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Rewriting integrals

  1. Jan 28, 2008 #1
    How does one evaluate [tex]\int (arctan(pi*x) - arctan(x))dx[/tex] from 0 to 2 by rewriting the integrand as an integral?
    Last edited: Jan 28, 2008
  2. jcsd
  3. Jan 28, 2008 #2


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    The point is that you have F(b)- F(a) which can be written as
    [tex]\int_a^b f(x) dx[/itex] where f is the derivative of F. The derivative of arctan(x) is
    [tex]\frac{x^2+ 1}[/itex] so this could be
    [tex]\int_0^2 \int_x^{\pi x} \frac{1}{t^2+ 1} dt dx[/itex]

    Now see what happens if you reverse the order of integration:
    For every x, t goes from x to [itex]\pi x[/itex] so boundaries on the region on which you are integrating are t= x, t= [itex]\pi x[/itex], and x= 2 (x= 0 just gives the intersection of the two straight lines).

    If you reverse the order of integration, you will need to take t going from 0 to [itex]2\pi[/itex]. The limits of integration for x are a bit more complicated. The left end will be at [itex]t= \pi x[/itex] or [itex]x= t/\pi[/itex]. For t between 0 and 2, the right end will be t= x or x= t, but for t between 2 and [itex]2\pi[/itex] the right end is at x= 2. The integral is
    [tex]\int_{t=0}^2 \int_{x= t/\pi}^t \frac{1}{1+ t^2} dx dt+ \int_{t= 2}^{2\pi} \int_{x= t/\pi}^2 \frac{1}{1+t^2} dx dt[/itex]
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