# Thread: [SOLVED] Double integral of floor function

1. ## [SOLVED] Double integral of floor function

Let denote the greatest integer that is smaller than or equal to x. Evaluate where D = {}.

Hints: The trick is to bring up the underlying region (which is a simple square) into two parts. D = D1 + D2

where

• D1 is the triangle with corners at (0,0), (1,0), and (0,1), excluding the diagonal from (0,1) to (1,0).
• D2 is the rest of the remaining region of D.

Therefore we have
= + .

2. The whole point of dividing D into two regions is that you can simplify the integrand $\lfloor{x+y}\rfloor$. In $D_1, 0\le{x+y}<1$, while in $D_2, 1\le{x+y}<2$ (except at the single point (1,1), which we can safely ignore), so $\lfloor{x+y}\rfloor = 0$ in $D_1$ and $\lfloor{x+y}\rfloor = 1$ in $D_2$. I think you can probably handle it from there.

Post again if you're still having trouble.

3. Thanks. I'm not understanding the integral of a floor function, though. We had a single integral of floor function x a few weeks ago. You guys around here helped me with that. I'm looking at those notes and I'm not able to work that out to the double integral of x+y. Something about adding that y in there is confusing me. I have no idea how to integrate this function.

4. $\int_{y}^{1+y}{\left\lfloor t \right\rfloor \,dt}=\sum\limits_{j=y}^{y}{\int_{j}^{j+1}{\left\l floor t \right\rfloor \,dt}}=y,$ now in the double integral put $t=x+y$ so its value is $\frac12.$

5. Think of the floor function as a piecewise constant function:

$\lfloor{x}\rfloor = 0$ when $0\le{x}<1, 1$ when $1\le{x}<2, 2$ when $2\le{x}<3$, etc.

When you integrate, you have to divide up the pieces:

$\int_{0}^{3}\lfloor{x}\rfloor dx = \int_{0}^{1}\lfloor{x}\rfloor dx + \int_{1}^{2}\lfloor{x}\rfloor dx + \int_{2}^{3}\lfloor{x}\rfloor dx = \int_{0}^{1}0 dx + \int_{1}^{2}1 dx + \int_{2}^{3}2 dx$

The double integral is the same idea. Now we're integrating over the square with vertices (0,0), (0,1), (1,1), (1,0) instead of a line, but it's the same thing - you divide the region into pieces. In this case, the triangle with vertices (0,0), (0,1), (1,0) where $\lfloor{x+y}\rfloor = 0$ is one region, and the triangle with vertices (1,1), (0,1), (1,0) where $\lfloor{x+y}\rfloor = 1$ is the other.

So in your last equation, you can substitute 0 for $\lfloor{x+y}\rfloor$ in the integral over $D_1$ and 1 for $\lfloor{x+y}\rfloor$ in the integral over $D_2$. The first integral will be zero, and the second integral is just the area of the triangle with vertices (1,1), (0,1), (1,0), which should come out to 1/2.

Don't be afraid to post again if you're still having trouble.

6. Hollywood, your last post made it click for me. I was writing down what you typed and literally said "oooohhh" out loud. It's not nearly as hard as I thought. I must have just been having trouble relating to the definition of this function because I'm not familiar with it. Once I drew the square instead of just looking at my professor's diagram, it all sunk in. Thank you!!!

7. ## Re: [SOLVED] Double integral of floor function

Originally Posted by ban26ana
Let denote the greatest integer that is smaller than or equal to x. Evaluate where D = {}.

Hints: The trick is to bring up the underlying region (which is a simple square) into two parts. D = D1 + D2

where

• D1 is the triangle with corners at (0,0), (1,0), and (0,1), excluding the diagonal from (0,1) to (1,0).
• D2 is the rest of the remaining region of D.

Therefore we have
= + .
if where D = {1<= x <= 3, 2<=y<=5} how do resolve this?

thanks

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# evaluate the double integration of greatest integer function x y

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