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Math Help - [SOLVED] Improper double integral

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Improper double integral

    D is the positive quadrant.

    \int \int_D \frac{xydxdy}{(1+x^2+y^2)^3}

    Switching to polar coordinates doesn't help much.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Spec View Post
    D is the positive quadrant.

    \int \int_D \frac{xydxdy}{(1+x^2+y^2)^3}

    Switching to polar coordinates doesn't help much.
    Using polar coordinates:

    \begin{aligned}\iint_D \frac{xy}{(1+x^2+y^2)^3}\,\mathrm{d}x\,\mathrm{d}y<br />
&=\int_0^{\frac{\pi}{2}}\int_0^{\infty} \frac{\rho \cos \theta \rho \sin \theta}{(1+\rho^2)^3}\,\rho\,\mathrm{d}\rho\,\math  rm{d}\theta\\<br />
&=\int_0^{\frac{\pi}{2}}\cos \theta \sin\theta \,\mathrm{d}\theta <br />
 \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho<br />
\end{aligned}

    To compute \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho<br />
you can try integration by parts (let u'=\frac{\rho}{(1+\rho^2)^3} and v=\rho ^2).
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by flyingsquirrel View Post

    To compute \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho<br />
you can try integration by parts (let u'=\frac{\rho}{(1+\rho^2)^3} and v=\rho ^2).
    Just let z=1+\rho^2, no need to apply partial integration.
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  4. #4
    Senior Member Spec's Avatar
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    It was easy when you integrated in polar coordinates using z = 1 + r^2

    I was just put off by the looks of that integral, so I didn't really attempt to solve it initially.
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