Contour integration and the central binomial coefficients I am trying to compute the integral $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx.$$ From computational evidence, it's very obvious that $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx = \frac{\pi}{4^n} {2n \choose n}.$$ Indeed, I can prove this via the generating function for the central binomial coefficients.
However, I want to prove this via contour integration. With $f(z) = z^{2n} / (z^2 + 1)^{n + 1}$, we can integrate over the semicircle of radius $R$ in the upper half-plane. Call this contour $\gamma_R$. The integral over the arc of $\gamma_R$ goes to zero as $R \to \infty$, which leaves $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx = 2\pi i \operatorname{res}_i f.$$
The residue of $f$ at $i$ is $g^{(n)}(i) / n!$, where $$g(z) = \frac{z^{2n}}{(z + i)^{n + 1}}.$$ Thus, we should have $$g^{(n)}(i) = \frac{-i n! {2n \choose n}}{2^{2n + 1}}.$$
How can I show that this equality holds? Or, more generally, How can I compute the residue of $f$ at $i$?
I tried using the series $$\frac{1}{(1 - z)^{n + 1}} = \sum_{k \geq 0} {k + n \choose k} z^k,$$ but couldn't really make it work.
Edit: I was asked to explain why my evaluation is "obvious." This is from using a computer algebra system to directly evaluate $g^{(n)}(i)$ for a few dozen $n$. This gives some rational expressions which, when looked up in the OEIS, suggest the closed form I have given here. Then it is a trivial matter to estimate the integral numerically and compare it for hundreds of terms.
 A: We             have             by            the             Leibniz
rule, that
$$\frac{1}{n!} \left(z^{2n} \frac{1}{(z+i)^{n+1}} \right)^{(n)}
\\ = \frac{1}{n!}
\sum_{q=0}^n {n\choose q} \frac{(2n)!}{(2n-q)!} z^{2n-q}
(-1)^{n-q} \frac{(n+n-q)!}{n!} \frac{1}{(z+i)^{n+1+n-q}}
\\ = {2n\choose n} \sum_{q=0}^n {n\choose q} z^{2n-q}
(-1)^{n-q} \frac{1}{(z+i)^{2n+1-q}}
\\ = {2n\choose n} \frac{z^{2n}}{(z+i)^{2n+1}}
\sum_{q=0}^n {n\choose q}
(-1)^{n-q} \frac{(z+i)^q}{z^q}
\\ = {2n\choose n} \frac{z^{2n}}{(z+i)^{2n+1}}
\left(\frac{z+i}{z} - 1\right)^n
\\ = {2n\choose n} \frac{i^n z^{n}}{(z+i)^{2n+1}}.$$
Returning to the main computation we set $z=i$ to obtain
$$2\pi i \times
{2n\choose n} \frac{i^{2n}}{(2i)^{2n+1}}
\\= 2\pi i \times {2n\choose n} \frac{1}{2^{2n+1}} \frac{1}{i}
= \frac{\pi}{4^{n}} {2n\choose n}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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A "simple" way is the Ramanujan-MT:
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
{x^{2n} \over \pars{x^{2} + 1}^{n + 1}}\,\dd x} =
\int_{0}^{\infty}
{x^{\color{red}{n + 1/2} - 1} \over \pars{1 + x}^{n + 1}}\,\dd x
\end{align}
\begin{align}
&\mbox{Note that}\quad
{1 \over \pars{1 + x}^{n + 1}} =
\sum_{k = 0}^{\infty}{-n - 1 \choose k}x^{k}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{n + k \choose k}
\pars{-1}^{k}x^{k}
=
\sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{n + 1 + k} \over \Gamma\pars{n + 1}}{\pars{-x}^{k} \over k!}
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
{x^{2n} \over \pars{x^{2} + 1}^{n + 1}}\,\dd x}
\\[5mm] = &\
\Gamma\pars{n + {1 \over 2}}\,
{\Gamma\pars{n + 1 - \bracks{n + 1/2}} \over \Gamma\pars{n + 1}}
\\[5mm] = &\
{\pars{n - 1/2}!\,\Gamma\pars{1/2} \over
n!\bracks{\pars{-1/2}!/\Gamma\pars{1/2}}} =
{n - 1/2 \choose n}\pars{\pi}
\\[5mm] = &\
\pi\
\underbrace{-1/2 \choose n}
_{\ds{\color{red}{\Large\S :}\
{2n \choose n}\pars{-4}^{-n}}}\pars{-1}^{n} =
\bbx{{\pi \over 4^{n}}{2n \choose n}} \\ &
\end{align}
$\ds{\color{red}{\Large\S :}}$ See this link.
