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Just for reference, F[eax]=1a+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 e23x times the constant 1/3?

 Jun 26, 2015

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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
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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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Alan Jun 27, 2015

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