# Integral of (1+[x^2])^(-n) from 0 to ∞

\begin{align*}& \int_0^{\infty} \dfrac{1}{(1+x^2)^n} dx \\& \\& \\&{\rm{Lemma \: (1)}}: \int_0^{\infty} \left[ f(x) \right] dx = \int_0^{\infty} \left[ \dfrac{f(\frac{1}{x})}{(x^2)} \right] dx \\& \\&{\rm{Lemma \: (2)}}: \int_0^{\frac{\pi}{2}} \left[\sin(x)\right]^{2g-1} \left[\cos(x)\right]^{2h-1} dx = \dfrac{\beta(g, h)}{2} \\& \\&{\rm{Lemma \: (3)}}: \beta(g, h) = \dfrac{(g-1)!(h-1)!}{(g+h-1)!} \\& \\&{\rm{Lemma \: (4)}}: (n-1)! = \dfrac{n!}{n} \\& \\&{\rm{Lemma \: (5)}}: \left(n-\dfrac{1}{2}\right)! = \dfrac{(2n-1)!!\sqrt{\pi}}{2^n} \: ; \: n \in \mathbb{Z} \\& \\& \\&\therefore {\rm{Apply \: the \: following \: substitutions}}: \\& \\&(1) \: x = \dfrac{1}{y} \\& (2) \: y=\tan(g) \\& \\&\therefore \int_0^{\frac{\pi}{2}} \left[\sin(x)\right]^{2(n-\frac{1}{2})-1} \left[\cos(x)\right]^{2(\frac{1}{2})-1} dx = \dfrac{\beta\left(n-\dfrac{1}{2}, \dfrac{1}{2}\right)}{2} \\& \\&= \dfrac{\sqrt{\pi}}{2} \left[ \dfrac{(n-1.5)!}{(n-1)!} \right] \: ; \:
0 < n \in \mathbb{R} \\& \\& \\& \\&{\rm{Special \: Cases}}: \\& \\&(1) \: 0 < n \in \mathbb{N} \\& \\&
\dfrac{\sqrt{\pi}}{2} \left[ \dfrac{(n-1.5)!}{(n-1)!} \right] =
\dfrac{\sqrt{\pi}}{2n-1} \left[ \dfrac{(n-0.5)!}{(n-1)!} \right]
\Longrightarrow \dfrac{\pi}{2^n} \left[ \dfrac{(2n-3)!!}{(n-1)!} \right] \\& \\&(2) \: n=1.5 \Longrightarrow
\int_0^{\infty} \dfrac{1}{(1+x^2)^{1.5}} dx = 1
\end{align*}

This site uses Akismet to reduce spam. Learn how your comment data is processed.