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# 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].

### 4 Answers

- rozeta53Lv 61 decade agoFavorite 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

- KevinMLv 71 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.