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

1. Apr 16, 2010

### Hoblitz

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

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

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 $\lambda > 0$ so large that (for given epsilon > 0) $\frac{1}{\lambda} < \frac{1}{2}\epsilon$.

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

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

Finally,
$$|\frac{\sin(b)}{\lambda} - \frac{\sin(a)}{\lambda}| \leq \frac{2}{\lambda} < \epsilon.$$

From there on it is clear that for $\lambda^{'} > \lambda > 0$, 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
$$\int_{a}^{b} f(x)\cos(\lambda x)dx$$
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 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. Apr 16, 2010

### Office_Shredder

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

3. Apr 16, 2010

### 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.