\[ I \da \int_{-\infty}^{\infty} \frac{d x}{\left(1+x^{2}\right)^{n+1}}=\frac{(2 n) !}{4^{n}(n !)^{2}} \pi . \]
Note that this solution can be written many ways: \[ I = {2n \choose n} {\pi \over 4^n} = {\falling{2n}{n} \over n!} {\pi \over 4^n} = {(2n)(2n-1)\cdots(n+1)\over n!} {\pi \over 4^n} .\]
Solution
The integrand is \(f\in \bigo\qty{1\over z^{2n+2}} \subseteq \bigo\qty{1\over z^{1+\eps}}\), so a semicircular contour will work:
Thus \[ I &= 2\pi i \sum_{z_k \in \HH} \Res_{z=z_k} {1\over (1+z^2)^{n+1}} \\ &= 2\pi i \Res_{z=i} {1\over (z+i)^{n+1}(z-i)^{n+1} } \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{\dd{}{z}}^n (z+i)^{-(n+1)} \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{\dd{}{z}}^{n-1} -(n+1)(z+i)^{-(n+2)} \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{\dd{}{z}}^{n-2} -(n+1) \cdot -(n+2) (z+i)^{-(n+3)} \\ &= \qquad \vdots \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{\dd{}{z}}^{n-k} (-1)^k (n+1)(n+2)\cdots(n+k) (z+i)^{-(n+k+1)} \\ &= \qquad \vdots \\ &= 2\pi i \lim_{z\to i} {1 \over n!}(-1)^n (n+1)(n+2)\cdots(2n)(z+i)^{-(2n+1)} \\ &= { \falling{2n}{n} \over n!} 2\pi i (-1)^n (2i)^{-(2n+1)} \\ &= { \falling{2n}{n} \over n!} 2\pi i (-1)^n {1\over 2^{2n+1} i^{2n+1}} \\ &= { \falling{2n}{n} \over n!} \pi (-1)^n {1\over 4^n i^{2n}} \\ &= { \falling{2n}{n} \over n!} \pi (-1)^n {1\over 4^n (-1)^n } \\ &= { \falling{2n}{n} \over n!} {\pi \over 4^n } .\]