Definite Integral from 0 to a of x^m by (a^n - x^n)^p

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Theorem

$\ds \int_0^a x^m \paren {a^n - x^n}^p \rd x = \frac {a^{m + 1 + n p} \, \map \Gamma {\frac {m + 1} n} \map \Gamma {p + 1} } {n \map \Gamma {\frac {m + 1} n + p + 1} }$


Proof

\(\ds \int_0^a x^m \paren {a^n - x^n}^p \rd x\) \(=\) \(\ds a \int_0^1 \paren {a u}^m \paren {a^n - \paren {a u}^n}^p \rd u\) substituting $x = a u$
\(\ds \) \(=\) \(\ds a \times a^m \times a^{n p} \int_0^1 u^m \paren {1 - u^n}^p \rd u\)
\(\ds \) \(=\) \(\ds \frac {a^{m + 1 + n p} } n \int_0^1 \frac {u^m} {u^{n - 1} } \paren {1 - t}^p \rd u\) substituting $t = u^n$
\(\ds \) \(=\) \(\ds \frac {a^{m + 1 + n p} } n \int_0^1 t^{\frac {m - n + 1} n} \paren {1 - t}^p \rd u\)
\(\ds \) \(=\) \(\ds \frac {a^{m + 1 + n p} } n \map \Beta {\frac {m + 1} n, p + 1}\) Definition 1 of Beta Function
\(\ds \) \(=\) \(\ds \frac {a^{m + 1 + n p} \, \map \Gamma {\frac {m + 1} n} \map \Gamma {p + 1} } {n \map \Gamma {\frac {m + 1} n + p + 1} }\) Definition 3 of Beta Function

$\blacksquare$


Sources

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