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

Given that $f\colon [-1,1] \to \mathbb{R}$ is a continual function such that $ \int_{-1}^{1} f(t) \ dt =1$, just how do I review the restriction of this indispensable : $$\lim_{n \to \infty} \int_{-1}^{1} f(t) \cos^{2}{nt} \,dt$$

What I did was to write $\cos^{2}{nt} = \frac{1+\cos{2nt}}{2}$ and also replace it in the indispensable to make sure that I can take advantage of the offered theory of $\int_{-1}^{1} f(t) \ dt =1$. So the indispensable comes to be,

\begin{align*} \int_{-1}^{1} f(t)\cos^{2}{nt} \ dt = \int_{-1}^{1} f(t) \biggl[\frac{1+\cos{2nt}}{2}\biggr] \ dt & \\ \hspace{3cm} = \frac{1}{2}\int_{-1}^{1}f(t) \ dt + \int_{-1}^{1} \frac{f(t)\cos{2nt}}{2} \ dt \end{align*}

But I do not actually recognize just how I can review the 2nd indispensable as well as additionally I can not understand regarding why that indispensable problem on $f$ has actually been thought. In addition without thinking that problem on $f$ is it feasible to review this indispensable? If of course, after that what would certainly the solution be?

2019-12-02 02:47:53
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Answers: 2

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

2019-12-03 03:30:56

\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 make use of $\sin^2 nt = \cos^2(nt + \pi/2)$ and also some adjustment 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}

2019-12-03 03:30:18