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

Existence of Limit with Integrals.

  1. Sep 13, 2011 #1
    Hi, I saw a proof/argument done today that I think was wrong:

    It is finding the limit as a->oo of the integral from 0 to b<oo:

    Int_(0..b) Sqr[x(1 +cos(ax))]dx , where Sqr is the square root

    Now, the argument given was that one could find a bound for the oscillation

    of (1+cos(ax)).

    The problem I have is that, no matter what the trick may be, cos(ax) will take

    values of 1 , and of -1 (at 2kPi and 2(k+1)Pi respectively; k an integer), so that

    the value of the limit will go from:

    i) 0 , when cos(ax)=-1 , to:

    ii) Int_
    (0..b) Sqr[2x]dx =(Sqr2)x^(-1/2), which is an improper integral at x=0, but does not

    go to zero.

    So the limit does _not_ exist, right?
  2. jcsd
  3. Sep 14, 2011 #2
    No. The limits does exits. It's still just the limit of a (bounded) Riemann sum. First write it clearly what we're talking about:

    [tex]\lim_{a\to\infty} \int_0^b \sqrt{x(1+\cos(ax))}dx,\quad b<\infty[/tex]

    Since b is finite and:

    [tex]0\leq|1+\cos(ax)|\leq 2[/tex]

    then cannot we say:

    [tex]0\leq \left|\int_0^b \sqrt{x(1+\cos(ax))}dx\right|\leq \int_0^b \sqrt{2x}dx[/tex]

    and therefore:

    [tex]0\leq \lim_{a\to\infty} \int_0^b \sqrt{x(1+\cos(ax))}dx\leq \int_0^b \sqrt{2x}dx[/tex]
  4. Sep 14, 2011 #3
    Well, yes, but I imagine the limit would exist if we can squeeze it from the right into
    being 0, right? If we could, e.g., only say that the limit is betwwm 0 and 3, can we
    then say the limit exists?
  5. Sep 14, 2011 #4
    I actually came up with numerical values near 80,000 for b near 2000; seems way too

    far from 0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook