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Homework Help: Changing integration limits from [0,a] to [0, inf)

  1. Oct 12, 2012 #1
    The problem statement, all variables and given/known data
    This isn't really homework, but a question I came upon when doing my homework.

    How can I go from an integral with limits 0 and a:

    [itex]
    \int_0^a f(x) dx
    [/itex]

    to something with limits 0 and [itex]\infty[/itex] (still giving the same answer)

    [itex]
    c\int_0^\infty f(u) du
    [/itex]

    , where [itex]c[/itex] is some kind of constant.

    The attempt at a solution
    I've tried substituting [itex]x[/itex] with [itex]tan(\frac{\pi x}{2a})[/itex], since this seems to give the correct limits, but I don't know if it makes any sense, and what to do next (how to find c for example).
     
  2. jcsd
  3. Oct 12, 2012 #2

    Ray Vickson

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    You can't, in general. The integral of f(x) for 0 ≤ x ≤ a depends only on the value of f(x) for x between 0 and a, while the integral of f(u) for 0 ≤ u < ∞ depends on the values of f(u) for all positive values of u, including values u > a. The behaviour of f on [0,a] may be essentially unrealated to the behaviour on [a,∞).

    Now, of course, if both integrals are finite, you can get c by dividing one by the other, but that is trivial and probably not what you had in mind. You cannot do that at all if the integral over [0,a] is finite while that over [0,∞) is infinite.

    RGV
     
  4. Oct 12, 2012 #3
    I think it is not possible to change limits from some value to infinity for an unkown function.... I can be wrong but I never saw any such thing before
     
  5. Oct 12, 2012 #4

    SammyS

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    If you make the substitution, [itex]\displaystyle u=\tan\left(\frac{\pi x}{2a}\right)\,,[/itex] then [itex]\displaystyle x=\frac{2a}{\pi }\tan^{-1}(u)\,,[/itex] so that [itex]\displaystyle dx=\frac{2a}{\pi }\frac{1}{1+u^2}du\ .[/itex]

    This gives: [itex]\displaystyle \int_0^a f(x)\,dx
    =\frac{2a}{\pi }\int_0^{\infty} f\left(\frac{2a}{\pi }\tan^{-1}(u)\right)\frac{1}{1+u^2}du\,,[/itex] which is probably not what you wanted.
     
  6. Oct 12, 2012 #5
    Thanks a lot guys. This is what happens when I sit and wonder about mathematics and physics into the night!
     
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