Question Number 28991 by abdo imad last updated on 02/Feb/18
$${prove}\:{that}\:{L}\left(\mathrm{1}\right)\left({s}\right)=\:\frac{\mathrm{1}}{{s}}\:\:{and}\:{L}\left({t}^{{n}} \right)\left({s}\right)=\:\frac{{n}!}{{s}^{{n}+\mathrm{1}} }\:.{L}\:{means} \\ $$$${laplace}\:{transform}. \\ $$
Commented by abdo imad last updated on 03/Feb/18
![L(1)(s)=∫_0 ^∞ e^(−st) dt =(1/s) with s>0 and L(t^n )(s)=∫_0 ^∞ t^n e^(−st) dt let use the ch. st=x L(t^n )(s)= ∫_0 ^∞ ((x/s))^n e^(−x) (dx/s) =∫_0 ^∞ (1/s^(n+1) ) x^n e^(−x) dx = (1/s^(n+1) )∫_0 ^∞ x^n e^(−x) dx let put A_n = ∫_0 ^∞ x^n e^(−x) dx by parts A_n = [−x^n e^(−x) ]_0 ^∞ +∫_0 ^∞ nx^(n−1) e^(−x) dx= n A_(n−1) A_n =n(n−1)A_(n−2) =....=n(n−1)....(n−p+1)LLA_(n−p) =n! A_(0 ) =n! (A_o =1) ⇒ L(t^n )(s)= ((n!)/s^(n+1) ) .](Q29071.png)
$${L}\left(\mathrm{1}\right)\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{st}} {dt}\:\:=\frac{\mathrm{1}}{{s}}\:\:{with}\:{s}>\mathrm{0}\:{and} \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{n}} \:{e}^{−{st}} \:{dt}\:\:{let}\:{use}\:{the}\:{ch}.\:{st}={x}\: \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\:\int_{\mathrm{0}} ^{\infty} \left(\frac{{x}}{{s}}\right)^{{n}} \:{e}^{−{x}} \:\frac{{dx}}{{s}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{s}^{{n}+\mathrm{1}} }\:{x}^{{n}} {e}^{−{x}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{{s}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−{x}} {dx}\:{let}\:{put}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−{x}} {dx}\:{by}\:{parts} \\ $$$${A}_{{n}} =\:\left[−{x}^{{n}} {e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:{e}^{−{x}} {dx}=\:{n}\:{A}_{{n}−\mathrm{1}} \\ $$$${A}_{{n}} ={n}\left({n}−\mathrm{1}\right){A}_{{n}−\mathrm{2}} =....={n}\left({n}−\mathrm{1}\right)....\left({n}−{p}+\mathrm{1}\right){LLA}_{{n}−{p}} \\ $$$$={n}!\:{A}_{\mathrm{0}\:} \:={n}!\:\:\:\:\:\:\:\:\:\:\:\:\:\left({A}_{{o}} =\mathrm{1}\right)\:\Rightarrow \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\:\frac{{n}!}{{s}^{{n}+\mathrm{1}} }\:. \\ $$
Answered by sma3l2996 last updated on 03/Feb/18
![L(1)(s)=∫_0 ^∞ e^(−st) dt=−(1/s)[e^(−st) ]_0 ^∞ =(1/s) L(t^n )(s)=∫_0 ^∞ t^n e^(−st) dt u=t^n ⇒u′=nt^(n−1) v′=e^(−st) ⇒v=−(1/s)e^(−st) L(t^n )(s)=(n/s)∫_0 ^∞ t^(n−1) e^(−st) dt so L(t^n )(s)=(n/s)L(t^(n−1) )(s) L(t^(n−1) )(s)=((n−1)/s)L(t^(n−2) )(s) so: L(t^n )(s)=(n/s)(((n−1)/s)(((n−2)/s)...(((n−(n−1))/s)L(t^(n−(n−1)) )...)) =((n!)/s^n )L(t)(s)=((n!)/s^(n−1) )((1/s)L(1)(s))=((n!)/s^n )((1/s)) L(t^n )=((n!)/s^(n+1) )](Q29011.png)
$${L}\left(\mathrm{1}\right)\left({s}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} {dt}=−\frac{\mathrm{1}}{{s}}\left[{e}^{−{st}} \right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{s}} \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{n}} {e}^{−{st}} {dt} \\ $$$${u}={t}^{{n}} \Rightarrow{u}'={nt}^{{n}−\mathrm{1}} \\ $$$${v}'={e}^{−{st}} \Rightarrow{v}=−\frac{\mathrm{1}}{{s}}{e}^{−{st}} \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\frac{{n}}{{s}}\int_{\mathrm{0}} ^{\infty} {t}^{{n}−\mathrm{1}} {e}^{−{st}} {dt} \\ $$$${so}\:\:{L}\left({t}^{{n}} \right)\left({s}\right)=\frac{{n}}{{s}}{L}\left({t}^{{n}−\mathrm{1}} \right)\left({s}\right) \\ $$$${L}\left({t}^{{n}−\mathrm{1}} \right)\left({s}\right)=\frac{{n}−\mathrm{1}}{{s}}{L}\left({t}^{{n}−\mathrm{2}} \right)\left({s}\right) \\ $$$${so}: \\ $$$${L}\left({t}^{{n}} \right)\left({s}\right)=\frac{{n}}{{s}}\left(\frac{{n}−\mathrm{1}}{{s}}\left(\frac{{n}−\mathrm{2}}{{s}}...\left(\frac{{n}−\left({n}−\mathrm{1}\right)}{{s}}{L}\left({t}^{{n}−\left({n}−\mathrm{1}\right)} \right)...\right)\right)\right. \\ $$$$=\frac{{n}!}{{s}^{{n}} }{L}\left({t}\right)\left({s}\right)=\frac{{n}!}{{s}^{{n}−\mathrm{1}} }\left(\frac{\mathrm{1}}{{s}}{L}\left(\mathrm{1}\right)\left({s}\right)\right)=\frac{{n}!}{{s}^{{n}} }\left(\frac{\mathrm{1}}{{s}}\right) \\ $$$${L}\left({t}^{{n}} \right)=\frac{{n}!}{{s}^{{n}+\mathrm{1}} } \\ $$$$ \\ $$