# The Lebesgue Theory basic Application , get stuck

Ok, I am working on a very easy question but I get stuck when I trying to justify my answer.

I know that, in order to use Lebesgue's dominated Convergence Theorem, there are two conditions that we need to satisfy:

First, ${\displaystyle \ f_n}$ need to be functions on $\mathbb{R}$ converging pointwise to a limit${\displaystyle \ f}$

Second, there must be a function ${\displaystyle \ g}$ with finite integral such that each |${\displaystyle \ f_n}$| ${\leqslant}$${\displaystyle \ g}, then: {\displaystyle \int \lim f_n\,dx}={\lim\displaystyle \int f_n\,dx} Now, I need to compute {\displaystyle \lim _{n\rightarrow\infty}\int_1^2 \! {x^{2-\sin(nx)/n}} \, \mathrm{d} x}. I am almost positive that it's the integration of {\displaystyle \ {x^2}} over the interval [1,2]. My question is, how do I justify this? I tried to prove it is pointwise convergence but failed to find a good N to do the job. And again, I apologize if this is a too-easy question for most of folks here, but I'd appreciate if you can help! - ## 1 Answer Since |\sin(x)|\leq 1 for all x, we have$$0\leq2-\frac{\sin(nx)}{n}\leq 2+\frac{|\sin(nx)|}{n}\leq 3\mbox{ for all }x\in [1,2]\mbox{ and }n\geq 1.$$This implies that$$\Big|x^{2-\frac{\sin(nx)}{n}}\Big|=x^{2-\frac{\sin(nx)}{n}}\leq x^3\mbox{ for all }x\in [1,2].$$Note that x^3 is integrable on [1,2] since \int_1^2x^3dx=\frac{2^4-1}{4}<\infty. Therefore, we can apply Lebesgue's dominated Convergence Theorem to conclude that$$\tag{1}\lim_{n\to\infty}\int_1^2x^{2-\frac{\sin(nx)}{n}}dx=\int_1^2\lim_{n\to\infty}x^{2-\frac{\sin(nx)}{n}}dx.$$Once again, since |\sin(x)|\leq 1 for all x, we have$$0\leq\lim_{n\to\infty}\frac{|\sin(nx)|}{n}\leq\lim_{n\to\infty}\frac{1}{n}=0$$which implies that$$\tag{2}\lim_{n\to\infty}\frac{\sin(nx)}{n}=0\mbox{ for all }x\in[1,2].$$Putting (1) and (2) together, we obtain$$\lim_{n\to\infty}\int_1^2x^{2-\frac{\sin(nx)}{n}}dx=\int_1^2\lim_{n\to\infty}x^{2-\frac{\sin(nx)}{n}}dx=\int_1^2x^{2-\lim_{n\to\infty}\frac{\sin(nx)}{n}}dx =\int_1^2x^{2}dx=\frac{2^3-1}{3}=\frac{7}{3}.$\$

-
Ugh! People are smart! I worked with sin(nx)/n directly and complaining the whole thing is too complicated to compute... But use x^3 is not something quite exceptional; there are "tricks" that I need to get familiar with. Haha... Thanks a lot @Paul, and also for these edits! My first time use Latex, much more to learn :) –  user48601 Feb 1 '13 at 1:42
You did a good job even it's your first time using Latex. Just keep practicing and you will get familiar with it. –  Paul Feb 1 '13 at 1:58