1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Changing integration limits from [0,a] to [0, inf)
  1. Integrate from 0 to b (Replies: 4)

  2. Inf(S) = 0 proof (Replies: 14)

Loading...