- #1
Hoblitz
- 11
- 0
Homework Statement
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].
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 [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 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!