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Homework Help: Integral of f times cosine, as period of cosine goes to zero.

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is as follows:

    Let f be a real valued function that is Riemann integrable on [a,b]. Show that

    [tex]

    \lim_{\lambda \rightarrow \infty} \int_{a}^{b} f(x)\cos(\lambda x)dx = 0

    [/tex].

    2. Relevant equations
    I am freely able to use the fact that the product of Riemann integrable functions is Riemann integrable, and that the integral of cosine is sine (+ constant).


    3. The attempt at a solution
    The first thing I tried was just to let f be the indicator function on the interval [a,b]. In that case, it wasn't so bad because then I could find [itex] \lambda > 0[/itex] so large that (for given epsilon > 0) [itex] \frac{1}{\lambda} < \frac{1}{2}\epsilon [/itex].

    Then for that lambda I can find a unique integer k such that
    [tex]
    \frac{2\pi k}{\lambda} \leq (b-a) < \frac{2\pi (k + 1)}{\lambda}
    [/tex].

    Then
    [tex]
    |\int_{a}^{b} \cos(\lambda x)dx| =

    |\int_{a}^{a + \frac{2 \pi k}{\lambda}} \cos(\lambda x) \,dx +

    \int_{a + \frac{2\pi k}{\lambda} }^{b} \cos(\lambda x)\,dx|

    = |\frac{\sin(b)}{\lambda} - \frac{\sin(a)}{\lambda}|
    [/tex]

    Finally,
    [tex]
    |\frac{\sin(b)}{\lambda} - \frac{\sin(a)}{\lambda}| \leq \frac{2}{\lambda} < \epsilon.
    [/tex]

    From there on it is clear that for [itex] \lambda^{'} > \lambda > 0 [/itex], we are still less than epsilon. (Find a new unique integer as above, and run through the integrals again).

    I've been trying to think of a way to apply this fact to the case with general [itex] f [/itex]. What I've tried to do is trap
    [tex]
    \int_{a}^{b} f(x)\cos(\lambda x)dx
    [/tex]
    between plus/minus some constant times [itex] \int_{a}^{b} \cos(\lambda x)dx [/itex], where the constant might be something like the max of the absolute values of the supremum and infemum of f over the interval [a,b], but I cant get it work out (problems arise with negatives multiplying to become positive, etc...). Any hints or comments would be very helpful, thank you!
     
  2. jcsd
  3. Apr 16, 2010 #2

    Office_Shredder

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    I think a bigger concern should be why f(x) is bounded on that interval (it doesn't have to be)
     
  4. Apr 16, 2010 #3
    Indeed I can think a few unbounded functions f which are integrable on an interval [a,b], so I can see why my idea would certainly fail. I'll try looking at this from a new perspective and see what I get. What I understand to be happening is that the rapidly quickening oscillations of cosine are going to force f to "cancel itself out" in the integral, eventually bringing that sucker close to zero...

    Thanks, I'll give this more thought.
     
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