$\lim_{n\to\infty} \sqrt{n} \int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^n}=\sqrt{\pi}$ solution doubts 
Prove $\lim_{n\to\infty}\sqrt{n}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^n}=\sqrt{\pi}$

I tried to solve the problem by checking if the function series was convergent and then pass the limit under the integral but I was not able to solve the integral so I turned to the book solution, which looked complex.
Solution:

We divide $(-\infty,\infty)$ into three parts as:
$(-\infty,-n^{-\frac{1}{3}})\cup [-n^{-\frac{1}{3}},n^{-\frac{1}{3}}]\cup[-n^{\frac{1}{3}},\infty]$
which we denote by $A_1,A_2,A_3$ respectively. For $1\leqslant k\leqslant 3$ >we put $I_k=\sqrt{n}\int_{A_k}\frac{dx}{(1+x^2)^n}$
Substituting $t=\sqrt{n}x$ we have
$I_2=\sqrt{n}\int_{A_2}\exp(-n\log(1+x^2)dx=\sqrt{n}\int_{-n^{-{1}/{6}}}^{n^{-{1}/{6}}}\exp(-n\log(1+\frac{t^2}{n})dt$
Since $-n\log(1+\frac{t^2}{n})=t^2+O(\frac{1}{n^{1/3}})$
uniformly in $|t|\leqslant n^{\frac{1}{6}}$, we obtain
$I_2=(1+O(\frac{1}{n^{1/3}}))(\sqrt{\pi}-\int_{-\infty}^{-n^{1/6}}e^{-t^2}dt-\int_{n^{1/6}}^{\infty}e^{-t^2}dt)=\sqrt{\pi}+O(\frac{1}{n^{1/3}})$
as $n\to\infty$. On the other hand it follows that
$0<I_1+I_2<\frac{\sqrt{n}}{(1+n^{-2/3})^{n-1}}\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^n}$
and right-hand side convergence to $0$ as $n\to\infty$

Questions:
1) How does the author establishes this equality $-n\log(1+\frac{t^2}{n})=t^2+O(\frac{1}{n^{1/3}})$?
2) How did he deduce the following $I_2=(1+O(\frac{1}{n^{1/3}}))(\sqrt{\pi}-\int_{-\infty}^{-n^{1/6}}e^{-t^2}dt-\int_{n^{1/6}}^{\infty}e^{-t^2}dt)$? Where does the $\sqrt{\pi}$ come from?
Sorry for the questions but it the first time I solve an integral like this.
Thanks in advance!
 A: As another answer suggested, the book's solution is a wild goose chase.
A nice way to see what's going on is to let $x=y/\sqrt n.$ The expression then becomes
$$\tag 1 \int_{\mathbb R} \frac{dy}{(1+y^2/n)^n}.$$
As $n\to \infty,$ $(1+y^2/n)^n\to e^{y^2}$ for any fixed $y.$ So it's reasonable to wonder if the answer is
$$\int_{\mathbb R} \frac{dy}{e^{y^2}} = \sqrt \pi.$$
This is easy to justify. Just recall that the denominators in $(1)$ increase with $n.$ This shows that a dominating functon for the integrands in $(1)$ is $1/(1+y^2) \in L^1(\mathbb R ).$ The DCT then finishes it off.
A: The book's solution is rather stupid, in my opinion. I propose a different approach. With residues we have:
$$\int_{-\infty}^{+\infty} \frac{\text{d}x}{(1+x^2)^n} \to \oint_{|z| = 1}\frac{\text{d}z}{(1+z^2)^n}$$
Which has poles at $z = \pm i$ of oder $n$.
Using the multi-poles residues formula 
$$2\pi i \frac{1}{(n-1)!}\frac{\text{d}^n}{\text{d}z^n}\lim_{z\to +i} \left((z-i)^n \frac{1}{(z+i)^n(z-i)^n}\right)$$
and arranging a bit we can easily get the answer:
$$\boxed{\frac{\sqrt{\pi } \Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}}$$
Where $\Gamma(\cdot)$ denotes the Euler Gamma Function.
The computation of
$$\lim_{n\to +\infty} \frac{\sqrt{n}\Gamma(n-1/2)}{\Gamma(n)}$$
Is rather easy, and gives you $1$, whence
$$\sqrt{\pi}\cdot 1 = \sqrt{\pi}$$
As wanted.
A: For (1),
since
$x-x^2/2
\lt \ln(1+x)
\lt x
$,
$\begin{array}\\
n \ln(1+t^2/n)
&\lt n(t^2/n)\\
&= t^2\\
\text{and}\\
n \ln(1+t^2/n)
&\gt n(t^2/n-t^4/(2n^2))\\
&= t^2-t^4/(2n)\\
&\gt t^2-(n^{1/6})^4/(2n)
\qquad\text{since }|t| \le n^{1/6}\\
&= t^2-n^{-1/3}/2\\
\end{array}
$
A: Mathematica knows 
$$ \int_{-\infty}^{\infty}dx \frac{1}{(1+x^2)^n}=\sqrt{\pi}\frac{\Gamma(n-1/2)}
{\Gamma(n)}. $$ Asymptotics of $\Gamma$ gets the answer.  This is how I'd work the problem.  If you need to prove the right hand side, you could work with the easy integral with integrand $1/(1+a\,x^2)$ and successively differentiate.
