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

[Hoblitz]

Homework Statement

The problem is as follows:

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

<br /> <br /> \lim_{\lambda \rightarrow \infty} \int_{a}^{b} f(x)\cos(\lambda x)dx = 0<br /> <br />.

Homework 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).

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 \lambda &gt; 0 so large that (for given epsilon > 0) \frac{1}{\lambda} &lt; \frac{1}{2}\epsilon.

Then for that lambda I can find a unique integer k such that
<br /> \frac{2\pi k}{\lambda} \leq (b-a) &lt; \frac{2\pi (k + 1)}{\lambda} <br />.

Then
<br /> |\int_{a}^{b} \cos(\lambda x)dx| = <br /> <br /> |\int_{a}^{a + \frac{2 \pi k}{\lambda}} \cos(\lambda x) \,dx +<br /> <br /> \int_{a + \frac{2\pi k}{\lambda} }^{b} \cos(\lambda x)\,dx| <br /> <br /> = |\frac{\sin(b)}{\lambda} - \frac{\sin(a)}{\lambda}|<br />

Finally,
<br /> |\frac{\sin(b)}{\lambda} - \frac{\sin(a)}{\lambda}| \leq \frac{2}{\lambda} &lt; \epsilon.<br />

From there on it is clear that for \lambda^{&#039;} &gt; \lambda &gt; 0, 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 f. What I've tried to do is trap
<br /> \int_{a}^{b} f(x)\cos(\lambda x)dx <br />
between plus/minus some constant times \int_{a}^{b} \cos(\lambda x)dx, 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 can't get it work out (problems arise with negatives multiplying to become positive, etc...). Any hints or comments would be very helpful, thank you!

 

[Office_Shredder]

I think a bigger concern should be why f(x) is bounded on that interval (it doesn't have to be)

 

[Hoblitz]

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.