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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!

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1 Answer 1

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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 $$\lim_{n\to\infty}\int_1^2x^{2-\frac{\sin(nx)}{n}}dx=\int_1^2\lim_{n\to\infty}x^{2-\frac{\sin(nx)}{n}}dx=\int_1^2x^{2-\lim_{n\to\infty}\frac{\sin(nx)}{n}}dx =\int_1^2x^{2}dx=\frac{2^3-1}{3}=\frac{7}{3}.$$

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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

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