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Homework Help: 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.

    INT [ f(x) ]


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

    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


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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


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    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]
    [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


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    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).

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

    Now do this again with double limits.
    \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

    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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