Why is this limit integrable?

Question

In the book Real Analysis by Bass, it asks to evaluate the Lebesgue integral

$$ \lim_{n\to\infty} \int_{0}^{n} \bigg(1 - \frac{x}{n} \bigg)^{n} \log(2+\cos(x/n)) \ \mathrm{d}x $$

I assume I should use one of the convergence theorems.

Define the integrand as $f_n(x)$.

Is it possible to evaluate this even if $f_n(x)$ is not defined for all $x \in \mathbb{R}$ for some values of $n \in \mathbb{R}$? To be specific, $f_n(x)$ is defined for all $x\in\mathbb{R}$, and the integral as well, if $n \in \mathbb{N}$. Is this enough for the limit to be integrable? Another thing that confuses me is that the upper limit of the integral is $n$, should I define the sequence of functions $g_n = \int_{0}^{n} \left(1 - \frac{x}{n} \right)^{n} \log(2+\cos(x/n)) \, \mathrm{d}x$ for each positive integer $n$?

My edit:

What I mean is that the integrand is real-valued on $(0,n)$ but I don't know how to tackle cases where the integration interval varies with the limit.

Answer 1

$HINT$

$1_{[0,n]}\leq 1, \forall n \in \Bbb{N}$

Note that $|\log{(2+\cos{(x/n)})}| \leq \log{3}$

Also since $1-x \leq e^{-x}$ we have that $$(1-\frac{x}{n})^ \leq e^{-\frac{x}{n}}$$

So by m DCT you have the limit.

Answer 2

$0 \leq I_{(0,n)}(x) (1-\frac xn)^{n} \log(2+cos (\frac x n))\leq (\log 3)e^{-x}$ and $e^{-x}$ is integrable on $(0,\infty)$. So we can apply DCT.

Answer 3

Here is another line of argument: Using the substitution $y = \frac{x}{n}$, the integral inside the limit reduces to

$$ \int_{0}^{1} n (1 - y) ^n \log(2+\cos y) \, \mathrm{d}y. $$

Now the following theorem kicks in:

Theorem. Let $\phi : [0, 1] \to \mathbb{R}$ be bounded and continuous at $0$. Then $$ \lim_{n\to\infty} \int_{0}^{1} n (1-y)^{n-1} \phi(y) \, \mathrm{d}y = \phi(0). $$

In our case, we can set $\phi(y) = (1-y) \log(2 + \cos y)$ to find out that the limit is $\phi(0) = \log 3$.

Proof of Theorem. For any $\epsilon > 0$, there exists $\delta \in (0, 1)$ such that $|\phi(y) - \phi(0)| < \epsilon$ whenever $|y| < \delta$. Also, let $M > 0$ be a bound of $\phi$. Then

\begin{align*} \left| \int_{0}^{1} n (1-y)^{n-1} \phi(y) \, \mathrm{d}y - \phi(0) \right| &\leq \int_{0}^{1} n (1-y)^{n-1} |\phi(y) - \phi(0)| \, \mathrm{d}y \\ &\leq 2M \int_{\delta}^{1} n (1-y)^{n-1} \, \mathrm{d}y + \epsilon \int_{0}^{\delta} n (1-y)^{n-1} \, \mathrm{d}y \\ &\leq 2M(1-\delta)^n + \epsilon, \end{align*}

and so,

$$ \limsup_{n\to\infty} \left| \int_{0}^{1} n (1-y)^{n-1} \phi(y) \, \mathrm{d}y - \phi(0) \right| \leq \epsilon. $$

But since the left-hand side is independent of $\epsilon > 0$, we can let $\epsilon \downarrow 0$ to find that this limsup is zero, which implies the desired claim. $\square$