Calculating a limit as a consequence of the Dominated Convergence Theorem

Question

Let $\lambda$ denote the Lebesgue measure. I have to calculate for all $a>0$ this: $$\lim_{n \rightarrow \infty} \int_a^n \frac{n^2 x e^{-n^2x^2}}{1+x^2} d\lambda(x)$$ as an excercise of my Measure Theory lessons.

As an instruction, the excercise says that it follows by delimiting the sequence below the integral and applying the Dominated Convergence Theorem. In order to do this, I have tried to use the inequality $\forall y \in \Bbb{R} : 1 + y \leq e^y$ but I am not reaching any point and I have no more ideas.

Any advice? Thanks in advance :)

Answer 1

For a positive $t$, one has $e^t\geqslant t $ hence $$ \frac{n^2 x e^{-n^2x^2}}{1+x^2} \mathbf{1}_{[a,n]}(x)\leqslant \frac{1}{1+x^2}n^2(n^2x^2)^{-1}\mathbf{1}_{[a,n]}(x)\leqslant \frac{1}{x^2(1+x^2)}\mathbf{1}_{[a,\infty)}(x). $$

Answer 2

$\textbf{Hint:}$ For $a > \frac{1}{n\sqrt{2}}$ and $n > \frac{1}{2}$ we have $$\frac{n^2xe^{-n^2x^2}}{1+x^2} < \frac{n^2ae^{-n^2a^2}}{1+x^2} < \frac{1}{a^2e}\frac{1}{1+x^2}$$

which is integrable on $[a,\infty)$

Answer 3

The limit can also be obtained as an application of the following version of dominated convergence:

If $|f_n|\leq g_n$, $(g_n:n\in\mathbb{N})\subset L_1$, $f_n\xrightarrow{n\rightarrow\infty}f$, $g_n\xrightarrow{n\rightarrow\infty}g$ and $\lim_n\int g_n=\int g$, then $$\lim_n\int f_n=\int f$$

In the OP's example $$0\leq I_n =\int^n_a\frac{n^2xe^{-(nx)^2}}{1+x^2}\,dx\leq\int^n_a n^2xe^{-(nx)^2}\,dx=-\frac{n}2e^{-(nx)^2}\Big|^n_a=\frac{ne^{-a^2n^2}-ne^{-n^4}}{2}\xrightarrow{n\rightarrow\infty}0$$