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

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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
\(\displaystyle \) \(=\) \(\displaystyle \frac {a^{m + 1} } {n a^{\paren {m + 1} \frac n {m + 1} } } \csc \paren {\frac {\paren {m + 1} \pi} n}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi a^{m + 1 - n} } {n \sin \paren {\paren {m + 1} \frac \pi n} }\)

$\blacksquare$

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