Calculate $ \lim _{n \rightarrow \infty }\int ^{n}_{-n} \left(\cos\left(\frac{x}{n}\right)\right)^{n^{2}}\,dx $

Question

Calculate this limit: $$ \lim _{n \rightarrow \infty }\int ^{n}_{-n} \left(\cos\left(\frac{x}{n}\right)\right)^{n^{2}}\,dx .$$

We're given the follow inequality: $$r^{k} \leq \exp(-k(1-r)),$$ with $0\leq r \leq 1$ and $k\in \mathbb{N}.$

I think the the value of the limit is $+\infty$ because I can't find a dominated function. But I'm not sure. Any ideas?

Answer 1

We first prove that for $|x|\leq n$ we then have $$\cos\left(\frac xn\right)^{n^2} \leq e^{-\frac{x^2}2}.$$ This inequality is equivalent to $$\ln \cos\left(\frac xn\right) + \frac{x^2}{2 n^2} \leq 0,\quad \forall x\in [-n,n]. \quad\quad\quad\quad (1)$$ It is enough to prove $(1)$ for $x\in [0,n]$. Let $t = x/n \in [0,1]$, and set $f(t) = \ln\cos t + t^2/2$. We have $f(0) = 0$ and $$f'(t) = -\frac{sin t - t \cos t}{\cos t} \leq 0,$$ since $\sin t - t\cos t \geq 0$ for $t\in [0,\pi/2]$. Since $1 < \pi/2$, then for any $t\in [0,1]$ we have $f(t) \leq f(0) =0$. Hence $(1)$ holds.

Using L'Hopital rule, we easily verify that $$\lim_{t\to 0} \frac{\ln \cos t}{t^2} = -\frac12.$$ Hence for any $x \in \mathbf{R}$, we have $$\lim_{n\to \infty} \cos\left(\frac xn\right)^{n^2} = e^{-\frac{x^2}2}.$$ Denoted $f_n(x) = \cos\left(\frac xn\right)^{n^2} \chi_{[-n,n]}(x)$ then we have $f_n(x) \leq e^{-x^2/2}$ for any $x$ and $\lim_{n\to\infty} f_n(x) = e^{-x^2/2}$ for any $x$. From the Lebesgue dominated convergence theorem, we have $$\lim_{n\to\infty} \int_{-n}^n \cos\left(\frac xn\right)^{n^2} dx = \lim_{n\to \infty} \int_{\mathbf{R}} f_n(x) dx = \int_{\mathbf{R}} e^{-x^2/2} dx = \sqrt{2\pi}.$$

Answer 2

Set $f_n(x)=\chi_{[-n,n]}(x)\cos (\frac{x}{n}) $, $r=\cos (\frac{x}{n})$ and use the fact that $\cos x\leq 1-\frac{x^2}{2}$, $|x|\leq \frac{\pi}{2},$(thus $1-cos x\geq \frac{x^2}{2}$). Take $k=n^2$ and this gives $\cos^{n^2} (\frac{x}{n}) \leq exp(-\frac{x^2}{2})$ and you can apply the dominated convergence theorem.

Answer 3

Let $y=x/n$ which gives $$\int_{-n}^{n}\left(\cos\frac{x}{n}\right)^{n^2}\,\text{d}x = n\cdot\int_{-1}^{1}(\cos y)^{n^2}\,\text{d}y\ .$$ Since $0<\cos y < 1$ when $-1<y<1$ the supplied inequality gives $$\int_{-1}^{1}n\cdot(\cos y)^{n^2}\,\text{d}y \leq \int_{-1}^{1}ne^{-n^2(1-\cos y)}\,\text{d}y = \int_{-1}^{1}ne^{-\frac{1}{2}n^2y^2}\cdot \underbrace{e^{-n^2o(y^2)}}_{\leq 1}\,\text{d}y\leq \int_{-1}^{1}ne^{-\frac{1}{2}n^2y^2}\,\text{d}y = 2ne^{-\frac{1}{2}n^2c^2}\ ,$$ where $o(y^{2})$ denotes a function such that $\lim_{y\to0}\frac{o(y^2)}{y^2} = 0$ and $c\in(-1,1)$ is some (unknown) number; in the final step you invoke the Lagrange Mean value theorem.

For every integer $n>0$ you now know that $$0 \leq \int_{-n}^{n}\left(\cos\frac{x}{n}\right)^{n^2}\,\text{d}x \leq 2ne^{-\frac{1}{2}n^2c^2}$$ and since the positive upper bound can be made as small as you like if you just make sure that $n$ is large enough, this seems to indicate that $$\lim_{n\to\infty} \int_{-n}^{n}\left(\cos\frac{x}{n}\right)^{n^2}\,\text{d}x = 0 \ .$$

Answer 4

$$I=\lim _{n \rightarrow \infty }\int ^{n}_{-n} \left(\cos\left(\frac{x}{n}\right)\right)^{n^{2}}\,dx$$ By continuity $$I=\int^{n}_{-n}\lim _{n \rightarrow \infty }(\cos\frac{x}{n})^{n^{2}}\,dx $$

$$(\cos\frac{x}{n})^{n^{2}}=\left(1+(\cos\frac{x}{n}-1)\right)^{n^{2}}=\left((1+(\cos\frac{x}{n}-1))\right)^{\frac{1}{\cos\frac{x}{n}-1}\cdot (n^2(\cos\frac{x}{n}-1))}$$ Hence $$I=\int ^{n}_{-n} e^{\lim _{n \rightarrow \infty }(n^2(\cos\frac{x}{n}-1))}\,dx$$ Calculation gives $$ n^2(\cos\frac{x}{n}-1)=-\frac{x^2}{2}+\frac{x^4}{24n^2}+…..\Rightarrow \lim _{n \rightarrow \infty }(n^2(\cos\frac{x}{n}-1)=-\frac{x^2}{2}$$ Hence $$I=\int ^{n}_{-n} e^{-\frac{-x^2}{2}}\,dx$$ As we know, this integral can not be calculated by elementary methods. Consulting Wolfram one has $$I=\int ^{n}_{-n} e^{-\frac{-x^2}{2}}\,dx=\sqrt{2\pi}\text{erf}(\frac{n}{\sqrt 2})$$ where erf denotes the error function defined by $$f(z)=\frac{2}{\sqrt{\pi}}\int^{z}_{0}e^{-t^2}\,dt$$ Thus $$\color{red}{I={\sqrt{2\pi}}}$$