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Obvious integral manipulation

  1. Mar 27, 2010 #1
    This wasn't obvious to me.

    From my book. We have,
    [tex]p(x)=\begin{cases}
    \frac{1}{2} & -2\leq x\leq-1\,\textrm{or}\;1\leq x\leq2,\\
    0 & \text{otherwise~}.\end{cases}[/tex]

    So, the kth moment is given by
    [tex]M_{k}=\frac{1}{2}\int_{-2}^{-1}x^{k}dx+\frac{1}{2}\int_{1}^{2}x^{k}dx[/tex]

    So, obviously,
    [tex]M_{k}=(1+(-1)^{k})\frac{1}{2}\int_{1}^{2}x^{k}dx[/tex]

    Is this a correct explanation of this step:
    Since p(x) is an even function, it follows that the left integral is exactly equal to the right integral when k is even, and exactly equal to -1 times the right integral when k is odd. Can it be explained better/ more rigorously ? Will this explanation always apply to any even real-valued function of x ?
     
  2. jcsd
  3. Mar 28, 2010 #2

    mathman

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    Science Advisor
    Gold Member

    Your analysis is correct. It will apply to any even function, implicitly assuming the integral exists. The function doesn't have be real-valued, just even. Also the integral being zero applies to any odd function, with the same implicit assumption.
     
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