Finding limit under integral

Question

Evaluate if $f \in C [-\pi,\pi]$

$$\lim_{n\to\infty} \int_{-\pi}^{\pi}f(t)\cos(nt)dt$$ and

$$\lim_{n\to\infty}\int_{-\pi}^{\pi} f(t)\cos^2(nt)dt$$

Answer 1

$f$ is uniform continuous on $[-\pi,\pi]$ so given $\epsilon>0 \quad \exists \delta>0$ such that $$|f(x)-f(y)|<\epsilon$$ whenever $|x-y|<\delta$.

Now using the fact that $f$ is integrable choose a partition $\{-\pi=x_0<x_1< \cdots<x_m=\pi\}$ of $[-\pi,\pi]$ having mesh less than $\delta$ then $$ \int^{\pi}_{-\pi} f(x) \cos nx dx\,=\sum^{m}_{k=1}\int^{x_k}_{x_{k-1}}f(x)\cos nx dx\,=\sum^{m}_{k=1}\left(f(x_k)\int^{x_k}_{x_{k-1}}\cos nx dx\,+\int^{x_k}_{x_{k-1}}(f(x)-f(x_k))\cos nx\right)dx\,$$ and $$\left| \int^{\pi}_{-\pi}f(x)\cos nx dx\,\right|\le\sum^{m}_{k=1}|f(x_k)|\left|\int^{x_k}_{x_{k-1}}\cos nx dx\,\right|+ \epsilon(\pi-(-\pi))|\le \frac{2}{n}Mm+\epsilon(\pi-(-\pi))$$ where $M=max{f(x)}$ on $[-\pi,\pi]$ Thus $$\lim_{n\to\infty} \int_{-\pi}^{\pi}f(t)\cos(nt)dt=0$$ For second part use $$\cos^2 nt = \frac{1}{2}(1 + \cos 2nt)$$you'll get $$\lim_{n\to\infty}\int_{-\pi}^{\pi} f(t)\cos^2(nt)dt=\frac{1}{2}\int^{\pi}_{-\pi}f(x) dx\,$$