Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then, 
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx 
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right)
\end{align}
The pole of the function at $i$ is of order $n+1$, so the residue is computed by
\begin{align}
\operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right) &= \frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left(\frac{1}{(z+i)^{n+1}}\right) \\
&= \frac{1}{n!}\lim_{z\to i}(-1)^n\frac{(n+1)(n+2)\cdots(2n+1)}{(z+i)^{2n+1}} \\
&=\frac{(2n+1)!}{i2^{2n+1}(n!)^2}
\end{align}
Hence, $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx = \pi\frac{(2n+1)!}{2^{2n}(n!)^2}$$ The answer provided is $\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\pi$. How do I manipulate my answer to obtain this answer?
 A: Your answer is actually
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}.$$
(Check the derivative.)
Then use
$$ (2n)!=(1 \cdot 3 \cdots (2n-1))\cdot 2^n \cdot (n!)$$ to get
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2^n \cdot (n!)}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)}.$$
A: Note that by residuals taking the upper semicircle as trajectory:
\begin{eqnarray*}
            \int_{-\infty}^{+\infty} \frac{1}{(1+x^{2})^{n+1}}
            & = &   2\pi i\text{Res}(f(z),i)\ i\text{ is a pole of order }n+1\\
            % & = &   2\pi i\left(\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}f(z)\right)\\
            & = &   2\pi i\left(\lim_{z\to i}\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}\frac{1}{(1+z^{2})^{n+1}}\right)\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\frac{(z-i)^{n+1}}{(-i^2+z^{2})^{n+1}}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{z-i}{z^{2}-i^2}\right)^{n+1}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{1}{z+i}\right)^{n+1}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\
            % & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-1}}{dz^{n-1}}(-n-1)(z+i)^{-(n+2)}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-2}}{dz^{n-2}}(-n-1)(-n-2)(z+i)^{-(n+3)}\\
            & = &   \dots\dots\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}(-n-1)(-n-2)\cdots(-n-n)(z+i)^{-(n+n+1)}\\
            & = &   \frac{2\pi i}{n!}\lim_{z\to i}(-1)^{n}(n+1)(n+2)\cdots(2n)(z+i)^{-(2n+1)}\\
            & = &   \frac{2\pi i}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-(2n+1)}\\
            & = &   \frac{\pi}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-2n}\\
            & = &   \frac{\pi(2n)!}{n!n!}(-1)^{n}(4i^2)^{-n}\\
            & = &   \frac{\pi(2n)!}{n!n!}\frac{1}{4^n}(-1)^{n}(-1)^{-n}=\frac{\pi(2n)!}{2^{2n}(n!)^2}=\pi\cdot\frac{1}{n!2^{n}}\cdot\frac{(2n)!}{n!2^{n}}\\
        \end{eqnarray*}
The important thing is to note that
\begin{equation*}
\frac{(2n)!}{n!2^{n}} = 1\cdot3\cdot5\cdots(2n-1)\text{ and }n!2^{n}=2\cdot4\cdot6\cdots(2n)
\end{equation*}
of what follows the identity sought
