Finding a dominating function to apply dominated convergence theorem on $\lim_{n\to \infty } \int_0^n (1-\frac{x}{n})^n \cos(\frac{x}{n})dx$

Question

Evaluate: $\lim_{n\to \infty } \int_0^n (1-\frac{x}{n})^n \cos(\frac{x}{n})dx$

For fixed $n$, I first rewrite my integral as $$\int_0^n \left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right)dx = \int_0^\infty \left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right) \chi_{[0,n]}(x) dx,$$ where $\chi_{[0,n]}(x)$ is the characteristic function.

We then have that $$\lim_{n\to \infty}\left(1-\frac{x}{n} \right)^n = e^{-x},$$ $$\lim_{n\to \infty}\cos\left(\frac{x}{n}\right) = 1,$$ $$\lim_{n\to \infty}\chi_{[0,n]}(x) = 1.$$

So that $$\lim_{n\to \infty}\left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right) \chi_{[0,n]}(x) = e^{-x}.$$

I'd now like to just quote the dominated convergence theorem and integrate $e^{-x}$ to get the limit of the integral. But I can't think of an integrable function that dominates the sequence of functions.

Any thoughts would be greatly appreciated.

Thanks in advance.

Answer 1

You can see that $\int_{0}^n (1-\frac{x}{n})^n\cos{(\frac{x}{n})}dx = \int_{0}^{1}n(1-t)^{n}\cos{(t)}dt$ and you can integrate by parts

Answer 2

$$ 0\leq 1-\frac{x}{n}\leq e^{-x/n}\qquad 0<x\leq n$$

and so $0\leq \Big(1-\frac{x}{n}\Big)^n\mathbb{1}_{(0,n]}(x)\leq e^{-x}$. The function $x\mapsto\cos(x/n)$ is controlled easily since $|\cos|\leq 1$.

By dominated convergence $$\lim_n\int^n_0\Big(1-\frac{x}{n}\Big)^n\cos(x/n)\,dx\xrightarrow{n\rightarrow\infty}\int^\infty_0 e^{-x}\,dx=1$$