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Is this valid for improper integrals.

  1. May 7, 2009 #1
    note: INT = integral and the above and bottom arguments are the bounds for the integral.


    infinity
    INT [ f(x) ]
    -infinity

    =

    lim a - >-infinity and lim b-> infinity of :
    b
    INT [f(x)] = pi
    a

    I am just wondering if the move having two limits in one integral is valid,
    as done above. Or would I have to separate it
     
    Last edited: May 7, 2009
  2. jcsd
  3. May 7, 2009 #2

    Cyosis

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    Homework Helper

    I suppose it is, but usually we write it as:

    [tex]\lim_{a \to \infty} \int_{-a}^a \arctan x dx[/tex]
     
  4. May 7, 2009 #3
    Oh, even better and more efficient, Thanks. To me its useless to split it into two integrals because there is not need.
     
  5. May 7, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Not, that is NOT how we usually write it! That is the "principal value" of the integral and can be very misleading.

    For example, if we write
    [tex]\int_{-\infty}^\infty \frac{1}{x} dx[/tex]
    as
    [tex]\lim_{a\rightarrow \infty} \int_{-a}^a \frac{1}{x}dx[/tex]
    it becomes
    [tex]\lim_{a\rightarrow \infty} ln(|a|)- ln(|-a|)= 0[/tex]

    The standard way to write it is, in fact,
    [tex]\lim_{a\rightarrow -\infty}\lim_{b\rightarrow \infty}\int_a^b f(x) dx[/tex]
    which, for f(x)= 1/x, does not exist.
     
  6. May 7, 2009 #5

    Cyosis

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    Homework Helper

    I'm confused myself now, while I see it going wrong for the limit to infinity this raises a question for the interval (-1,1).

    [tex]
    \lim_{a \to 1} \int_{-a}^a \frac{1}{x}dx=\lim_{a \to 1} \ln|a|-\ln|-a|=0
    [/tex]

    Now do this again with double limits.
    [tex]
    \lim_{a \to -1} \lim_{b \to 1} \int_a^b \frac{1}{x} dx=\lim_{a \to -1} \lim_{b \to 1} (\ln|b|-\ln|a|)=\ln|1|-\ln|-1|=0
    [/tex]

    But this integral should diverge should it not? So given that the function has a singularity somewhere within its integration interval should we not also split the integral up in, in this case -1,0 and 0,1?

    For example:

    [tex]\lim_{a \to -1} \lim_{\epsilon \uparrow 0} \int_a^\epsilon \frac{1}{x}dx+\lim_{b \to 1} \lim_{\epsilon \downarrow 0} \int_\epsilon^b \frac{1}{x} dx[/tex]
     
    Last edited: May 7, 2009
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