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!

Can a positive integrand oscillate fast enough so that the integral is finite?

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    If $$f(x)>0$$ is continuous for all $$x\ge0$$ and the improper integral $$\int_0^{\infty}f(x) dx$$ exists, then $$\lim_{x\rightarrow\infty}f(x)=0.$$

    2. Relevant

    I think this assertion is false. A counterexample can be constructed along the following lines of thought: Suppose a function oscillates between 0 and 2. Then the integral computes a positive area under the curve. However by increasing the frequency, the oscillations can be made so rapid that each peak is too sharp and too close to another peak for the increment in area along the $x$-axis to avoid becoming negligible.


    3. The attempt at a solution

    An example is the function $$f(x)=1+\sin e^x.$$ Clearly, $$f>0$$ because $$0<1+\sin(x)<2.$$ Also integrating $$f(x)-1$$,
    \begin{align*}
    \int_0^\infty\sin e^x dx&= \int_1^\infty \frac{\sin t}{t}\,dt \\
    &=\frac{\pi}{2} - \text{Si}(1)\\
    &<\infty
    \end{align*}
    where $$\text{Si}(x)$$ is the sine integral function. However,
    \begin{equation*}
    \lim_{x\rightarrow\infty}\sin(e^x)\ne0.
    \end{equation*}

    The problem is can such an integrand be constructed?
     
  2. jcsd
  3. May 17, 2012 #2

    sharks

    User Avatar
    Gold Member

    [tex]-1 \le \sin e^x \le 1
    \\0 \le 1+ \sin e^x \le 2[/tex]I would have suggested using the pinching theorem to find if the middle term converges or diverges, but the latter is not applicable in this case, as the limits of the two boundaries are not equal. In this case, f(x) diverges: [tex]0 \le \lim_{n \to \infty} (1+ \sin e^x) \le 2[/tex]But maybe i'm wrong.
     
    Last edited: May 17, 2012
  4. May 17, 2012 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    This is false. The function f becomes 0 many times. Not that it is very hard to fix this problem.
     
  5. May 17, 2012 #4
    Your intuition is correct regarding the validity of the claim. The easiest-to-construct counterexamples aren't "nice" functions, though. I like to keep a nice stock of "spiky" functions on hand to disprove these kinds of claims.

    For the record, there is a continuous function ##f## defined on ##[0,\infty)## with ##f(x)>0## for all ##x## such that ##\int_0^{\infty}f(x)dx## converges while ##\limsup_{x\rightarrow\infty}f(x)=\infty##.
     
  6. May 17, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For the record, what is this function?

    RGV
     
  7. May 17, 2012 #6
    Define ##g## piecewise as follows:

    ##g(x)=0## for ##x## in any interval of the form ##(n+\frac{1}{n^3},n+1)## for ##n=1, 2, 3, ...## as well as on ##[0,1)##.

    Let ##x_n## be the midpoints of intervals of the form ##[n, n+\frac{1}{n^3}]## for ##n=1,2,3,...##. Put ##g(n)=g(n+\frac{1}{n^3})=0##, and ##g(x_n)=n##. Connect the dots with lines on the remainder of ##[n, n+\frac{1}{n^3}]##. So we have a triangular spike of height ##h_n=n## and base ##b_n=\frac{1}{n^3}##.

    Then ##g## is (obviously?) continuous and non-negative. ##\int_0^{\infty}g(x)dx=\sum_1^{\infty}\frac{1}{2}b_nh_n=\sum_1^{\infty}\frac{1}{2n^2}=\frac{\pi^2}{12}##. And ##\limsup_{x\rightarrow\infty} g(x)\geq \lim_{n\rightarrow\infty}g(x_n)=\infty##.

    Let ##h(x)=e^{-x}##, put ##f=g+h##. ##f## has the desired properties.
     
  8. May 17, 2012 #7
    gopher_p:

    That was very clever! Thanks a lot!
     
  9. May 17, 2012 #8
    I take no credit for the idea. Like I said before, the spiky functions are pretty useful as counterexamples to naive notions of what we think should happen.

    As a follow-up exercise, replace continuity of ##f## with uniform continuity. Now is the statement true?

    Or keep the original setup, and ask yourself if it must be true that ##\liminf_{x\rightarrow\infty} f(x)=0##.
     
  10. May 17, 2012 #9
    Does that not affect the answer? If every part of the curve must cover some area, my intuition says the total area covered must be infinite. You could make it oscillate fast enough so that the function is effectively equal to some [itex]\epsilon > 0[/itex] for all [itex]x[/itex], but the integral of that constant small number diverges.
     
  11. May 17, 2012 #10
    gopher_p, thanks for the suggestions. By the way, this question came up in a proseminar packet which have to master before I can qualify for PhD.

    Also Steely_Dan's objection to my previous incorrect example makes sense. Infinite epsilons do add up and the integral diverges. So instead of adding 2, I took the absolute value. Now does this integral diverge:

    $$
    \int_{0}^\infty\lvert\sin(e^x)\rvert dx?
    $$
     
  12. May 18, 2012 #11
    That is still problematic, because the sine function will always have zeros at some locations, so it doesn't satisfy the requirement that the function must be strictly positive.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can a positive integrand oscillate fast enough so that the integral is finite?
Loading...