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Prove the convergence of the following integral: $$ \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx $$

I tried to use the obvious Weierstrass comparison test in the following way:

$$ \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx \leqslant \lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2}dx = \lim\limits_{n\to\infty}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{nx^2}dx $$ However after thinking about a variable substitution I concluded that it would complicate the expression.

Question:

How should I prove the integral converges? What are your suggestions?

Edit: Tried a new approach but I would like someone to check it out. $\lim\limits_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx=\lim\limits_{n\to\infty}\frac{2}{n}\int_\limits{0}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx$ since the function is even. Substitution $\sqrt{n}x=y$ So we get after some adjustments: $\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{1-e^{-y^2}}{y^2(1+y^2)}dy$ Then I am not sure about the following step: $1-e^{-y^2}\sim y^2$ So we would get $\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{{y^2}}{y^2(1+y^2)}dy=$$\lim\limits_{n\to\infty}\frac{2}{\sqrt{n}}\int_\limits{0}^{n\sqrt{n}}\frac{1}{(1+y^2)}dy=0$

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  • $\begingroup$ I think it works by using the dominated convergence theorem, by considering the function : $\frac{1-e^{-nx^2}} {nx^2(1+nx^2)}1_{[-n;n]}$ which tends almost everywhere to 0 and which is dominated by a function on $L^1(\mathbb{R})$. $\endgroup$ – ChocoSavour Apr 22 '18 at 11:45
  • $\begingroup$ @ChocoSavour I am sorry but I am not supposed to use the Dominated Convergence Theorem. Anyway thanks for your input! $\endgroup$ – Pedro Gomes Apr 22 '18 at 12:24
  • $\begingroup$ @ChocoSavour Have you got any opinion about my edit? If you have one feel free to comment. $\endgroup$ – Pedro Gomes Apr 22 '18 at 12:25
  • $\begingroup$ In my opinion, there is two things : first, using asymptotic analysis is useful to determine if the integral converges or not, but not to determine the limit, cause if $f \sim g$ (and $f$, $g$ positive), I don't think that it permits to say that if the integral converges, then it converges to the same limit. However, here we have $1-e^{-y^2} \leq y^2$, so we have : $| \frac{2}{\sqrt{n}} \int_{0}^{n\sqrt{n}} \frac{1-e^{-y^2}}{y^2(1+y^2} \, \mathrm{d}y | \leq \frac{2}{\sqrt{n}} \int_{0}^{n\sqrt{n}} \frac{y^2}{y^2(1+y^2)} \, \mathrm{d}y$, and from here we can conclude (the second one tends to 0). $\endgroup$ – ChocoSavour Apr 22 '18 at 13:08
  • $\begingroup$ @ChocoSavour I guess you can say that $\lim_{n\to\infty}\frac{1}{n}\int_\limits{-n}^{n}\frac{1-e^{-nx^2}}{x^2(1+nx^2)}dx=0$ once $ \frac{1-e^{-nx^2}}{x^2(1+nx^2)}>0$. Since it cannot be negative and the original integral is lower than $0$ according to my resolution we could state it equals $0$. $\endgroup$ – Pedro Gomes Apr 22 '18 at 13:50
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As in the comments indicated, We can just use the change of variables $x \sqrt{n} =y$ to get $$ \frac{1}{n} \int_{-n}^n \frac{1-\exp(-nx^2)}{x^2(1+nx^2)} =\frac{2}{\sqrt{n}} \int_0^{n^{3/2}} \frac{1-\exp(-x^2)}{x^2(1+x^2)} \mathrm{d} x. $$ Noting that $$0 \leq 1- \exp(-x^2) = \int_{0}^{x^2} \exp(-t) \, \mathrm{d} t \leq \min(x^2, \int_0^\infty \exp(-t) \mathrm{d} t) = \min(x^2,1)$$ we see that the last integral is bounded by $$\int_0^\infty \frac{\min(1,x^2)}{x^2(1+x^2)} \leq \int_0^1 \frac{1}{1+x^2} \mathrm{d} x+ \int_1^\infty \frac{1}{x^2(1+x^2)} < \infty.$$ Thus, we can conclude that the limes is $0$ and the convergence rate is $1/\sqrt{n}$.

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