\[I= \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^n} dx\]
consider \[F_n( \alpha )= \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^n} dx\]
.\[ F_1( \alpha )= \int^{\infty}_0 \frac{1}{x^2+ \alpha^2} dx= \frac{\pi}{2 \alpha }\]
\[\begin{equation} \begin{aligned} \frac{dF_1(\alpha)}{d \alpha}& =\int^{\infty}_0 \frac{-2 \alpha}{(x^2+ \alpha^2)^2} dx= - \frac{\pi}{2 \alpha^2} \rightarrow \\ & F_2 (\alpha )= \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^2} dx \\ &= \frac{\pi}{2 \times 2 \alpha^3} \end{aligned} \end{equation}\]
.\[\begin{equation} \begin{aligned} \frac{dF_2 (\alpha)}{d \alpha} &= \int^{\infty}_0 \frac{(-2)2 \alpha}{(x^2+ \alpha^2)^3} dx= - \frac{3 \pi}{4 \alpha^4} \rightarrow \\ & F_3 (\alpha )= \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^3} dx = \frac{3 \pi}{2 \times 2 \times 4 \alpha^5} \end{aligned} \end{equation}\]
.\[\begin{equation} \begin{aligned} \frac{dF_3 (\alpha)}{d \alpha} &= \int^{\infty}_0 \frac{(-3)2 \alpha}{(x^2+ \alpha^2)^4} dx= - \frac{3 \times 5 \pi}{2 \times 2 \times 4 \times 6 \alpha^6} \rightarrow & F_4 (\alpha ) = \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^4} dx= \frac{3 \times 5 \pi}{2 \times 2 \times 4 \times 6 \alpha^7} \end{aligned} \end{equation}\]
.We can prove by induction
\[F_n ( \alpha ) = \int^{\infty}_0 \frac{1}{(x^2+ \alpha^2)^{n+1}} dx = \frac{1 \times 3 \times 5 \times ... \times (2n-3) \pi }{2 \times 2 \times 4 \times 6 \times ... \times (2n-3) \alpha^{2n-1}} \]
.