Question Number 37364 by math khazana by abdo last updated on 12/Jun/18
$${calculate}\:\:{L}\left\{\:\frac{{x}^{{n}−\mathrm{1}} \:{e}^{−{ax}} }{\left({n}−\mathrm{1}\right)!}\right\}\:{then}\:{conclude} \\ $$$${L}^{−\mathrm{1}} \left\{\:\:\frac{\mathrm{1}}{\left({a}+{x}\right)^{{n}} }\right\} \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
![we have L{((x^(n−1) e^(−ax) )/((n−1)!))}=∫_0 ^∞ ((t^(n−1) e^(−at) )/((n−1)!)) e^(−xt) dt =(1/((n−1)!))∫_0 ^∞ t^(n−1) e^(−(a+x)t) dt =_((a+x)t=u) (1/((n−1)!))∫_0 ^∞ (u^(n−1) /((a+x)^(n−1) )) e^(−u) (du/(a+x)) = (1/((n−1)!(a+x)^n )) ∫_0 ^∞ u^(n−1) e^(−u) du but A_n =∫_0 ^∞ u^(n−1) e^(−u) du =[(1/n)u^n e^(−u) ]_0 ^∞ +∫_0 ^∞ (1/n) u^n e^(−u) du = (1/n) A_(n+1) ⇒ A_(n+1) =n A_n ⇒ Π_(k=1) ^(n−1) A_(k+1) =(n−1)! Π_(k=1) ^(n−1) A_k ⇒ A_n = (n−1)! A_1 =(n−1)! ⇒ L{ ((x^(n−1) e^(−ax) )/((n−1)!))} = (1/((a+x)^n )) ⇒ L^(−1) ((1/((x+a)^n ))) = ((x^(n−1) e^(−ax) )/((n−1)!)) .](https://www.tinkutara.com/question/Q37561.png)
$${we}\:{have}\:{L}\left\{\frac{{x}^{{n}−\mathrm{1}} \:{e}^{−{ax}} }{\left({n}−\mathrm{1}\right)!}\right\}=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{n}−\mathrm{1}} \:{e}^{−{at}} }{\left({n}−\mathrm{1}\right)!}\:{e}^{−{xt}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \:\:\:{t}^{{n}−\mathrm{1}} \:{e}^{−\left({a}+{x}\right){t}} {dt} \\ $$$$=_{\left({a}+{x}\right){t}={u}} \:\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{{n}−\mathrm{1}} }{\left({a}+{x}\right)^{{n}−\mathrm{1}} }\:{e}^{−{u}} \:\frac{{du}}{{a}+{x}} \\ $$$$=\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!\left({a}+{x}\right)^{{n}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{u}^{{n}−\mathrm{1}} \:{e}^{−{u}} \:{du}\:\:{but} \\ $$$${A}_{{n}} \:=\int_{\mathrm{0}} ^{\infty} \:{u}^{{n}−\mathrm{1}} \:{e}^{−{u}} \:{du}\:=\left[\frac{\mathrm{1}}{{n}}{u}^{{n}} \:{e}^{−{u}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$+\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}}\:{u}^{{n}} \:{e}^{−{u}} \:{du}\:=\:\frac{\mathrm{1}}{{n}}\:{A}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$${A}_{{n}+\mathrm{1}} ={n}\:{A}_{{n}} \:\:\Rightarrow\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{A}_{{k}+\mathrm{1}} =\left({n}−\mathrm{1}\right)!\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{A}_{{k}} \\ $$$$\Rightarrow\:{A}_{{n}} \:=\:\left({n}−\mathrm{1}\right)!\:{A}_{\mathrm{1}} =\left({n}−\mathrm{1}\right)!\:\Rightarrow \\ $$$${L}\left\{\:\frac{{x}^{{n}−\mathrm{1}} \:{e}^{−{ax}} }{\left({n}−\mathrm{1}\right)!}\right\}\:=\:\:\frac{\mathrm{1}}{\left({a}+{x}\right)^{{n}} }\:\Rightarrow \\ $$$${L}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\left({x}+{a}\right)^{{n}} }\right)\:=\:\frac{{x}^{{n}−\mathrm{1}} \:{e}^{−{ax}} }{\left({n}−\mathrm{1}\right)!}\:. \\ $$