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Integrating over piecewise functions

  1. Feb 15, 2013 #1

    joshmccraney

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    given a function [itex]f(x)[/itex] that is piecewise smooth on interval [itex]-L<x<L[/itex] except at [itex]N-1[/itex] points, is [itex]\int_{-L}^L f'(x)dx [/itex] legal or would i have to [tex]\sum_{i=1}^N \int_{x_i}^{x_{i+1}} f'(x)dx‎‎[/tex]
    where [itex]x_{N+1}=L[/itex] and [itex]x_{1}=-L[/itex]

    also, am i correct that if [itex]f(x)[/itex] is piecewise smooth, then [itex]f'(x)[/itex] is piecewise continuous but not necessarily piecewise smooth?

    thanks in advance!
     
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  3. Feb 15, 2013 #2

    jbunniii

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    The latter. Suppose for example that ##f(x) = |x|^{1/2}##. This is smooth (infinitely differentiable) everywhere except at ##x = 0##, and
    $$f'(x) = \begin{cases}
    \frac{1}{2|x|^{1/2}} & \textrm{ if }x > 0 \\
    \frac{-1}{2|x|^{1/2}} & \textrm{ if }x < 0 \\
    \end{cases}$$
    As ##f'## is unbounded on ##[-L,L]##, it's necessary to use two (improper) integrals to integrate it:
    $$\lim_{a \rightarrow 0^-} \int_{-L}^{a} f'(x) dx + \lim_{b \rightarrow 0^+}\int_{b}^{L} f'(x) dx$$
    Both limits exist and the answers have opposite signs, so the result is 0.
    Assuming "smooth" means infinitely differentiable, ##f'## will be piecewise smooth. If by "smooth" you merely mean (once) differentiable, then ##f'## is not necessarily even piecewise continuous.
     
  4. Feb 17, 2013 #3

    joshmccraney

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    thanks for the reply. think i have it now.
     
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