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Question Number 42489 by maxmathsup by imad last updated on 26/Aug/18

calculate L (sinxe^(−ax) ) with a>0 L means laplace transform .

calculateL(sinxe−ax)witha>0Lmeanslaplacetransform.

Commented bymaxmathsup by imad last updated on 27/Aug/18

L(sinx e^(−ax) )=∫_0 ^∞ sint e^(−at) e^(−xt) dt =L(f)(x) = ∫_0 ^∞ sint e^(−(a+x)t) dt =Im (∫_0 ^∞ e^(it −(a+x)t) dt) but ∫_0 ^∞ e^((i−a−x)t) dt = [(1/(i−a−x)) e^((i−a−x)t) ]_(t=0) ^(+∞) =(1/(i−a−x))(−1) =(1/(a+x −i)) =((a+x+i)/((a+x)^2 +1)) ⇒ L(sinx e^(−ax) ) = (1/((a+x)^2 +1)) .

L(sinxe−ax)=∫0∞sinte−ate−xtdt=L(f)(x) =∫0∞sinte−(a+x)tdt=Im(∫0∞eit−(a+x)tdt)but ∫0∞e(i−a−x)tdt=[1i−a−xe(i−a−x)t]t=0+∞=1i−a−x(−1) =1a+x−i=a+x+i(a+x)2+1⇒L(sinxe−ax)=1(a+x)2+1.

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