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Thread: improper integral

  1. #1
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    improper integral

    Hello.
    improper integral-improper_integral.jpg

    I don't know how to solve the above improper integral.
    I tried to get rid of dy, but turn out to be improper integral-improper_integral2.jpg
    which is even more difficult to me.

    I don't know how to get rid of x^2y^2 in the denominator.
    Can any teach me how to solve this question?
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  2. #2
    MHF Contributor
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    Re: improper integral

    If you could change the order of integration, you would have $\displaystyle \int_5^7 \int_0^\infty \frac{1}{1+x^2y^2}\,dy\,dx$. Can you integrate that?

    - Hollywood
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  3. #3
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    Re: improper integral

    Consider the inner integral first that is integral of 1 /(1 + x^2y^2) w r t y, thereafter proceed to integrate w r t x.
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  4. #4
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    Re: improper integral

    Hi Hollywood and ibdutt,
    Sorry, I tried.
    But I still don't know to integrate even I know I should solve the inner dy integral first.
    Can you show me how to do it?
    .
    When I tried to solve it, I let xy=tan(theatre),
    then the question becomes integral limit from zero to infinity [Arctan(7x)-arctan(5x)]/x dx.
    Which when I tried to continue to solve it,
    It becomes the starting question (the question with 1/(1+(xy)^2) )
    Oh............
    It becomes a recurring problem -.-
    Last edited by happymatthematics; Feb 9th 2013 at 10:32 PM.
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  5. #5
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    Re: improper integral

    Once you change the order of integration, you can integrate the inner integral first, treating x as a constant:

    $\displaystyle \int_0^\infty \frac{1}{1+x^2y^2}\,dy$

    substitute $\displaystyle y=\frac{1}{x}\tan{\theta}$, $\displaystyle dy=\frac{1}{x}\sec^2{\theta}\,d\theta$, $\displaystyle \frac{1}{1+x^2y^2}=\cos^2{\theta}$, and converting the limits of integration: $\displaystyle y=0$ to $\displaystyle \infty$ means $\displaystyle \theta=0$ to $\displaystyle \frac{\pi}{2}$.

    $\displaystyle \int_0^\frac{\pi}{2} \cos^2{\theta}\frac{1}{x}\sec^2{\theta}\,d\theta =$

    $\displaystyle \frac{1}{x}\int_0^\frac{\pi}{2} \,d\theta =$

    $\displaystyle \frac{\pi}{2x}$.

    Now we can do the outer integral:

    $\displaystyle \int_5^7 \int_0^\infty \frac{1}{1+x^2y^2}\,dy\,dx =$

    $\displaystyle \int_5^7 \frac{\pi}{2x}\,dx =$

    $\displaystyle \frac{\pi}{2}\int_5^7 \frac{dx}{x} =$

    $\displaystyle \frac{\pi}{2}(\ln{7}-\ln{5})$.

    - Hollywood
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