Question Number 30527 by abdo imad last updated on 22/Feb/18
$${find}\:\:{I}_{{n},{p}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:{e}^{−{px}} \:\:\:\:\:{with}\:{n}\:{and}\:{p}\:{from}\:{N}^{\bigstar} \:. \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
![I=∫_0 ^∞ x^n e^(−px) dx u=x^n ⇒u′=nx^(n−1) v′=e^(−px) ⇒v=((−1)/p)e^(−px) I=−(1/p)[x^n e^(−px) ]_0 ^∞ +(n/p)∫_0 ^∞ x^(n−1) e^(−px) dx (lim_(x→∞) x^n e^(−px) =0) =(n/p)∫_0 ^∞ x^(n−1) e^(−px) dx u=x^(n−1) ⇒u′=(n−1)x^(n−2) v′=e^(−px) ⇒v=((−1)/p)e^(−px) I=((n(n−1))/p^2 )∫_0 ^∞ x^(n−2) e^(−px) dx . . . k times I=((n(n−1)(n−2)...(n−k+1))/p^k )∫_0 ^∞ x^(n−k) e^(−px) dx . . . n times I=((n(n−1)(n−2)...(n−n+1))/p^n )∫_0 ^∞ e^(−px) dx I=((n!)/p^n )[−(1/p)e^(−px) ]_0 ^∞ =((n!)/p^(n+1) )(−(lim_(x→∞) e^(−px) −e^0 )) I=((n!)/p^(n+1) )](Q30531.png)
$${I}=\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{px}} {dx} \\ $$$${u}={x}^{{n}} \Rightarrow{u}'={nx}^{{n}−\mathrm{1}} \\ $$$${v}'={e}^{−{px}} \Rightarrow{v}=\frac{−\mathrm{1}}{{p}}{e}^{−{px}} \\ $$$${I}=−\frac{\mathrm{1}}{{p}}\left[{x}^{{n}} {e}^{−{px}} \right]_{\mathrm{0}} ^{\infty} +\frac{{n}}{{p}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{1}} {e}^{−{px}} {dx}\:\:\left(\underset{{x}\rightarrow\infty} {{lim}x}^{{n}} {e}^{−{px}} =\mathrm{0}\right) \\ $$$$=\frac{{n}}{{p}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{1}} {e}^{−{px}} {dx} \\ $$$${u}={x}^{{n}−\mathrm{1}} \Rightarrow{u}'=\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \\ $$$${v}'={e}^{−{px}} \Rightarrow{v}=\frac{−\mathrm{1}}{{p}}{e}^{−{px}} \\ $$$${I}=\frac{{n}\left({n}−\mathrm{1}\right)}{{p}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{2}} {e}^{−{px}} {dx} \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$${k}\:{times} \\ $$$${I}=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\left({n}−{k}+\mathrm{1}\right)}{{p}^{{k}} }\int_{\mathrm{0}} ^{\infty} {x}^{{n}−{k}} {e}^{−{px}} {dx} \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$${n}\:{times} \\ $$$${I}=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\left({n}−{n}+\mathrm{1}\right)}{{p}^{{n}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{px}} {dx} \\ $$$${I}=\frac{{n}!}{{p}^{{n}} }\left[−\frac{\mathrm{1}}{{p}}{e}^{−{px}} \right]_{\mathrm{0}} ^{\infty} =\frac{{n}!}{{p}^{{n}+\mathrm{1}} }\left(−\left(\underset{{x}\rightarrow\infty} {{lim}e}^{−{px}} −{e}^{\mathrm{0}} \right)\right) \\ $$$${I}=\frac{{n}!}{{p}^{{n}+\mathrm{1}} } \\ $$