Show that \[ \int_{-\infty}^{\infty} \frac{d x}{\left(1+x^{2}\right)^{n+1}}=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \cdot \pi .\]
Solution
By evenness and the substitution \(x=\tan\theta\), \[ I_n:=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^{n+1}} =2\int_0^{\pi/2}\cos^{2n}\theta\,d\theta. \] Integration by parts gives the Wallis recursion \[ \int_0^{\pi/2}\cos^{2n}\theta\,d\theta =\frac{2n-1}{2n} \int_0^{\pi/2}\cos^{2n-2}\theta\,d\theta. \] Since \(\int_0^{\pi/2}1\,d\theta=\pi/2\), induction yields \[ \int_0^{\pi/2}\cos^{2n}\theta\,d\theta =\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\frac\pi2. \] Therefore \[ \boxed{I_n= \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\pi.} \]