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Prove that a function must have only x dependency

  1. Jun 12, 2010 #1
    If I have this equation:

    \int_0^{+\infty}{\frac{e^{-x}}{\int_{\Delta T}f(x,t)}f(x,t)xdx}=Ag(t)


    I am interested to show that f(x,t) must be g(t). This condition must be necessary and sufficient. Is this possible to prove?
  2. jcsd
  3. Jun 12, 2010 #2
    First, what does [tex]\int_{\Delta T}f(x,t)[/tex] mean?

    Second, is the problem to show that the only way for this statement to be true is for f(x,t) to be equal to g(t) (for all x)?
    Last edited: Jun 12, 2010
  4. Jun 13, 2010 #3
    [tex]\int_{\Delta T}f(x,t)=\int_0^{\Delta T}f(x,t)dt[/tex]
    sorry, I skip the "dt"
    yes, for all positive x.
  5. Jun 13, 2010 #4
    What is ΔT?
  6. Jun 13, 2010 #5
    a positive real number, sorry
  7. Jun 13, 2010 #6
    Once you do the outer integration, all instances of x disappear, so you are left with a function of t. Of course, it's completely possible this may equal ±∞ for some or all values of t.
  8. Jun 13, 2010 #7
    if the integrandus if choose to be positive for all x and t, can't you say nothing?

    furthermore,if [tex]f(x,t)=g(t)+h(x)[/tex]
    is now possibile to prove that h(x)=0 for all x??
  9. Jun 13, 2010 #8
    I don't understand your line of questioning. What does f(x,t) being positive have to do with anything? Moreover, you can not say that f(x,t) = g(t) + h(x) without further constraints on f(x,t); who knows how mixed up x and t are. The process of integration gets rid of the x.
  10. Jun 13, 2010 #9
    of course it does.
    sorry, I wasn't as clear as possibile.
    If now I fix the x dependency by means of f(x,t)=g(t)+h(x), it can be proved that h(x)=0 or I'm fancy about it?
  11. Jun 13, 2010 #10


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    [tex]\int_a^b f(t)dt= \int_a^b f(u)du= \int_a^b f(x)dx= \int_a^b f(y)dy= F(b)- F(a)[/tex].

    Whatever variable the integral is "with respect to" is a "dummy" variable and does not appear in the result. "t" is the only "true" variable in that equation.
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