Prove $\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}} = \frac{(1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)} \pi \ \ \ \forall n \in \mathbb{N}$ My attempt starts with a contour integral in the half disk, I let the radius -> infinity and so the contour integral
\begin{equation}
\int_{\gamma} \frac{dz}{(1+z^2)^{n+1}} = 2 \pi i \ res_{z_0 = i} f
\end{equation}
reduces to
\begin{equation}
\int^{\infty}_{-\infty} \frac{dx}{(1+x^2)^{n+1}} = 2 \pi i \ res_{z_0 = i} f
\end{equation}
Since the poles are of order n+1, the residual is
\begin{equation}
res_{z_0 = i} f = \lim_{z\to i} \frac{1}{n!} \bigg ( \frac{d}{dz} \bigg)^n (z+i)^{-[n+1]} = \frac{(-1)^n (2n)! (2i)^{-[2n+1]}}{(n!)^2} = \frac{(2n)!}{(n!)^2 2i 2^{2n}}
\end{equation}
and so I have 
\begin{equation}
\frac{(2n)! \pi}{(n!)^2 2^{2n}}
\end{equation}
now the part that I'm struggling with, showing that
\begin{equation}
\frac{(2n)!}{(n!)^2 2^{2n}} = \frac{(1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}
\end{equation}
 A: Another way to use residue calculus is to look at the generating function (for very small $t$) $$\sum_{n=0}^{\infty} \int_{-\infty}^{\infty} \frac{t^{n+1}}{(1+x^2)^{n+1}} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{t / (1+x^2)}{1 - \frac{t}{1+x^2}} \, \mathrm{d}x = t \int_{-\infty}^{\infty} \frac{1}{1+x^2 - t} \, \mathrm{d}x.$$ This has its simple pole at $z = i \sqrt{1-t}$ with residue $-\frac{i}{2\sqrt{1-t}}$, so $$\sum_{n=0}^{\infty} \Big(\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^{n+1}} \, \mathrm{d}x \Big) t^{n+1} = \frac{\pi t}{\sqrt{1-t}} = \sum_{n=0}^{\infty} \pi \binom{-1/2}{n} (-1)^n t^{n+1}$$ and you get your integral by comparing coefficients.
A: $$\begin{equation}
\frac{(2n)!}{(n!)^2 2^{2n}} = \frac{(1)(2)(3)...(2n)}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2} \times \frac{1}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2}
\end{equation}$$
Now, rewrite 
$$1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2=2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)$$
in the first denominator, and cancel.
