Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across the following problem in my self-study, and wanted to know how to use Lebesgue Dominated Convergence to compute any of the following limits:

(a) $\lim\limits_{n \rightarrow \infty}$ $\int_0^\infty$ $(1+(x/n))^{-n} \sin (x/n)dx$

(b) $\lim\limits_{n \rightarrow \infty}$ $\int_0^1$ $(1+nx^{2})(1+x^2)^{-n}dx$

(c) $\lim\limits_{n \rightarrow \infty}$ $\int_0^\infty$ $n \sin (x/n) [x(1+x^2)]^{-1}dx$

Any help is greatly appreciated.

Update: I think I have successfully worked out arguments for each of (a) and (b), so I am less concerned about answers/strategies to those parts. However, (c) seems more tricky than the others, so if anyone visiting today sees how to handle (c) (in particular, a nice-enough dominating function!), let me know as it would be greatly appreciated.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

Hint to (c):

  • What can we say about the function $$g(t)=\frac{\sin t}{t}\qquad\text{?}$$

  • What can we say about the function $$h(x)=\frac{1}{1+x^2}\qquad\text{?}$$

Edit: Adding an extra hint... $$n\sin(x/n)[x(1+x^2)]^{-1}=g(x/n)h(x).$$

share|improve this answer

I'm assuming you are using Folland in your study.

(a) Let $f_n(x)= (1+(x/n))^{-n}\sin(x/n)$ and $g_n(x)= (1+(x/n))^{-n}$. Observe that $\lim_{n\to\infty}(1+(x/n))^{-n}=e^{-x}$ and $\lim_{n\to\infty} f_n(x)=0$. Since $|f_n(x)|\leq g_n(x)$ and it is easy to see that $\int g_n\to 1$, then by LDCT (the version of exercise 21 chapter 2) it follows that $\int f_n\to 0$.

share|improve this answer
    
Since the sequence is decreasing, your inequality won't be true. –  Robert Israel Feb 12 '12 at 6:16
    
You right, It is not as straightforward as I though. I'll edit with more details. –  azarel Feb 12 '12 at 6:45
    
You can use $(1+x/2)^{-2}$ as your bound. –  Robert Israel Feb 12 '12 at 6:50
    
@azarel How can you apply dominated convergence theorem on $g_n(x)$ to show that $\int g_n(x) \to 1$? It is not true that: $$\dfrac{1}{\left(1+\frac{x}{n}\right)^n} \le \dfrac{1}{e^x}$$ hence which function do you take to bound $g_n(x)?$ –  Ale Jun 5 at 12:54

For part (c), you can use the following inequality (which is not hard to verify) $$ |\sin(x)| \leq x\quad \forall x\in[0,\infty) $$ Then, we can get a dominating function as follows $$ \left|\frac{n\sin(x/n)}{x(1+x^2)}\right| \leq \frac{n(x/n)}{x(1+x^2)} = \frac{1}{1+x^2} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.