Show that $\lim_{n\to\infty}\int_{-\pi}^{\pi}e^{ix}\cos(nx)dx = 0$

Question

I now that I can prove this by the definition of $\epsilon$ $\delta$, but I was playing and I found that if I use the substitution

$\cos(nx) = \frac{1}{2}(e^{inx} + e^{-inx})$

Hence, for any $n \in \mathbb{N}$ we have $$ \frac{1}{n+1}\sin([n+1]x) + \frac{1}{n-1}\sin([n-1]x)\biggr|_{-\pi}^{\pi} = 0$$

It's something wrong with my solution?

Answer 1

Doing integration by parts

$$A_n=||\int_{-\pi}^{\pi}e^{ix}\cos(nx)dx||=||\frac{e^{ix}(n\sin{nx}+i \cos{nx})}{n^2-1}]_{-\pi}^{\pi}||_2=|-\frac{2n \sin{n\pi}}{n^2-1}|=0$$

where $||.||_2$ denotes the complex absolute value.

Thus $\limsup_{n \rightarrow \infty}A_n=0$