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.

Need to prove $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy =0 $ and $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $

Can someone solve using dominated convergence theorem? I want to know how LDC is applied.

share|improve this question
    
Dominated convergence theorem definitely gives you a single-line proof for the first one. Just observe that the integrands are dominated by an integrable function $(1 + y^{2})^{-1}$. But 5xum's idea is much easier to utilize in both of the problems. –  sos440 Apr 22 at 9:18
    
Do you mean I just do Riemann integral? I am confused if Lebesgue is used in this problem. –  user220055 Apr 22 at 9:28
    
@user220055 For continuous functions, there is no difference. –  5xum Apr 22 at 9:32
    
Right, thank you –  user220055 Apr 22 at 9:33

1 Answer 1

up vote 8 down vote accepted

How about this:

$$0<\frac{1}{1+y^2}<1,$$ meaning that $$\frac{1}{1+y^2}e^{-ay} < e^{-ay}$$ and

$$\int _0^\infty\frac{1}{1+y^2}e^{-ay}dy < \int_0^\infty e^{-ay}dy$$

share|improve this answer
    
$\lim\limits_{a \rightarrow + \infty} cos (a)\int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy$ is in another problem. Is it legal to say it is 0? And do I need to prove continuity for the question I asked? –  user220055 Apr 22 at 10:42
    
If $$\lim_{a\to\infty} f(a) = 0$$ and $g$ is a bounded function, then it is easy to see that $$\lim_{a\to\infty} f(a)g(a) = 0.$$ –  5xum Apr 22 at 10:50
    
I kind remember this but was not sure.Thank you. –  user220055 Apr 22 at 10:54
    
@user220055 If you are not sure, I advise you to try and prove the statement, it's not hard. –  5xum Apr 22 at 11:15
    
Could you please show me how LDC works in this question? –  user220055 Apr 22 at 22:16

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.