1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this Riemann Integrable?

  1. Oct 22, 2005 #1
    let f(x) = 1 when x in in [0,1)
    f(x) = -1/2 when x is in [1,2)
    f(x) = 1/3 when x is in [2, 3)

    and so on, in othe words its the sequence (1/n)(-1)^n, whose series obviously converges to log 2. However is f(x) Riemann integrable and equal to this series?
    If so, how to give an upper sum lower sum proof?, just choose a good partion?


  2. jcsd
  3. Oct 22, 2005 #2
    that function is riemann integrable but i'm not sure what it has to do with that sequence/series?
  4. Oct 22, 2005 #3
    Ok, but how to prove it?

    It has to do with that sequence because on each interval [n, n+1) it is equal to (1/n+1)(-1)^n.
  5. Oct 22, 2005 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Perhaps it would help if you clarified what your function is. In your original post you had
    "let f(x) = 1 when x in in [0,1)
    f(x) = -1/2 when x is in [1,2)
    f(x) = 1/3 when x is in [2, 3)"

    Which is a piecewise constant function that has trivially integrable. Yes, it is equal to (-1)^n/(n+1) on the interval [n,n+1) but what does that have to do with the series 1- 1/2+ 1/3+....

    Ah, wait- you aren't asking whether the function is Riemann integrable- you are asking whether the definite integral from 0 to infinity exists! I think you've answered your own question. The improper integral [tex]\int_0^\infty f(x)dx[/tex] exists if and only if the limit [tex]lim_{A->0}\int_0^A f(x) dx[/tex] exists.

    With f as you give it, [tex]\int_0^A f(x)dx= \Sigma_{i=1}^{N}\frac{(-1)^i}{i+1}[/tex] where N is the largest integer less than or equal to A and, as you say, that series converges to log 2.
    Last edited: Oct 22, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Is this Riemann Integrable?
  1. Riemann integration (Replies: 2)

  2. Riemann integrability (Replies: 1)

  3. Riemann integrability (Replies: 1)

  4. Riemann Integrability (Replies: 2)