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Just for reference, $$F[e^{-ax}]=\frac{1}{a+iw}$$. Now I've got no idea why in the second step the Heaviside function is introduced(underlined), since here a=2/3 wouldn't the inverse transform just be $$e^{-\frac{2}{3}x}$$ times the constant 1/3?

 Jun 26, 2015

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 #1
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The exponential expression doesn't hold if x is negative, that's why the Heaviside function is there.

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 Jun 27, 2015
 #1
avatar+30355 
+15
Best Answer

The exponential expression doesn't hold if x is negative, that's why the Heaviside function is there.

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Alan Jun 27, 2015

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