if ∫_0 ^∞ e^(−ax) dx = (1/a) show that ∫_0 ^∞ e^(−ax) x^n dx = ((n!)/a^(n+1) )


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Question Number 188384 by universe last updated on 28/Feb/23

$if \int_{0}^{\infty} e^{- a x} d x = \frac{1}{a}$ $s h o w t h a t \int_{0}^{\infty} e^{- a x} x^{n} d x = \frac{n!}{a^{n + 1}}$
	Answered by qaz last updated on 28/Feb/23

	$I (n) = \int_{0}^{\infty} x^{n} e^{- a x} d x = - \frac{1}{a} x^{n} e^{- a x} ∣_{0}^{\infty} + \frac{n}{a} \int_{0}^{\infty} x^{n - 1} e^{- a x} d x$ $= \frac{n}{a} I (n - 1) = \frac{n (n - 1)}{a^{2}} I (n - 2) = . . . = \frac{n (n - 1) . . . (n - n)}{a^{n}} I (n - n) =$ $= \frac{n!}{a^{n}} I (0) = \frac{n!}{a^{n + 1}}$
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