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Show that $\int_{-\infty}^{\infty}(1+\frac{x^2}{n})^{-n}dx\rightarrow \int_{-\infty}^{\infty} e^{-x^2}dx$

I thought the best way to go about this is to show uniform convergence of $f_n(x)=(1+\frac{x^2}{n})^{-n}$ to $f(x)=e^{-x^2}$, which would justify swapping the limit and integral (I think? This would hold on a bounded interval, not so sure about this). Dini's theorem would work had it been a closed interval, but since we're after uniform convergence in all of $\Bbb R$ I can't seem to find a way to work this out.

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    $\begingroup$ I'm not sure that even uniform convergence would help, given we're not integrating over a bounded interval (I could be wrong though). $\endgroup$ – Theo Bendit Jul 1 '18 at 15:11
  • $\begingroup$ Yeah, I'm not too sure about this either now that I think of it. $\endgroup$ – Noa Jul 1 '18 at 15:12
  • $\begingroup$ It's not something I'm particularly comfortable with, but the Dominated Convergence Theorem probably applies, as the sequence $(1 + x^2 / n)^{-n}$ is monotone increasing, dominated by the function $e^{-x^2}$ (which I think is Lebesgue integrable). $\endgroup$ – Theo Bendit Jul 1 '18 at 15:18
  • $\begingroup$ @ComplexYetTrivial Yeah, that's a typo. Thanks $\endgroup$ – Noa Jul 1 '18 at 15:19
  • $\begingroup$ @TheoBendit The sequence is actually decreasing and we have $\mathrm{e}^{-x^2} \leq f_{n+1} (x) \leq f_n (x)$ for $n \in \mathbb{N}$ and $x \in \mathbb{R}$, but we can take $f_1$ as the dominating function and then apply the dominated convergence theorem. $\endgroup$ – ComplexYetTrivial Jul 1 '18 at 15:51
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As discussed in the comments you can apply the dominated convergence theorem to prove the convergence of the integrals. Here's an alternative solution using uniform convergence:

Let $f(x) = \mathrm{e}^{-x^2} \, , \, x \in \mathbb{R}$ . For $R \geq 1$ and $n \in \mathbb{N}$ we have \begin{align} \left|~ \int \limits_{\mathbb{R}\setminus [-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| &\leq 2 \int \limits_R^\infty \left[\mathrm{e}^{-x^2} + \frac{1}{\left(1+\frac{x^2}{n}\right)^n}\right] \, \mathrm{d} x \\ &\leq 2 \int \limits_R^\infty \left[\mathrm{e}^{-x} + \frac{1}{x^2}\right] \, \mathrm{d} x \\ &= 2 \left(\mathrm{e}^{-R}+\frac{1}{R}\right) \, . \end{align} Now let $\varepsilon > 0$. Choose $R \geq 1$ such that $2(\mathrm{e}^{-R} + \frac{1}{R}) < \frac{\varepsilon}{2}$ . Since $[-R,R]$ is compact, we can use Dini's theorem to conclude that $f_n \rightarrow f$ uniformly on $[-R,R]$ as $n \rightarrow \infty$ . But then the sequence of integrals over this interval converges as well, so we can find an $N \in \mathbb{N}$ such that $$ \left|~ \int \limits_{[-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| < \frac{\varepsilon}{2} $$ holds for every $n \geq N$ . This implies \begin{align} \left|~ \int \limits_{\mathbb{R}} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| &\leq \left|~ \int \limits_{\mathbb{R}\setminus [-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| + \left|~ \int \limits_{[-R,R]} (f(x) - f_n (x)) \, \mathrm{d} x ~\right| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align} for $n \geq N$ as was to be shown.

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