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Math Help - square of a function is not integrable

  1. #1
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    square of a function is not integrable

    Is there a function f: R->R is integrable but f^2 is not? why?
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    Re: square of a function is not integrable

    Quote Originally Posted by parklover View Post
    Is there a function f: R->R is integrable but f^2 is not? why?
    Theorem: If f is an integrable function in [a,b] then f^2 is an integrable function in [a,b].

    P.S. In response to the to reply #3.
    It depends on how one defines integral.
    Most often Riemann integrals require bounded functions.
    That example is not a bounced function.
    The term is improper integral.
    Last edited by Plato; August 16th 2011 at 01:47 PM.
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    Re: square of a function is not integrable

    Consider f(x) =\begin{cases} \frac 1{\sqrt{|x|}}&\mbox{if } -1\leq x<0 \mbox{ or }0<x\leq 1\\0&\mbox{ if } |x|>1 \end{cases} and choose any value you want for x=0.
    Last edited by girdav; August 16th 2011 at 01:57 PM.
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    Re: square of a function is not integrable

    Quote Originally Posted by girdav View Post
    Consider f(x) =\begin{cases} \frac 1{\sqrt |x|}&\mbox{if } -1\leq x<0 \mbox{ or }0<x\leq 1\\0&\mbox{ if } |x|>1 \end{cases} and choose any value you want for x=0.
    sorry, I am slow. why is it not integrable?i think ln (x)
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    Super Member girdav's Avatar
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    Re: square of a function is not integrable

    The function f is integrable but not its square. As you noticed, a primitive of \frac 1x is \ln x, but if you compute \int_{\varepsilon}^1\frac 1xdx, you will find -\ln \varepsilon, and it has an infinite limit as \varepsilon\to 0.
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    Re: square of a function is not integrable

    Quote Originally Posted by parklover View Post
    sorry, I am slow. why is it not integrable?i think ln (x)
    Read my P.S.
    It depends on what we mean by integrable.
    The function \frac{1}{\sqrt{|x|}} is not Riemann integrable on [-1,0].
    However as an improper integral \int_{ - 1}^0 {\frac{{dx}}{{\sqrt {\left| x \right|} }}}=2 .

    But \int_{ - 1}^0 {\frac{{dx}}{{ {\left| x \right|} }}} does not exist even as an improper integral.
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