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Calculate $$\lim\limits_{n\to\infty}\int_a^{\infty}\frac{n}{1+n^2x^2}\,d\mu$$ where $\mu$ is the Lebesgue measure and $a\geq0$.

First is easy to see that $\arctan(nx) ' = \frac{n}{1+n^2x^2}$ so I concluded that the limit is $\begin{cases}0 &\text{if $a>0$}\\\pi/2&\text{if $a=0$}\end{cases}$.

But the idea is to use the dominated convergence theorem, so if $a>0$ is easy to see that $\frac{n}{1+n^2x^2} < \frac{1}{x^2}$, so we conclude that the limit is $0$. My problem is when $a=0$. I integrate by parts getting

$$\int_0^{\infty}\frac{n}{1+n^2x^2} = \int_0^{\infty}\frac{2n^3x^2}{(1+n^2x^2)^2}$$

On the right side separate the domain on $[0,1]\cup(1,\infty)$ y I demonstrate that $\frac{2n^3x^2}{(1+n^2x^2)^2} < \frac{1}{x}\in L^1([0,1],\mu)$ and is easy to notice that $\frac{2n^3x^2}{(1+n^2x^2)^2} < \frac{2}{x^2}\in L^1([1,\infty),\mu)$, but the dominated convergence theorem implies that the limit is $0$, not $\pi/2$ as I concluded before.

Where is my mistake?

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    $\begingroup$ $\frac1x \notin L^1([0,1],\mu)$. $\endgroup$ – Daniel Fischer Oct 31 '13 at 17:02
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Do a change of variables.
$$\int_a^\infty{n\,dx\over 1 + n^2x^2} = \int_{na}^\infty {n\cdot (dx/n) \over 1 + x^2} = \int_{na}^\infty {dx\over 1 + x^2} = {\pi\over 2} - \arctan(na). $$ The last term goes to $0$ at $\infty$. If you wish to use dominated convergence, it applies in the integral just before the final evaluation.

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by using fatou lemma and dividing by 2 part I think integral=0

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Here is some $f_n(x)$

I think $f_n(x)\to f$ uniformly in (0,infinity)

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  • $\begingroup$ No, but it goes uniformly on intervals of the form $[b,\infty), where $b > 0$. $\endgroup$ – ncmathsadist Nov 1 '13 at 1:19

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