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The Laplace transform of exp(at), where a > 0, is defined only for the Laplace parameter, s > a since
the function is exponential.
the Laplace transform of integral of exp(at) has finite values only for s > a.
the Laplace transform integral of exp(at) has initial values only for s > a.
the function exp(at) is piece-wise continuous only for s > a.
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Correct Answer:
the Laplace transform of integral of exp(at) has finite values only for s > a.
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