 Power-complement integral recurrence

ID: power-complement-integral-recurrence

For

p > 0

and a nonnegative integer

n

n > 0

J_{n} (p) = p n \int_{0}^{1} s^{p} (1 - s^{p})^{n - 1} d s = p n (J_{n - 1} (p) - J_{n} (p)) .

The endpoint term

[s (1 - s^{p})^{n}]_{0}^{1}

vanishes. Therefore

J_{n} (p) = \frac{p n}{p n + 1} J_{n - 1} (p), J_{0} (p) = 1.

Repeated application yields

J_{n} (p) = \prod_{j = 1}^{n} p j / (p j + 1)

w = s^{p}

also identifies

J_{n} (p) = B (1/ p, n + 1) / p

with an Euler beta function. A related weighted integral follows immediately from the same substitution:

\int_{0}^{1} s^{p - 1} (1 - s^{p})^{n} d s = \frac{1}{p ( n + 1 )} .

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