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Thread: double integral

  1. #1
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    double integral

    I'm trying to show that

    $\displaystyle \int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$

    so that I can use Fubini's theorem to deduce that the iterated integrals of the integrand(without the absolute value function) are not the same.

    But I got

    $\displaystyle \int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy + \frac{\pi}{4}_{.}$

    I guess it does not really make any difference in the end because the right hand sides of both of the equalities are $\displaystyle \infty$.

    But which one is right??
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  2. #2
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    Quote Originally Posted by willy0625 View Post
    I'm trying to show that

    $\displaystyle \int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$

    so that I can use Fubini's theorem to deduce that the iterated integrals of the integrand(without the absolute value function) are not the same.

    But I got

    $\displaystyle \int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy + \frac{\pi}{4}_{.}$

    I guess it does not really make any difference in the end because the right hand sides of both of the equalities are $\displaystyle \infty$.

    But which one is right??
    Have you tried converting to polars?
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  3. #3
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    First, because of the absolute value, the obvious thing to do is to write this as two integrals. As long as 0< y< x, the fraction is postive:
    $\displaystyle \int_{x=0}^1\int_{y= 0}^x \frac{x^2- y^2}{(x^2+ y^2)^2}dy dx$
    and as long as 0< x< y, the fraction is negative:
    $\displaystyle -\int_{x=0}^1\int_{y= x}^1 \frac{x^2- y^2}{(x^2+ y^2)^2}dy dx$

    Then do as Prove It suggested, convert to polar coordinates:
    The line x= 1 is given by $\displaystyle r cos(\theta)= 1$ so $\displaystyle r= \frac{1}{cos(\theta)}= sec(\theta)$

    $\displaystyle \int_{\theta= 0}^{\pi/4}\int_{r= 0}^{sec(\theta)} \frac{r^2(cos^2(\theta)- sin^2(\theta))}{r^4} r drd\theta$$\displaystyle =\int_{\theta= 0}^{\pi/4}\int_{r= 0}^{sec(\theta)} \frac{cos(2\theta)}{r} drd\theta$

    $\displaystyle -\int_{\theta= \pi/4}^{\pi/2}\int_{r= 0}^{sec(\theta)} \frac{r^2(cos^2(\theta)- sin^2(\theta))}{r^4} r drd\theta$$\displaystyle =-\int_{\theta= \pi/4}^{\pi/2}\int_{r= 0}^{sec(\theta)} \frac{cos(2\theta)}{r} drd\theta$

    The original integral is the sum of those, of course.
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