Prove $\int_{-\infty}^\infty\frac{dx}{(1+x^2/a)^n}$ converges 
For $n\geq 1, a>0:$
  $$I_{n,a}=\int_{-\infty}^\infty\frac{dx}{(1+x^2/a)^n}$$
  $(a)$ Prove $I_{n,a}$ convergs and
$$I_{n+1,a}=\frac{2n-1}{2n}I_{n,a}$$
$(b)$ Prove 
  $$I_{n,n} \xrightarrow {n \to \infty} \int_{-\infty}^\infty e^{-x^2} dx$$

My attempt:
$\frac{1}{(1+x^2/a)^n}$ is positive and even, therefore by Comparision Test:
$$I_{n,a}=2\int_0^\infty\frac{dx}{(1+x^2/a)^n}\leq2\int_0^\infty\frac{dx}{(1+x^2/a)}<\infty$$
Is this correct?
I tried to do some manipulations, but couldn't get to the desired expression of $I_{n+1,a}$.
As for $(b),$ I know $\frac{1}{(1+x^2/n)^n} \xrightarrow {n \to \infty} e^{-x^2}.$ Is that enough for the proof, or should anything be added?
Any help appreciated.
 A: Hint. One may integrate by parts,
$$
\begin{align}
I_{n,a}&=\int_{-\infty}^\infty\frac{dx}{(1+x^2/a)^{n}}\qquad (n\ge1)
\\\\&=\int_{-\infty}^\infty 1 \cdot\frac{1}{(1+x^2/a)^{n}}\:dx
\\\\&=\left[ x\cdot  \frac{1}{(1+x^2/a)^{n}}\right]_{-\infty}^\infty -\int_{-\infty}^\infty  x\cdot  \frac{-n\cdot\frac{2x}{a}}{(1+x^2/a)^{n+1}}\; dx
\\\\&=\color{red}{0}+2n\int_{-\infty}^\infty  \frac{\frac{x^2}{a}}{(1+x^2/a)^{n+1}}\; dx 
\\\\&=2n\int_{-\infty}^\infty  \frac{1+\frac{x^2}{a}-1}{(1+x^2/a)^{n+1}}\; dx 
\\\\&=2n\int_{-\infty}^\infty  \frac{1}{(1+x^2/a)^{n}}\; dx-2n\int_{-\infty}^\infty  \frac{1}{(1+x^2/a)^{n+1}}\; dx 
\\\\&=2nI_{n,a}-2n I_{n+1,a}
\end{align}
$$ giving $(a)$.
One may apply the dominated convergence theorem to obtain $(b)$.
A: Using successively the change of variables  $x = \sqrt{a}t$, $u = t^2$ and $ s =\frac{1}{1+u}$ we obtain
$$ I_{n,a}  = 2\sqrt{a}\int_{0}^{\infty} \frac{dt}{(1+t^2)^n} = \sqrt{a}\int_{0}^{\infty} \frac{ u^{-1/2}du}{(1+u )^n}  =  \sqrt{a}\int_{0}^{1}s^{n-1/2-1} (1-s)^{-1/2} ds $$ 
Then we obviously see that $0\le I_{n,a}\le I_{n-1,a}$ i.e a nondecreasing and bounded sequence this give the convergence of $(I_{n,a})_n$. 
On other hand integration by path give 
$$ I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$.
Also we can from the given formula above that 
$$I_{n,a}=\sqrt{a} B\left(1/2,n-1/2\right)=\sqrt{a}\frac{\Gamma\left(1/2\right)\Gamma\left(n-1/2\right)}{\Gamma\left(n\right)}$$
Using the following famous Stirling formula: Given $x>0$
$$ \lim_{x\to +\infty} \frac{\Gamma(x+1)}{\left(\frac{x}{e}\right)^x \sqrt{2\pi x} }=1. $$
Where $\Gamma$ is the Gamma function of Euler and $n! =\Gamma(n+1)$ for $n\in \mathbb{N}$ .
we find out that, 
\begin{split}
 \lim_{n\to \infty} I_{n+1,n+1} &=&\Gamma\left(1/2\right)\lim_{n\to \infty}  \sqrt{n+1}\frac{\Gamma\left(n+1-1/2\right)}{\Gamma\left(n+1\right)}\\
 &= & \Gamma\left(1/2\right)\lim_{n\to \infty} \sqrt{n+1}\left(\frac{e}{n}\right)^n\left(\frac{n-1/2}{e}\right)^{n-1/2} \\
&=& \Gamma\left(1/2\right)\lim_{n\to \infty} \sqrt{\frac{n+1}{n}}\left(\frac{n-1/2}{n}\right)^{n-1/2}e^{1/2} \\
&=& e^{1/2} \Gamma\left(1/2\right)\lim_{n\to \infty} \left(\frac{n-1/2}{n}\right)^{n-1/2}
\end{split}
But (easy tocheck)
$$\lim_{n\to \infty} \left(\frac{n-1/2}{n}\right)^{n-1/2} =e^{-1/2} $$
Finally we have $$ \lim_{n\to \infty} I_{n,n} = \Gamma(1/2) := \int_0^{\infty} e^{-t} t^{1/2-1} dt = \int_{-\infty}^{\infty}e^{-x^2}dx$$
