Primitive of Power of x over Even Power of x minus Even Power of a

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Theorem

\(\ds \int \frac {x^{p - 1} \rd x} {x^{2 m} - a^{2 m} }\) \(=\) \(\ds \frac 1 {2 m a^{2 m - p} } \sum_{k \mathop = 1}^{m - 1} \map \cos {\frac {k p \pi} m} \map \ln {x^2 - 2 a x \map \cos {\frac {k \pi} m} + a^2}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 {m a^{2 m - p} } \sum_{k \mathop = 1}^{m - 1} \map \sin {\frac {k p \pi} m} \map \arctan {\frac {x - a \map \cos {\dfrac {k \pi} m} } {a \map \sin {\dfrac {k \pi} m} } }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 {2 m a^{2 m - p} } \paren {\map \ln {x - a} + \paren {-1}^p \map \ln {x + a} }\)

where $0 < p \le 2 m$.


Proof

The integrand is a rational function.

It has simple poles at $x = \omega_k a$ where $\omega_k = e^{\pi i k /m} $, $k = 0, 1, \ldots, 2 m - 1$ are the $2m$'th roots of unity:

\(\ds \map f x\) \(=\) \(\ds \dfrac {x^{p - 1} } {x^{2 m} - a^{2 m} }\)
\(\ds \) \(=\) \(\ds \dfrac {x^{p - 1} } {\ds \prod_{k \mathop = 0}^{2 m - 1} \paren {x - \omega_k a} }\)

Its residue at $x = \omega_k$ is then:

$\Res f {\omega_k a} = \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m}$

so that:

$\ds \map f x = \sum_{k \mathop = 0}^{2 m - 1} \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m \paren {x - \omega_k a} }$

which has a primitive:

$\ds \map F x = \sum_{k \mathop = 0}^{2 m - 1} \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m} \map \ln {x - \omega_k a}$


For positive real $x, a \in \R_{>0}$ we can write:

$\map \ln {x - \omega_k a} = \dfrac 1 2 \map \ln {x^2 - 2 a x \map \cos {\dfrac {2 \pi k} m} + a^2} - i \map \arctan {\dfrac {a \map \sin {\dfrac {2 \pi k} m} } {x - a \map \cos {\dfrac {2 \pi k} m} } }$
$\paren {\omega_k a}^{p - 2 m} = a^{p - 2 m} \paren {\map \cos {\dfrac {p k \pi} m} + i \map \sin {\dfrac {p k \pi} m} }$

and the formula follows.




Sources