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calculate-L-sinxe-ax-with-a-gt-0-L-means-laplace-transform-




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(sinxeax)witha>0Lmeanslaplacetransform.
Commented by maxmathsup 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(sinxeax)=0sinteatextdt=L(f)(x)=0sinte(a+x)tdt=Im(0eit(a+x)tdt)but0e(iax)tdt=[1iaxe(iax)t]t=0+=1iax(1)=1a+xi=a+x+i(a+x)2+1L(sinxeax)=1(a+x)2+1.

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