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Integrating functions with absolute values

  1. Jul 16, 2014 #1
    To find [itex]E |X|[/itex] of a cauchy random variable, I need to integrate
    [itex]\int_{-\infty}^{\infty}\frac1{\pi}\frac{|x|}{1+x^2}dx [/itex].

    From the definition of absolute value, we have
    [itex]\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx + \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx[/itex] (I think).

    But, the very next step in the textbook, which I'm trying to follow along, is
    [itex]\frac2{\pi}\int_0^{\infty}\frac{x}{1+x^2}dx[/itex].

    Unless I made some mistake, the reasoning here is that
    [itex]\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx = \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx[/itex].

    If this is true, why is this?
     
  2. jcsd
  3. Jul 16, 2014 #2
    What happens if you do the substitution ##-x=t##?
     
  4. Jul 17, 2014 #3
    Same idea behind $$\int_{a}^{b}\omega=-\int_{b}^{a}\omega.$$
     
  5. Jul 17, 2014 #4
    Thanks for your help guys. Understanding accomplished.

    Let [itex]t=-x[/itex], then [itex]-dx=dt[/itex], [itex]\frac{-x}{1+x^2}=\frac{t}{1+t^2}[/itex], and the bottom limit of integration changes from -∞ to ∞. Thus

    [itex]\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx=(-1)\cdot \int_{\infty}^0\frac1{\pi}\frac{t}{1+t^2}dt = \int_0^{\infty}\frac1{\pi}\frac{t}{1+t^2}dt [/itex].

    Since t is a dummy variable, the integrand can be rewritten with x=t.
     
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