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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.
     
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