Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two improper integrals

  1. Oct 19, 2014 #1
    I read that the improper Riemann integral ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx## converges and that ##\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx## does not.
    I have tried comparison criteria for ##\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx##, but I cannot find a function ##f## with a divergent integral such that ##0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|##.
    I heartily thank you for any help!

    EDIT: I have managed to use complex analysis to show that ##\int_0^{\infty} \frac{1}{x}\sin xdx=\frac{\pi}{2}##, but, with a change of variable, I see that ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx=\int_1^{\infty}\frac{1}{x}\sin x dx##. ##\frac{1}{x}\sin x## is integrable on ##[0,1]## (it can be "made continuous" in 0 where it approaches 1), therefore ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx## converges.
     
    Last edited: Oct 19, 2014
  2. jcsd
  3. Oct 19, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    A lower bound for the integrand is "f(x)=0.1/x if |sin(1/x)| > 0.1, f(x)=0 otherwise". The first case always has larger intervals than the second, so you can combine those intervals to find another lower bound that diverges.
     
  4. Oct 19, 2014 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Let y = 1/x, for both. You will see that the absolute value case can be easily studied. |sinx/x| < c/x for most of the domain of integration.
     
  5. Oct 20, 2014 #4
    Thank you both!
    Forgive my ununderstanding: how can we use the fact to show that ##\int_{0}^{1}\frac{1}{x}|\sin(\frac{1}{x}|) dx##, which I do see to be the same as ##\int_{1}^{+\infty}\frac{1}{x}|\sin(x)| dx##, diverges?

    Excuse me: how do we rigourously see that the intervals where ##f(x)=0.1/x## are larger than where ##f(x)=0##, and how do we see that it diverges? I do know that ##\frac{1}{10}\int_{0}^{1}\frac{1}{x}dx## does, but "some pieces" are lacking here and I am not having any idea to verify that...

    I heartily thank you both again!!!
     
  6. Oct 20, 2014 #5

    mathman

    User Avatar
    Science Advisor
    Gold Member

    All you need to check is one cycle of sinx.
     
  7. Oct 20, 2014 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    One half-cycle will do the job. If you want to do it in detail, it will take some steps, but it is possible. The value of 0.1 is chosen in a way to give enough room for useful approximations (but you can also use 0.01 if you want).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Two improper integrals
  1. Improper integral (Replies: 1)

  2. Improper Integration (Replies: 5)

  3. Improper integrals (Replies: 4)

  4. Improper Integrals (Replies: 1)

  5. "Improper" Integral (Replies: 3)

Loading...