3
$\begingroup$

We want to evaluate the double integral $$ \int \int_{R} f(x,y) ~ \text{d}x \text{d}y $$ over the rectangle $R$ given by $0 \leq x \leq 2, 0 \leq y \leq 1$, with the function $f$ defined by $$ f(x,y) = \dfrac{xy(x^2 - y^2)}{(x^2 + y^2)^3} \quad \text{ for } (x, y) \neq (0,0) $$ and $f(0,0) = 0$. We integate first in the $y$-direction. For any $x \in [0,2]$, we define $A(x) = \int^{1}_{0} f(x,y) ~ \text{d}y$. For now, we assume that $x \neq 0$. We make the substitutions $u = x^2 + y^2, \text{d}u = 2y \text{d}y$, use $x^2 - y^2 = x^2 - (u - x^2) = 2x^2 - u$, and compute \begin{align*} A(x) &= \int^{x^2 + 1}_{x^2} \dfrac{x(2x^2 - u)}{2u^3} ~ \text{d}u = \int^{x^2 + 1}_{x^2} \left( \dfrac{x^3}{u^3} - \dfrac{x}{2u^2} \right) ~ \text{d}u = \left. - \dfrac{x^3}{2u^2} + \dfrac{x}{2u} \right|^{u = x^2 + 1}_{u = x^2} \\ &= - \dfrac{x^3}{2(x^2 + 1)^2} + \dfrac{x}{2(x^2 + 1)} + \dfrac{1}{2x} - \dfrac{1}{2x} = \dfrac{x}{2(x^2 + 1)^2} \end{align*} We note that this formula remains valid for $x = 0$, as $f$ vanishes on the whole $y$-axis. Then we integrate in the $x$-direction $$ \int^{2}_{0} A(x) ~ \text{d}x = \int^{2}_{0} \dfrac{x}{2(x^2 + 1)^2} ~ \text{d}x = \left. \dfrac{-1}{4(x^2 + 1)} \right|^{x = 2}_{x = 0} = - \dfrac{1}{20} + \dfrac{1}{4} = \dfrac{1}{5} $$ In the next step we switch the order of integration, i.e. we integrate first in the $x$-direction. For any $y \in [0,1]$, we define $B(y) = \int^{2}_{0} f(x,y) ~ \text{d}x $. If $y \neq 0$, we again make the substitution $u = x^2 + y^2$, except now $\text{d}u = 2x \text{d}x$. We obtain $$ B(y) = \left. - \dfrac{y}{2u} + \dfrac{y^3}{2u^2} \right|^{u = 4 + y^2}_{u = y^2} = - \dfrac{y}{2(4 + y^2)} + \dfrac{y^3}{2(4 + y^2)^2} = \dfrac{-2y}{(4+y^2)^2} $$ This formula too remains valid for $y = 0$, as $f$ vanishes on the $x$-axis. Then we integrate in the $y$-direction $$ \int^{1}_{0} B(y) ~ \text{d}y = \int^{1}_{0} \dfrac{-2y}{(4+y^2)^2} ~ \text{d}y = \left. \dfrac{1}{4 + y^2} \right|^{y=1}_{y=0} = \dfrac{1}{5} - \dfrac{1}{4} = - \dfrac{1}{20} $$ which is not the same as our first answer.

What happened here? The fact that all the functions we integrated are continuous functions of one variable offers no clue that anything is wrong. The point is that Fubini's Theorem $\textbf{does not apply}$, because the function $f$ is not integrable over $R$, indeed, it is not even bounded on $R$.

At the end we should conclude that $f$ is not continuous in $(0,0)$. I don't see how to do that. Can anyone give me a hint?

$\endgroup$
2
  • $\begingroup$ I think you can do this take a small ball around $(0,0)$ and integrate the same function over a rectangle which is inside the ball. Then you should get both integrals same and If you do not then it is not continuous at (0,0) $\endgroup$
    – happymath
    Apr 23, 2015 at 10:32
  • $\begingroup$ If $f$ were continuous at the origin, then $f$ would be globally continuous. And then Fubini's theorem would be applicable. $\endgroup$ Apr 23, 2015 at 10:53

1 Answer 1

0
$\begingroup$

using polar coordinates f(x,y)=f(r,a)=(sin4a)/r^2 if a is not zero ans 0 otherwise.If we take the ray a=pi/8 we can see that f is not bounded near the origen and so it is not continuous: f(0,0)=0

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .