Using the Riemann-Lebesgue lemma to find the integral of a very oscillating function

Question

I'm a bit confused with how the Riemann-Lebesgue lemma is used to show that the integral of a very oscillating function approaches zero as the number of oscillation increases. To be more specific, let $f(x)=\cos(kx)$ for all $x\in [0,1]$ and zero everywhere else, and we want to find \begin{equation} \int_{0}^{1} f(x) {\rm d}x \end{equation} how does one use the Riemann-Lebesgue lemma to show that this definite integral approaches zero as $k \rightarrow \infty$?

Answer 1

Let $f(x)=\chi_{[0,1]}(x)$. Then the Fourier transform of $f$ is given by $$\hat{f}(k)=\int_\mathbb{R} e^{-ikx}f(x)\,dx=\int_\mathbb{R}(\cos(-kx)+i\sin(-kx))\chi_{[0,1]}(x)\,dx$$ (perhaps also multiplying by $\pi$ somewhere depending on your convention). Thus, the real part of the Fourier transform of $f$, $\Re \hat{f}(k)$ is just equal to $$\int_\mathbb{R}\cos(-kx)\chi_{[0,1]}(x)\,dx=\int_0^1\cos(kx)\,dx$$ where we have used that $cos(-kx)=cos(kx)$. The result then follows from the usual Riemann-Lebesgue lemma.

However, I would suggest thinking about this result "by hand" by going back to the proof of the Riemann-Lebesgue lemma. The essence of the lemma is to use oscillation to obtain cancellation and thus decay. This situation occurs often in analysis and the usual way to make this precise is via integration by parts. In this instance, we would have to approximate $f$ by a smooth function close to it in the $L^1$ norm, and then proceed as in the proof of Riemann-Lebesgue.