Integration over a set in n-dimensional space

  • Thread starter Thread starter sphere1
  • Start date Start date
  • Tags Tags
    Integration Set
sphere1
Messages
2
Reaction score
0
Homework Statement
Not a homework problem, but self-studying Example 1.2.24 from Hogg et al.'s Introduction to Mathematical Statistics (7th ed).
Relevant Equations
N/A
Example 1.2.24: Let C be a set in n-dimensional space, and let Q(C) = ∫...∫Cdx1dx2...dxn. If C= {(x1,x2,...,xn : 0 ≤ x1 x2 ≤ ... ≤ xn ≤ 1}, then Q(C) = ∫010xn0xn-1...∫0x30x2dx1dx2...dxn = 1/n!. (Apologies for the poor formatting of the integration bounds.)

I haven't the foggiest idea how to get to 1/n!. My solution for the latter integral is x2x3...xn. Even if I were to assume that [0, 1] were divided into intervals of equal length, giving x1 = 1/n, x2 = 2/n, etc., I get n!/nn, which I'm pretty sure doesn't reduce to 1/n!. My assumption is that I am somehow integrating the second integral incorrectly, but I'm not sure how. Any suggestions would be appreciated!
 
Physics news on Phys.org
I suggest you start by solving the more general integral
$$
I_n(x) = \int_0^{x} \int_0^{x_{n}} \cdot \int_0^{x_2} dx_1\, dx_2\ldots dx_{n}.
$$
You can solve this by induction. The base case is $$I_1(x) = \int_0^x dx_1 = x$$.

Compute the integral ##I_n(x)## for the first few values of ##n## and try to see the pattern. Then show that it is the correct pattern by induction applying$$I_{n+1}(x) = \int_0^x I_n(x_{n+1}) dx_{n+1}$$
 
Last edited:
  • Like
Likes Gavran and FactChecker
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top