# Riemann Lebesgue Lemma Clarification

If $f$ is continuous real-valued function, does the Riemann Lebesgue Lemma give us that $\int_{m}^k f(x) e^{-inx}\,dx \rightarrow 0\text{ as } n\rightarrow \infty$ for all $m\le k$? Specifically, is this true for any continuous function, whether it’s periodic or not?

Edited.

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If $f$ is Lebesgue integrable on $[r,s]$, then the result is still true. You can show it by approximation. For $\epsilon > 0$, there exists a continuously differentiable function $g$ on $[r,s]$ such that $\int_{m}^{k}|f-g|dx < \epsilon/2$. And, as $n\rightarrow\pm\infty$,
$$\int_{r}^{s}e^{-inx}g(x)dx = \left.\frac{e^{-inx}}{-in}g(x)\right|_{x=r}^{s}-\int_{r}^{s}\frac{e^{-inx}}{-in}g'(x)dx \rightarrow 0$$
This really has nothing to do with integers. You can assume $m,k,n$ are all real if you like, and you get the same result:
$$\left|\int_{r}^{s}e^{-inx}f(x)dx\right| \le \int_{r}^{s}|f(x)-g(x)|dx + \left|\int_{r}^{s}e^{-inx}g(x)dx\right| < \epsilon$$
if $|n|$ is chosen large enough that the second integral on the right is bounded by $\epsilon/2$.