(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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

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