How to prove $$\lim_{n \rightarrow \infty} \left|\int_{-\pi} ^\pi f(x)e^{inx} dx\right|=0$$
I know the Riemann-Lebesgue lemma is $$\lim_{n \rightarrow \infty} \hat{f}(n)=\lim_{n \rightarrow \infty}\frac{1}{2\pi}\int_{-\pi} ^\pi f(x)e^{-inx}dx=0$$
Will it be used here? Am I on the right track?
Let $f$ be a continuous periodic function, it means that $f(-\pi)=f(\pi)$, then $f$ is bounded and uniformly continuous, (there is $0<M<\infty$ such that $|f| \leq M$)
$$\int_{-\pi} ^\pi |f(x)e^{inx}| dx \leq \int_{-\pi} ^\pi |f(x)| dx \leq 2 \pi M \Rightarrow f \in L^1[-\pi,\pi].$$
Now, (as commented) use (to make it looks like Fourier transform) $g(x):= f(-x),\quad x \in [-\pi,\pi]$ and use Riemann-Lebesgue Lemma directly. You will have
$$\lim_{n\to \infty} | \hat{g}(n) | = \left| \int_{-\pi}^\pi g(x) e^{-i n x} dx \right| \to 0 \quad (n \to \infty) $$