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Calculate the Lebesgue integral of the function

$$ f(x,y)=\left\lbrace\begin{array}{ccl}[x+y]^{2} &\quad&|x|,|y| <12 ,\quad xy \leq 0\\ 0 &\quad&\text{otherwise}\end{array} \right.$$

in $\mathbb{R}^2$.

Can anyone help with this? I can't find a way to make the expression of $f$ more simply to calculate the integral.

edit: $[\cdot]$ is the integer part.

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In general, we use $\lfloor \cdot \rfloor$ to denote the floor function. That's "\lfloor" and "\rfloor". –  Isaac Solomon Jan 22 '12 at 19:06
2  
This function has finite range. It can be integrated by drawing a picture. –  ncmathsadist Jan 22 '12 at 19:16
    
@ncmathsadist: Can you explain this a little more? –  passenger Jan 22 '12 at 19:18
    
The integrand breaks up into a finite number of cases. It is piecewise constant on strips between lines of the form $y = a - x$ and $y = a + 1 - x$, where $a$ is an integer. Draw the slices; the function is constant between the parallel lines that result. Multiply the area of each strip by the value of the function on the strip. –  ncmathsadist Jan 22 '12 at 19:29
    
The Riemann integral gives the same value (why?) –  AD. Jan 22 '12 at 19:54

2 Answers 2

up vote 8 down vote accepted

Denote $$ A_{m,n}=\{(x,y):m\leq x<m+1,\quad n\leq y<n+1\}\qquad a_{mn}=\int_{A_{m,n}}f(x,y)d\mu(x,y) $$ then $$ \int_{\mathbb{R}^2}f(x,y)d\mu(x,y)=\sum_{(m,n)\in\mathbb{Z}^2}a_{mn} $$ From definition of $f$ it follows that $a_{mn}\neq 0$ only for pairs $(m,n)\in\mathbb{Z}^2$ such that $-N\leq m\leq N-1$, $-N\leq n\leq N-1$ and $mn\leq 0$, because $f$ is non zero only on this sets. Hence $$ \int_{\mathbb{R}^2}f(x,y)d\mu(x,y)=\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}a_{mn}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}a_{mn} $$ It is remains to get the formula for $a_{mn}$. Consider sets $$ B_{mn}=\{(x,y)\in A_{mn}:x+y<m+n+1\}\qquad C_{mn}=\{(x,y)\in A_{mn}:x+y\geq m+n+1\} $$ It is easy to see that $A_{mn}=B_{mn}\cup C_{mn}$, $B_{mn}\cap C_{mn}=\varnothing$ and $$ f(x,y)=(m+n)^2\quad\text{for}\quad(x,y)\in B_{mn} $$ $$ f(x,y)=(m+n+1)^2\quad\text{for}\quad(x,y)\in C_{mn} $$ So, $$ \begin{align} a_{mn}=\int_{A_{m,n}}f(x,y)d\mu(x,y) &=\int_{B_{m,n}}f(x,y)d\mu(x,y)+\int_{C_{m,n}}f(x,y)d\mu(x,y)\\ &=(m+n)^2\mu(B_{mn})+(m+n+1)^2\mu(C_{mn})\\ &=\frac{1}{2}(m+n)^2+\frac{1}{2}(m+n+1)^2\\ &=m^2+n^2+2mn+m+n+0.5 \end{align} $$ Now we can find our integral $$ \int_{\mathbb{R}^2}f(x,y)d\mu(x,y)=\sum_{(m,n)\in\mathbb{Z}^2}a_{mn}= \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(m^2+n^2+2mn+m+n+0.5) $$ This is a labour computation to get this sum, so we will find it by parts $$ \begin{align} \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(m^2) &=\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}m^2+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}m^2\\ &=N\sum\limits_{m=-N}^{-1}m^2+N\sum\limits_{m=0}^{N-1}m^2\\ &=N\sum\limits_{m=1}^{N}m^2+N\sum\limits_{m=1}^{N-1}m^2\\ &=N\frac{N(N+1)(2N+1)}{6}+N\frac{N(N-1)(2N-1)}{6}\\ &=\frac{2N^4+N^2}{3}\\ \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(m) &=\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}m+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}m\\ &=N\sum\limits_{m=-N}^{-1}m+N\sum\limits_{m=0}^{N-1}m\\ &=N\sum\limits_{n=-N}^{N-1}m=N\cdot(-N)=-N^2 \end{align} $$ Similarly, $$ \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(n^2)=\frac{2N^4+N^2}{3} $$ $$ \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(n)=-N^2 $$ Then $$ \begin{align} \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(mn) &=\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}mn+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}mn\\ &=\sum\limits_{m=-N}^{-1}m\sum\limits_{n=0}^{N-1}n+\sum\limits_{m=0}^{N-1}m\sum\limits_{n=-N}^{-1}n\\ &=-\sum\limits_{m=1}^{N}m\sum\limits_{n=0}^{N-1}n-\sum\limits_{m=0}^{N-1}m\sum\limits_{n=1}^{N}n\\ &=-\frac{N(N+1)}{2}\frac{N(N-1)}{2}-\frac{N(N-1)}{2}\frac{N(N+1)}{2}\\ &=-\frac{N^2(N^2-1)}{2}\\ \left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(0.5) &=\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}0.5+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}0.5\\ &=0.5N^2+0.5N^2\\ &=N^2 \end{align} $$ Finally, we get $$ \begin{align} \int_{\mathbb{R}^2}f(x,y)d\mu(x,y) &=\left(\sum\limits_{m=-N}^{-1}\sum\limits_{n=0}^{N-1}+\sum\limits_{m=0}^{N-1}\sum\limits_{n=-N}^{-1}\right)(m^2+n^2+2mn+m+n+0.5)\\ &=\frac{2N^4+N^2}{3}+\frac{2N^4+N^2}{3}-2\frac{N^2(N^2-1)}{2}-N^2-N^2+N^2\\ &=\frac{N^4+2N^2}{3} \end{align} $$ If we take $N=12$ we will obtain $7008$

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Hint:

  1. The function is non-negative, and hence one may apply Tonelli's theorem (sometimes cited as Fubini-Tonelli's or even Fubini' theorem).

  2. Draw the domain of integration (that is the set where $f(x,y)\ne0$). Split up the domain in order to adopt step 1.

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Кого я вижу! Ты ушел с dxdy? –  userNaN Jan 22 '12 at 21:36
    
@Norbert What do you mean? –  AD. Jan 23 '12 at 7:08
    
May be I've confused you with my friend from MSU. Artem is it you? –  userNaN Jan 23 '12 at 14:35
    
@Norbert Sorry that is not me. :) –  AD. Jan 23 '12 at 16:42
    
What a nuisance... The reason of this confusion is that he have the same nick and the same range of mathematical interests as yours! –  userNaN Jan 23 '12 at 17:11

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