# Finding a dominating function for this sequence of functions

Problem:

Find the limit $$\lim_{n\to\infty} \int_0^n \left( 1 + \frac{x}{n}\right )^{-n} \log(2 + \cos(x/n))dx$$ and justify your reasoning.

My Solution:

Let $f_n = \left( 1 + \frac{x}{n}\right )^{-n} \log(2 + \cos(x/n))\chi_{(0,n)}$. To find this limit, we use the Dominated Convergence Theorem where our dominating function is $g(x) = 2e^{-x}\chi_{(0,\infty)}$.

First, we show that $|f_n| \leq g$. Notice that by the first derivative test, $h(t) = \left(1+\frac{x}{t}\right)^{-t}$ is increasing for $x,t>0$. So this part of our sequence of functions is increasing and has limit $e^{-x}$. Thus, $$\left|\left( 1+\frac{x}{n}\right)^{-n}\right| \leq e^{-x}.$$ Also, since $|\cos(x/n)| \leq 1$ and $\log$ is increasing, we know $|\log(2 + \cos(x/n))|\leq \log(3) \leq 2$. So, $$|f_n| = \Big|\left( 1 + \frac{x}{n}\right )^{-n}\Big|\cdot\Big|\log(2 + \cos(x/n)) \Big|\cdot \Big| \chi_{(0,n)}\Big| \leq 2e^{-x}\chi_{(0,\infty)}.$$ Now we prove that $g$ is integrable: $$\int 2e^{-x}\chi_{(0,\infty)} dx = \sum_{k=1}^\infty \int 2e^{-x}\chi_{(k-1,k)}dx \leq 2e\sum_{k=1}^\infty \frac{1}{e^k} \leq \infty,$$ where the first equality holds by the Monotone Convergence Theorem, and thhe last inequality holds because of geometric series.

Now that all conditions are met, we can apply the Dominated Convergence Theorem: \begin{align*} \lim_{n\to\infty} \int_0^n \left( 1 + \frac{x}{n}\right )^{-n} \log(2 + \cos(x/n))dx &= \int\lim_{n\to\infty}\left[\left( 1 + \frac{x}{n}\right )^{-n} \log(2 + \cos(x/n))\cdot \chi_{(0,n)}\right]dx\\ &=\int e^{-x}\log(3)\chi_{(0,\infty)}dx\\ &= \log(3)\int_0^\infty e^{-x} dx\\ &= \log(3). \end{align*}

My Questions:

At this point in the book I'm using, we have not been told that Lebesgue integrable implies Riemann integrable. In fact, we don't really have any methods for evaluating $\int_0^\infty e^{-x} dx$. Is there a (reasonable) way to evaluate this without using facts about improper Riemann integrals?

I'm curious if I'm taking the wrong approach to these types of problems. I don't feel comfortable switching between Riemann and Lebesgue integration. I've looked around here to see what conditions have to be true of an improper Riemann integral to be equivalent to its Lebesgue counter part, but couldn't find much.

Sorry about the length of this post.

• See this <math.stackexchange.com/questions/67198/…> – coffeebelly Jul 10 '14 at 0:03
• Lebesgue integrable does not imply Riemann integrable. To evaluate $\int_0^{\infty} e^{-x}$, use the monotone convergence theorem: $$\int_0^{\infty} e^{-x} = \lim_{n \to \infty} \int_0^n e^{-x}$$ – user61527 Jul 10 '14 at 0:07
• The bound $\left|\left( 1+\frac{x}{n}\right)^{-n}\right| \leq e^{-x}$ is incorrect. You might want to double check your argument. – user940 Jul 10 '14 at 0:07
• Thanks Byron. T.Bongers, are you saying to use this since, on a bounded interval, Riemann and Lebesgue integrals agree? – dannum Jul 10 '14 at 0:36
• It isn't positive and when integrated its positive and negative parts cancel each other just right for the improper Riemann integral to converge. But for the Lebesgue integral to exist the absolute value also has to be integrable, so no cancellations. – Conifold Jul 10 '14 at 1:15

You could first perform the change of variables $w=\left(1+{x\over n}\right)^{-1}$ $$\int^n_0\left(1+{x\over n}\right)^{-n} \, \log(2+\cos(x/n))\,dx=n\int_{1/2}^1w^{n-2} \log\left(2+\cos\left({1-w\over w}\right)\right)\,dw,$$ followed by the result here to show that the limit is $\log(3)$.