# Definite Integral to Infinity of Power of x over Power of x plus Power of a

## Theorem

$\displaystyle \int_0^\infty \dfrac {x^m \rd x} {x^n + a^n} = \frac {\pi a^{m + 1 - n} } {n \sin \left({\left({m + 1}\right) \frac \pi n}\right)}$

for $0 < m + 1 < n$.

## Proof

 $\displaystyle \int_0^\infty \dfrac {x^m \rd x} {x^n + a^n}$ $=$ $\displaystyle \int_0^\infty \dfrac {x^m \rd x} {\paren {x^{m + 1} }^{\frac n {m + 1} } + \paren {a^{m + 1} }^{\frac n {m + 1} } }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {m + 1} \int_0^\infty \dfrac 1 {u^{\frac n {m + 1} } + \paren {a^{m + 1} }^{\frac n {m + 1} } } \rd u$ substituting $u = x^{m + 1}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\paren {m + 1} \paren {\frac n {m + 1} \paren {a^{m + 1} }^{\frac n {m + 1} - 1} } } \csc \paren {\frac {\paren {m + 1} \pi} n}$ Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$: Corollary $\displaystyle$ $=$ $\displaystyle \frac {a^{m + 1} } {n a^{\paren {m + 1} \frac n {m + 1} } } \csc \paren {\frac {\paren {m + 1} \pi} n}$ $\displaystyle$ $=$ $\displaystyle \frac {\pi a^{m + 1 - n} } {n \sin \paren {\paren {m + 1} \frac \pi n} }$

$\blacksquare$