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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.

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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. – Sangchul Lee Apr 22 '14 at 9:18
Do you mean I just do Riemann integral? I am confused if Lebesgue is used in this problem. – user220055 Apr 22 '14 at 9:28
@user220055 For continuous functions, there is no difference. – 5xum Apr 22 '14 at 9:32
Right, thank you – user220055 Apr 22 '14 at 9:33
up vote 9 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

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

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$\lim\limits_{a \rightarrow + \infty} cos (a)\int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy$ 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 '14 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 '14 at 10:50
I kind remember this but was not sure.Thank you. – user220055 Apr 22 '14 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 '14 at 11:15
Could you please show me how LDC works in this question? – user220055 Apr 22 '14 at 22:16

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