Each of m and n is a positive integer with m < n. Evaluate this double definite integral in terms of m and n.

∫ ∫ [x+y]*{x+y}*(x+y) dx dy for x = 1 to m and, y = 1 to n

*** [x] denotes the smallest integer ≤ x, and {x} is the decimal part of x, ie {x}=x-[x].

Consider the unit square, A, with opposite vertices at (a, b) and (a+1, b+1).
Now consider the elemental strip within A, made up of all points (x, y), for which

                        a + b + s < x + y < a + b + s + ds

For 0 <= s <= 1  (i.e. in the triangular region, A1, below the diagonal)

Integrand = [x + y]{x + y}(x + y) = (a + b)s(a + b + s)
                                                = cs(c + s)        where c = a + b

Area of elemental strip = s ds

So the integral over A1 = Int(s=0 to 1) of cs(c + s) s ds

            = c(cs³/3 + s⁴/4)            between 0 and 1

            = c(4c + 3)/12               (1)

For 1 <= s <= 2  (i.e. in the triangular region, A2, above the diagonal)

Integrand = [x + y]{x + y}(x + y) = (a + b + 1)(s - 1)(a + b + s)

                                                = (c + 1)(s - 1)(c + s)

Area of elemental strip = (2 - s) ds

Integral over A2 = Int(s=1 to 2) of (c + 1)(s - 1)(c + s)(2 - s) ds

            = (c + 1) Int(c(3s - s² - 2) + 3s² - s³ - 2s) ds

            = (c + 1)(c(3s²/2 -s³/3 -2s) +s³ - s⁴/4 -s²) between 1 and 2

            = (c + 1)(2c + 3)/12                  (2)

Adding (1) and (2) gives:

Integral over A = (6c² + 8c + 3)/12

                        =  (6a² + 6b² + 12ab + 8a + 8b + 3)/12               (3)

To find the required integral, over the region x = 1 to m, y = 1 to n, we must carry out a double summation of (3) over all included squares. Thus:

Sum(a = 1 to m-1, b = 1 to n-1) of   (6a² + 6b² + 12ab + 8a + 8b + 3)/12

Standard formulae for sums of integers and their squares can be used and, after much algebra, the following result is obtained.

Integral  =  (m - 1)(n - 1)(2m² + 2n² + 3mn +3m + 3n + 3)/12

So, for example, m = n = 2 gives:           Integral = 43/12

The formula is symmetrical in m and n, as would be expected, so the given condition that m < n is not needed.

Posted by Harry on 2011-07-25 23:35:03