How can I find this definite integral: ∫∫∫...∫ [x1 + x2 + x3 + ... + xN] dx1 dx2 dx3 ... dxN?

Square brackets [x] denote "integer part of x", that is largest integer number less or equal to x.

The limits of integration for all N variables x1,x2,x3, ... xN are [0..1].

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  • 1 decade ago
    Favorite Answer

    J[1] = 0

    J[2] = 1/2

    J[3] = 1

    - - - - - - - -

    J[n]=(n-1)/2

    J[1] = ∫[0,1] 0*dx = 0

    J[2] = ∫[0,1] ∫0,1-x] 0*dy*dx +∫[0,1] ∫1-x,1] 1*dy*dx =

    0+∫[0,1] y*dx [y from 1-x to 1] = ∫[0,1] [1-(1-x)]*dx =

    ∫[0,1] x*dx = x²/2 [x from 0 to 1] = 1/2

    J[3] = 0*1/6+1*4/6+2*1/6=1

    Because below the plane x+y+z=1 is 1/6 of the volume of integration (a unit cube), where x1+x2+x3+...+xN=0, which gives 0*1/6=0.

    Between the planes x+y+z=1 and x+y+z=2 are 4/6 of the volume of integration, where x1+x2+x3+...+xN=1, which gives 1*4/6=4/6.

    Above the plane x+y+z=2 is 1/6 of the volume of integration, where x1+x2+x3+...+xN=2, which gives 2*1/6=2/6.

    J[n]=(n-1)/2

    Based on a unit cube for solving J[3], I "see" the domain of integration as a unit "cube" in an n-dimensional space, made of pairs of equal "volumes".

    For even n there are n/2 pairs, such that if in one element of the pair the value of [x1+x2+x3+...+xN] is k, then in the corresponding element of the pair the same value is (n-1-k) ==>

    the mean value of [x1+x2+x3+...+xN] for each pair is (n-1)/2, so the integral becomes

    0.5(n-1)*∫∫∫...∫ dx1 dx2 dx3 ... dxN (over a unit n-dimensional "cube") = (n-1)/2.

    For odd n the number of the pairs is (n-1)/2 and there is a central "volume" where [x1+x2+x3+... +xN]=(n-1)/2, while for the pairs, if in one element of the pair the value of [x1+x2+x3+...+xN] is k, then in the corresponding element of the pair the same value is (n-1-k) ==>

    the mean value of [x1+x2+x3+...+xN] for each pair is (n-1)/2, the same as for the central element, so again the integral becomes

    0.5(n-1)*∫∫∫...∫ dx1 dx2 dx3 ... dxN (over a unit n-dimensional "cube") = (n-1)/2.

  • 1 decade ago

    Based on your descripton, it is a step-function. There is no area beneath the curve btw 0 and 1

    Answer: 0

    *********** Addendum **********

    I didn't read the question correctly. lol. It is a step function, but it will be the integral of [f(n) ... 1] where f(n) is based on the round-down step function.

    Rozetta may have the correct answer -- I'll have to review his answer before I attempt to derive the answer. However, on its face, Rozetta's answer looks correct.

    .........

    Just playing around, it looks like f(n) = 1/n which means the answer is that you integrate from [1/n ...1] which gives you

    (n-1)/2 ... Congrats to Rozetta

  • KevinM
    Lv 7
    1 decade ago

    Assuming that all of the xN are independent, you'll get:

    1/2 x1^2 * [x2*x3*x4...*xN] + 1/2 (x2)^2 * [x1*x3*x4...*xN]... etc.

    = [x1*x2*x3....*xN] * (1/2 x1 + 1/2 x2 + 1/2 x3 .... + 1/2 xN)

    = 1/2 [x1*x2*x3....*xN] * (x1 + x2 + x3 + x4 ... + xN)

    Taken from 1 to 0 on all ranges, this will yield N/2.

  • Retsum
    Lv 6
    1 decade ago

    The variables are separable so you can integrate term by term.

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