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Question Number 188384 by universe last updated on 28/Feb/23
        if    ∫_0 ^∞ e^(−ax) dx  =  (1/a)          show that  ∫_0 ^∞ e^(−ax)  x^n dx  =  ((n!)/a^(n+1) )
$$\:\:\: \\ $$$$\:\:\:\mathrm{if}\:\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}\:\:=\:\:\frac{\mathrm{1}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} \:{x}^{{n}} {dx}\:\:=\:\:\frac{{n}!}{{a}^{{n}+\mathrm{1}} } \\ $$
Answered by qaz last updated on 28/Feb/23
I(n)=∫_0 ^∞ x^n e^(−ax) dx=−(1/a)x^n e^(−ax) ∣_0 ^∞ +(n/a)∫_0 ^∞ x^(n−1) e^(−ax) dx  =(n/a)I(n−1)=((n(n−1))/a^2 )I(n−2)=...=((n(n−1)...(n−n))/a^n )I(n−n)=  =((n!)/a^n )I(0)=((n!)/a^(n+1) )
$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{ax}} {dx}=−\frac{\mathrm{1}}{{a}}{x}^{{n}} {e}^{−{ax}} \mid_{\mathrm{0}} ^{\infty} +\frac{{n}}{{a}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{1}} {e}^{−{ax}} {dx} \\ $$$$=\frac{{n}}{{a}}{I}\left({n}−\mathrm{1}\right)=\frac{{n}\left({n}−\mathrm{1}\right)}{{a}^{\mathrm{2}} }{I}\left({n}−\mathrm{2}\right)=…=\frac{{n}\left({n}−\mathrm{1}\right)…\left({n}−{n}\right)}{{a}^{{n}} }{I}\left({n}−{n}\right)= \\ $$$$=\frac{{n}!}{{a}^{{n}} }{I}\left(\mathrm{0}\right)=\frac{{n}!}{{a}^{{n}+\mathrm{1}} } \\ $$

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