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I have to use lebesgue dominated convergence theorem to prove that

$$ \lim_{n\rightarrow\infty}\int_0^\infty \left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda\ = 5 $$

Do I have to find the function $g$ which bounds all $f_n$? [since it already says to prove using the dominated convergence theorem]

What I have done so far is:

$$\lim_{n\rightarrow\infty}\int_0^\infty \left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda $$ $$ = \int_0^\infty \lim_{n\rightarrow\infty}\left[1+ \frac{\ln(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\frac{1}{n+x}\right)\right] e^{-x/5} \, d\lambda$$

Then,

$$\frac{\ln(x+n^2)}{n^{1/2}}=\frac{\ln\left(n^2\left(1+\frac{x}{n^2}\right)\right)}{n^{1/2}} = \frac{2\ln n +\ln\left(1+\frac{x}{n^2}\right)}{n^{1/2}}$$

I used the maclaurin expansion of $\ln(1+X)$ to prove that $\lim_{n\rightarrow\infty}\frac{\ln(1+\frac{x}{n^2})}{n^{1/2}}=0$

I compared $\lim_{n\rightarrow\infty}\frac{\ln{n}}{n^{1/2}}$ to $\lim_{x\rightarrow\infty}\frac{\ln{x}}{x^{1/2}}$ and used l'Hopital's rule to prove that the limit is zero since I found such an example on youtube.

I don't know whether it is good. Also, when I use maclaurin's expansion of $\cos(X)$ and take $\lim_{n\rightarrow\infty}\cos(\frac{1}{n+x})$, I get it to be $1$.

Then I get $10$ as final answer.

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Is $d\lambda$ a typo? –  littleO Nov 14 '12 at 10:26
    
It is the integral with respect to a measure $\lambda$ –  Bhavish Suarez Nov 14 '12 at 10:55

1 Answer 1

up vote 4 down vote accepted

Your evaluation of the integral assuming the DCT applies is correct (I also arrive at $10$ instead of $5$) so all you have left to do is find a function $g$ such that $f_n(x) \leq g(x)$ for all $n\in \mathbb{N}, x\in (0,\infty)$ and $\int^{\infty}_0 g(x) dx < \infty.$

Since $e^{-x/5}$ decreases quite rapidly, even a relatively weak bound on $$h_n(x) = 1+ \dfrac{\log(x + n^2)}{n^{1/2}}\sin(x^2) + \cos\left(\dfrac{1}{n+x}\right)$$

should be sufficient to ensure the required conditions. The middle term can be estimated $$\dfrac{2\log n + \log(1+ x/n^2)}{\sqrt{n}}< 3+x$$

since $\dfrac{2\log n}{\sqrt{n}} < 3$ for all $n\in \mathbb{N}$ and $\dfrac{\log(1+x/n^2)}{\sqrt{n}}< \dfrac{x/n^2}{\sqrt{n}}< x.$ Thus we can see that $h_n(x)< 5+x$ so $g(x) = (5+x)e^{-x/5}$ is a suitable dominating function.

share|improve this answer
    
thanks a lot for your answer! Then the final answer as 5 is incorrect? –  Bhavish Suarez Nov 14 '12 at 10:53
    
@BhavishSuarez Yes I think so. –  Katie Dobbs Nov 14 '12 at 10:57
    
Thanks again for your help! –  Bhavish Suarez Nov 14 '12 at 10:59

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