Evaluating the integral $\mathop{\lim}\limits_{n \to \infty} \int_{-1}^{1} f(t)\cos^{2}(nt) \ dt $

Solution 1:

Define $g(t):[-\pi,\pi]\to \mathbb{R}$ such that $g(t)=f(t)$ for $t\in [-1,1]$ and $g(t)=0$ elsewhere. Then $g$ is integrable and by Lebesgue-Riemann lemma $$\lim_{n\to\infty}\int_{-\pi}^\pi g(t)\cos 2nt dt=0.$$ But this is just the same as $$\lim_{n\to\infty}\int_{-1}^1 f(t)\cos 2nt dt=0.$$

Solution 2:

\begin{equation} \int_{-1}^1 f(t) \cos^2 nt dt = 1 - \int_{-1}^1 f(t) \sin^2 nt dt \end{equation}

You can use $\sin^2 nt = \cos^2(nt + \pi/2)$ and some manipulation to show that

\begin{equation} \lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \lim_{n \to \infty} \int_{-1}^1 f(t) \sin^2 nt dt \end{equation}

Combining these, you get

\begin{equation} \lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \frac{1}{2} \end{equation}