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I need help calculating these double integrals (in order to show they are not equal): $$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx\,dy$$

George2014L
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  • These are not really double integrals; they are iterated integrals. If you find $\int_a^b f(x,y),dx$, getting a function of $y$, and then integrate that with respect to $y$ and get a number, that's an iterated integral. A double integral, on the other hand, is defined with respect to two-dimensional measure in the $(x,y)$-plane. Funbini's theorem implies that the double integral is equal to both of the iterated integrals unless the double integral $\displaystyle\iint\limits_{[\text{plane region}]} |f(x,y)|,dx,dy$ (the integral of the ABSOLUTE VALUE) is infinite. ${}\qquad\qquad{}$ – Michael Hardy Jun 24 '14 at 20:35
  • (Hence in this case we can deduce that $\displaystyle\iint\limits_{[0,1]^2} \left| \frac{x^2-y^2}{(x^2+y^2)^2} \right|,dy,dx=\infty$.) ${}\qquad{}$ – Michael Hardy Jun 24 '14 at 20:36
  • Huh? You have $I = -I$, so why would they be the same? – IAmNoOne Jun 24 '14 at 20:58
  • @Michael Hardy You're right, I study Mathematics in Greek so the definitions in English is something I have to figure on my own. – George2014L Jun 24 '14 at 21:22
  • @user159942 : Many English-speaking students miss this point as well. – Michael Hardy Jun 24 '14 at 22:16
  • @MichaelHardy What's the difference? Isn't it basically the same as finding the volume under $f(x,y)$, bound by the region $0≤x≤1, 0≤y≤1$? – user3932000 Jan 23 '16 at 03:21
  • @Nameless : They would be the same if $\displaystyle \int_0^1\int_0^1 f(x,y),dx,dy$ were always the same as $\displaystyle \int_0^1\int_0^1 f(x,y),dy,dx$, as in the conclusion of Fubini's theorem. In this case the hypotheses of Fubini's theorem are not satisfied, since the integral of the absolute value is infinite. $\qquad$ – Michael Hardy Jan 23 '16 at 06:06
  • @user3932000 : No --- it's not the area under the graph. Notice that the graph is below the $(x,y)$-plane where the value of the function is negative and above it where the value is positive. If it were the same as finding the volume, then the values of the two iterated integrals would be the same. But they are not, as you will see if you simply evaluate them. The difference between a double integral and an iterated integral is this: In a double integral one integrates with respect to area in the $(x,y)$-plane, and that means if one does things as in${},\ldots\qquad$ – Michael Hardy Jan 23 '16 at 06:09
  • $\ldots,{}$Riemann's theory, one's Riemann sums are sums of values of $f$ multiplied by areas $\Delta x,\Delta y$ in the $(x,y)$-plane, and in Lebesgue's theory the measures of sets in the domain are $2$-dimensional measures in the plane. With an iterated integral, on the other hand, one first integrates with respect to one variable, thus: $\displaystyle \int_0^1 f(x,y),dx$, and one gets a function of the other variable (in this example, $y$), and then one integrates that function of the other variable. @user3932000 ${}\qquad{}$ – Michael Hardy Jan 23 '16 at 06:12
  • @user3932000 : The values of the two iterated integrals and the double integral are the same if the integral of the absolute value is finite. But in cases like this example, the integral of the absolute value is infinite and the double integral is undefined, and the two iterated integrals have different values. $\qquad$ – Michael Hardy Jan 23 '16 at 06:14
  • @user3932000 : If you take a course on measure theory, you'll learn this stuff. $\qquad$ – Michael Hardy Jan 23 '16 at 06:17

1 Answers1

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Consider the double integral over the rectangle $[0,1]\times[0,1]$

$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x.$$

Partial fraction decomposition yields $$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}y \, \mathrm{d}x,$$

$$=\int\limits_{[0,1]} \! \frac{y}{x^2+y^2} \Bigg|_{y=0}^{y=1} \, \mathrm{d}x = \int\limits_{[0,1]} \! \frac{\mathrm{d}x}{x^2+1} = \frac{\pi}{4}.$$

We then consider the double integral over the rectangle $[0,1]\times[0,1]$

$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y.$$

$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}x \, \mathrm{d}y,$$

$$=-\int\limits_{[0,1]} \! \frac{x}{x^2+y^2} \Bigg|_{x=0}^{x=1} \, \mathrm{d}y = -\int\limits_{[0,1]} \! \frac{\mathrm{d}y}{y^2+1} = -\frac{\pi}{4}.$$

Therefore,

$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x \neq\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y$$

Gahawar
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