- Show that
\[ \int_ {- \infty} ^ {\infty} \frac {d x}{(1 + x ^ {2}) ^ {n + 1}} = \frac {1 \cdot 3 \cdot 5 \cdots (2 n - 1)}{2 \cdot 4 \cdot 6 \cdots (2 n)} \cdot \pi . \]
Solution
For \(m\ge1\), set \[ J_m=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^m}. \] We have \(J_1=\pi\). For \(m\ge2\), differentiate \[ \frac{x}{(1+x^2)^{m-1}}. \] Since this function tends to \(0\) at both ends, \[ 0=\int_{-\infty}^{\infty}\left[ \frac1{(1+x^2)^{m-1}}- \frac{2(m-1)x^2}{(1+x^2)^m} \right]dx. \] Using \[ \frac{x^2}{(1+x^2)^m} =\frac1{(1+x^2)^{m-1}}-\frac1{(1+x^2)^m}, \] we get \[ J_{m-1}=2(m-1)(J_{m-1}-J_m), \] so \[ J_m=\frac{2m-3}{2m-2}J_{m-1}. \] Iterating from \(J_1=\pi\) gives \[ J_{n+1} =\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\,\pi. \] Hence \[ \boxed{ \int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^{n+1}} =\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\pi}. \]