Evaluating $ \lim_{n\rightarrow\infty} \int_a^\infty \frac{n}{1+n^2x^2}dx$.

Question

Evaluate$$ \lim_{n\rightarrow\infty} \int_a^\infty \frac{n}{1+n^2x^2}dx$$ where $a \in \mathbb R$.

I know for $a>0$, we can take the limit inside of the integral to get the answer 0 by checking the conditions for the standard theorem on "putting the limit inside".(integrable majorant $1/x^2$ etc...)

However for the cases $a=0$ and $a<0$, the conditions break down, namely the integrand blows up as $n \rightarrow \infty$ for $x=0$.

How can I resolve this problem?

Any hints is appreciated.

Answer 1

Letting $u = nx$ , we have

$$\lim_{n \to \infty} \int_{an}^\infty \frac{1}{1+u^2} = \lim_{n \to \infty} \int_{-\infty}^\infty \mathbb{1}_{[an,\infty]}\frac{1}{1+u^2} $$ Where $\mathbb{1}_A$ is the characteristic function on $A$. The last integral is bounded by $\pi$, so we may apply the dominated convergence theorem and check the cases $a < 0$ , $a > 0$, and $a = 0$

Answer 2

By change of variables, you have \begin{align} \int^\infty_a \frac{n}{1+n^2x^2}\ dx=\int^\infty_{na} \frac{du}{1+u^2} \end{align}

Answer 3

It's really not necessary to do too much work with interchanging limits, since the integral can be explicitly computed. After all,

$$\int \frac{n}{1 + n^2 x^2} \, dx = \int \frac{d(nx)}{1 + (nx)^2} = \arctan(nx)$$

so that

$$\int_a^{\infty} \frac{n}{1 + n^2 x^2} dx = \frac{\pi}{2} - \arctan(an).$$

Now there are three cases, each easily handled.