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Let $f(x,y) = \dfrac{1}{(y+1)^2}$ and let $A$ and $B$ be the open subsets

$A = \{(x,y)\,|\,x > 0 \text{ and } x < y < 2x\}$

$B = \{(x,y)\,|\,x > 0 \text{ and } x^2 < y < 2x^2\}$

of $\mathbb{R}^2$.

How to show that the $\int_A f$ does not exist but the $\int_B f$ does and find its value?

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Where are you stuck? –  Rasmus Feb 20 '12 at 10:04
    
OK, I see. Made some errors in the integration. –  James R. Feb 20 '12 at 10:49
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2 Answers

up vote 0 down vote accepted

$\bf Hint:$ For the first part $\int_A f=\int^\infty_0(\int_x^{2x} \frac 1 {(1+y)^2}dy)dx=\int_0^\infty \frac{-1}{1+2x}+\frac 1{1+x}dx=\lim_{x\to\infty}-\frac 1 2\log(1+2x)+\log(1+x)=? $
The second part is similar.

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It is almost a directly calculation, the only thing you need to know is that the set $A$ and $B$ has the property that make the following calculation valid.

  1. \begin{align} \int_0^{+\infty}\int_x^{2x}\frac{1}{(y+1)^2}d y d x=\int_0^{+\infty}\left.\frac{-1}{3(y+1)^3}\right|^{2x}_x d x\ &=\int_0^{+\infty}\frac{7x^3+9x^2+3x}{3(1+x)^3(1+2x)^3} dx \end{align} Now note that for sufficient large $x>N$, we will have $$ \frac{7x^3+9x^2+3x}{(1+x)^3(1+2x)^3}>1/6. $$ which will give $$ \int_0^{+\infty}\int_x^{2x}f(x,y) dx dy\geq\int_N^{+\infty}1/6 dx=+\infty. $$
  2. $$ \int_0^{+\infty}\int_x^{2x}\frac{1}{(y+1)^2}d y d x=\int_0^{+\infty}\left.\frac{-1}{3(y+1)^3}\right|^{2x^2}_{x^2} d x\\ $$ Now $$ \left.\frac{-1}{3(y+1)^3}\right|^{2x^2}_{x^2}=\frac{7 x^6+o(x^6)}{3x^{12}+o(x^{12})}<7/3x^6 $$ for sufficient large $x>0$, the result obtained by observe that $$ \int_0^N \left.\frac{-1}{3(y+1)^3}\right|^{2x^2}_{x^2} d x<+\infty,\quad\int_N^{+\infty}7/3x^6 dx<+\infty. $$
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Sorry to make a mistake for the "Primitive Function", I think you can give the result by a similar method. –  van abel Feb 20 '12 at 10:47
    
van abel, it looks like the answer is wrong, above, but the analysis is correct –  James R. Feb 24 '12 at 8:55
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