Obvious integral manipulation

  • Thread starter longrob
  • Start date
  • #1
52
0
This wasn't obvious to me.

From my book. We have,
[tex]p(x)=\begin{cases}
\frac{1}{2} & -2\leq x\leq-1\,\textrm{or}\;1\leq x\leq2,\\
0 & \text{otherwise~}.\end{cases}[/tex]

So, the kth moment is given by
[tex]M_{k}=\frac{1}{2}\int_{-2}^{-1}x^{k}dx+\frac{1}{2}\int_{1}^{2}x^{k}dx[/tex]

So, obviously,
[tex]M_{k}=(1+(-1)^{k})\frac{1}{2}\int_{1}^{2}x^{k}dx[/tex]

Is this a correct explanation of this step:
Since p(x) is an even function, it follows that the left integral is exactly equal to the right integral when k is even, and exactly equal to -1 times the right integral when k is odd. Can it be explained better/ more rigorously ? Will this explanation always apply to any even real-valued function of x ?
 

Answers and Replies

  • #2
mathman
Science Advisor
7,858
446
Your analysis is correct. It will apply to any even function, implicitly assuming the integral exists. The function doesn't have be real-valued, just even. Also the integral being zero applies to any odd function, with the same implicit assumption.
 

Related Threads on Obvious integral manipulation

  • Last Post
Replies
1
Views
923
Replies
5
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
619
Replies
4
Views
2K
  • Last Post
Replies
2
Views
509
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
6
Views
667
Replies
8
Views
1K
Top