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Using convergence theorems, I am trying to compute the value of

$$ \lim_{n\to\infty}\int_a^\infty \frac n{1+n^2x^2}\,\mathbb{d}x $$

for $a \in \mathbb{R}$, and with respect to the Lebesgue measure. Firstly I tried using the dominated convergence theorem but it turns out that I can't find a dominating function (since $f_n(0) = n$).

The other theorem I have at hand is the monotone convergence theorem, but it is not clear that we have $f_1(x) \leq f_2(x) \leq \mathbb{ ...}$ , so I don't see how I can apply that one either. Can anyone point me in the right direction?

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    $\begingroup$ Use a substitution before you invoke the convergence theorem(s). $\endgroup$ – Daniel Fischer Dec 4 '14 at 17:18
  • $\begingroup$ Realize that $1+n^2x^2\leq 1+(n+1)^2x^2$ $$\frac{n}{1+n^2x^2}\geq \frac{n+1}{1+(n+1)^2x^2}$$ which gives monotonically decreasing $\endgroup$ – David Dec 4 '14 at 17:32
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    $\begingroup$ @gloom: The function sequence cannot be decreasing because of $f_n (0)=n$. $\endgroup$ – PhoemueX Dec 4 '14 at 17:46
  • $\begingroup$ ya,, its true.. but for all other values it works..@PhoemueX $\endgroup$ – David Dec 4 '14 at 17:49
  • $\begingroup$ So the function sequence is decreasing for almost every x, so I suppose I can apply the monotone convergence theorem. What about the substitution method? I don't see which substitution will help me. Sorry... $\endgroup$ – Tom Offer Dec 4 '14 at 18:08
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The substitution $y = nx$ yields

$$\int_a^\infty \frac{n}{1+n^2x^2}\,dx = \int_{na}^\infty \frac{dy}{1+y^2} = \int_\mathbb{R} \chi_{[na,\infty)}(y)\cdot\frac{1}{1+y^2}\,dy.$$

In that form, both, the dominated convergence theorem and the monotone convergence theorem, can be easily applied to find the limit.

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  • $\begingroup$ Ah, thanks. It was a simpler substitution than I thought! $\endgroup$ – Tom Offer Dec 5 '14 at 11:23
  • $\begingroup$ If we take $f_n(y) = \chi_{[na,\infty)}(y)\cdot\frac{1}{1+y^2}$, then $\lim_{n \to \infty}f_n(y)$ depends on whether $a=0, a<0, a>0$ right? So we get three different answers for the integral (namely $\pi /2, \pi, 0$ respectively)? $\endgroup$ – Tom Offer Dec 5 '14 at 13:30
  • $\begingroup$ Right on all counts. It's not at all obvious before the substitution, is it? $\endgroup$ – Daniel Fischer Dec 5 '14 at 13:32
  • $\begingroup$ Not at all! Cheers. $\endgroup$ – Tom Offer Dec 5 '14 at 13:33

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