Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
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 $7008$

share|improve this answer


  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.

share|improve this answer
Кого я вижу! Ты ушел с dxdy? –  Norbert 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? –  Norbert 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! –  Norbert Jan 23 '12 at 17:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.