find I_n (x)= ∫_0 ^∞ t^n e^(−xt) dt x>0 n∈ N.


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Question Number 31095 by abdo imad last updated on 02/Mar/18

$f i n d I_{n} (x) = \int_{0}^{\infty} t^{n} e^{- x t} d t x > 0 n \in N .$
Commented by abdo imad last updated on 04/Mar/18

$c h . x t = u g i v e I_{n} (x) = \int_{0}^{\infty} {(\frac{u}{x})}^{n} e^{- u} \frac{d u}{x}$ $= \frac{1}{x^{n + 1}} \int_{0}^{\infty} u^{n} e^{- u} d u l e t p u t A_{n} = \int_{0}^{\infty} u^{n} e^{- u} d u b y p a r t s$ $A_{n} = {[- u^{n} e^{- u}]}_{0}^{\infty} + \int_{0}^{\infty} n u^{n - 1} e^{- u} d u = {n A}_{n - 1} \Rightarrow$ $\prod_{k = 1}^{n} A_{k} = n! \prod_{k = 1}^{n} A_{k - 1} \Rightarrow A_{n} = n! A_{0} = n! (A_{0} = 1) \Rightarrow$ $I_{n} (x) = \frac{n!}{x^{n + 1}} .$
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