Define:
$$ I(n) = \sqrt{n}\int_{-\infty }^{\infty} \frac{1}{( 1+x^{2})^n} \, dx$$
I have to show that:
$$ \lim\limits_{n\to\infty}\sqrt{n} I(n) = \sqrt{\pi} $$
Classical solution:
Based on this answer we have a closed form for the integral:
$$ I(n) = \sqrt{\pi}\,\frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)}$$
Also from here we get:
$$\sqrt{n}\,\frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)} \to 1 \quad \text{as} \quad n \to \infty $$ Any other methods?
(Migrated from comment) Using the substitution $\sqrt{n}x=u$, we get
$$ \sqrt{n} I(n) = \int_{-\infty}^{\infty} \left(1 + \frac{u^2}{n}\right)^{-n} \, \mathrm{d}u.$$
Since $(1 + u^2/n)^n \geq 1 + u^2$, the above integrands are dominated by the integrable function $1/(1+u^2)$, and so, by the dominated convergence theorem,
$$\lim_{n\to\infty}\sqrt{n} I(n) = \int_{-\infty}^{\infty} \lim_{n\to\infty} \left(1 + \frac{u^2}{n}\right)^{-n} \, \mathrm{d}u = \int_{-\infty}^{\infty} e^{-u^2} \, \mathrm{d}u = \sqrt{\pi}.$$
Note that $I_n=2\int_0^{\infty}{\frac{dx}{(1+x^2)^n}}$
Use substitution $t=\tan u$ and obtain: $$\int_0^{\infty}{\frac{dt}{(1+t^2)^n}}=\int_{0}^{\frac{\pi}{2}}{({\cos{u})^{2n-2}}du}=W_{2n-2}\sim\frac12\sqrt{\frac{\pi}{n-1}}$$ see Wallis’s Integral. Thus: $$\sqrt{n}I(n)\sim\sqrt{\pi\frac{n}{n-1}}$$
