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calculate-L-x-n-1-e-ax-n-1-then-conclude-L-1-1-a-x-n-

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Question Number 37364 by math khazana by abdo last updated on 12/Jun/18
calculate L{ ((x^(n−1) e^(−ax) )/((n−1)!))} then conclude L^(−1) { (1/((a+x)^n ))}
calculateL{xn−1e−ax(n−1)!}thenconcludeL−1{1(a+x)n}
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)!)) .
wehaveL{xn−1e−ax(n−1)!}=∫0∞tn−1e−at(n−1)!e−xtdt=1(n−1)!∫0∞tn−1e−(a+x)tdt=(a+x)t=u1(n−1)!∫0∞un−1(a+x)n−1e−udua+x=1(n−1)!(a+x)n∫0∞un−1e−udubutAn=∫0∞un−1e−udu=[1nune−u]0∞+∫0∞1nune−udu=1nAn+1⇒An+1=nAn⇒∏k=1n−1Ak+1=(n−1)!∏k=1n−1Ak⇒An=(n−1)!A1=(n−1)!⇒L{xn−1e−ax(n−1)!}=1(a+x)n⇒L−1(1(x+a)n)=xn−1e−ax(n−1)!.

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