Real Analysis, Lebesgue, limit of an integral

[purpleehobbit]

I am absolutely lost. I had to take Advanced Calculus as independent study in a one month class and this book has very few examples, if any. I'm not even sure where to start on this one.

I have to compute the limit of an integral and then justify my methods according to the Lebesgue theory. I can also use Fubini's Theorem.

I have not started a solution on this yet because I don't know where to start. So perhaps I can just post the problem and then you can lead me somewhere where I might be able to see a similar example as to not violate the rules? I know I'm extremely new here but I'm not looking for someone to do it for me, I'm just having a horrible time finding a concrete example.

Thanks in advance, any help is appreciated.

(sorry, I don't know how to do the fancy equation editors)

Homework Statement

Compute the limit as n approaches infinity of the integral (from 1 to 2) of {x^(2-(sin nx)/n) dx}

 

[micromass]

So you have a sequence of functions (fn), and you want to find

$$\lim_{n\rightarrow+\infty}{\int{f_n}}$$

What you want to do of course is exchange limit and integral, thus

$$\lim_{n\rightarrow+\infty}{\int{f_n}}=\int{\lim_{n\rightarrow +\infty}{f_n}}$$

but we cannot always do that. The good thing about Lebesgue integrals is that there are theorems that do alow us to do this (in certain cases): these are the monotone convergence theorem and the dominated convergence theorem. So I suggest you use this.

Let me give you an easy example: consider $$f_n(x)=x/n$$. We want to calculate

$$\lim_{n\rightarrow +\infty}{\int_0^1\frac{x}{n}dx}$$

naively interchanging limit and integrals gives us:

$$\int_0^1{\lim_{n\rightarrow +\infty}{\frac{x}{n}}dx}=\int_0^1{0dx}=0$$

But can we interchange limit and integral. Yes: by applying the dominated convergence theorem. Since on [0,1], we have $$\left|\frac{x}{n}\right|\leq 1$$. So 1 is a dominating function, and we have
$$\int_0^1{1dx}=1$$

Thus 1 is also integrable. This implies that the dominated convergence theorem is applicable!

 

[purpleehobbit]

micromass,

Thanks for your quick response.

I think I'm starting to see some understanding.

So I need to look at my fn. And as n approaches infinity, (sin nx)/n will approach 0 and I would be left with just x^2.

For the dominated convergence, I would have |x^2| <= 4 on [1,2]

And from there I take the integral from 1 to 2 of x^2.

Am I on the right track?

Thanks again for getting back so quickly.

 

[micromass]

micromass,

Thanks for your quick response.

I think I'm starting to see some understanding.

So I need to look at my fn. And as n approaches infinity, (sin nx)/n will approach 0 and I would be left with just x^2.

This is already correct, so you have found that

$$\lim_{n\rightarrow +\infty}{\int_1^2{x^{2-\sin(nx)/n}dx}}=\int_1^2{x^2dx}$$

IF the dominated convergence theorem applies! So, we must check now that the dominated convergence theorem applies. For this, you need to find an integrable function f such that

$$|f_n|\leq f$$ on [1,2]

What you have done is this:

For the dominated convergence, I would have |x^2| <= 4 on [1,2]

This is not sufficient! You'll need to find a function f such that

$$|x^{2-\sin(nx)/n}}|\leq f(x)$$ on [1,2]

So our function f must be independent of n! Furthermore, this function must be integrable!

 

[purpleehobbit]

Ok, I see what you mean.

Thanks.

 

[dirk_mec1]

This is already correct, so you have found that
So our function f must be independent of n! Furthermore, this function must be integrable!

Did you meant n or x?