
  2. jcsd
  3. May 16, 2017 #2

    fresh_42

    Staff: Mentor

    Can you say something about the integration limits? As it stands, the integral is a function of ##t## and only of ##t##. Thus ##g(x,t)=\int f(x,t)dx = F(t)## and ##F(0) = g(x,0)## as ##t## is treated as a constant in the integral.
     
  4. May 16, 2017 #3
    The integration is from ##-\infty## to ##\infty##.
    Screen Shot 2017-05-17 at 3.51.14 AM.png
     
  5. May 16, 2017 #4

    fresh_42

    Staff: Mentor

    So the integration limits don't have a hidden dependency on ##t## and the above remains valid. The equation has to hold for all values of ##t## for which it is defined and thus especially for ##t=0##.
     
  6. May 16, 2017 #5
    Is the equation true because integration is a linear operator, i.e., ##\int t\,f(x)\,dx = t\int f(x)\,dx##? If so, how can we proof it using linearity?
     
  7. May 16, 2017 #6

    fresh_42

    Staff: Mentor

    No, it doesn't have anything to do with linearity and as ##t## is in the exponent, it even isn't linearly depending on ##t##
    It is much simpler: We have a function ##\int f(x,t) dx##. But what if we wrote ##f(x,t_0)=:h(x)##. As long as ##t_0## is constant, this is allowed. We haven't changed anything. The original equation can be thought of as: ##\textrm{For all values } t=t_0 \textrm{ we have } g(x,t_0)=\int f(x,t_0)dx## and therefore also for ##t=t_0=0##. The integration has nothing to do with ##t## so ##t## is constant for the integration. One could even redefine ##h(x)=f(x,t)## and the integral would be the same, only that ##t## is now somewhere hidden in the definition of ##h(x)##, say ##h_t(x)##. Then ##g(x,t)= \int h_t(x)dx## again for all defined values ##t##, e.g. for ##t=0##.

    I'm getting the feeling that the more I explain the more it might confuse you. At least it starts to confuse me. Maybe we should simply write ##t=c## and the entire equation looks a lot better: ##g(x,c)=g_c(x)=\int f(x,c)dx = \int f_c(x)dx##.
     
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