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I'm messing around with Laplace, and was trying to find the transform of $e^{t}$ and I have to evaluate $$\lim_{h \to \infty} e^{h(1-s)}$$ I figure if $s=1$, the limit is $1$. If $0≤s<1$, the limit is $\infty$. If $s>1$, the limit is $0$.

WolframAlpha tells me the answer is complex infinity. I don't know what that is.

When I ask WolframAlpha for the transform of $e^t$, it says $\frac{1}{s-1}$, which would assume that s>1, though it doesn't mention this. What if $s$ isn't greater than 1?

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    When you compute Laplace transforms, you should assume that `s` is real and sufficiently large, or is complex and has real part sufficiently large. Try to compute the Laplace transform of 1, for example: you need to know that `s` is positive.2012-12-17
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    Where does this assumption that s is large come from?2012-12-17
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    The integral won't converge if $s$ is too small: the factor $e^{-st}$ in the integrand, as $t$ goes from 0 to $\infty$, can be problematic if $s<0$.2012-12-17
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    I understand that if $s$ is negative, then it won't converge. But if $s$ is tiny, it still converges, just much slower. Obviously I'd like to use the one that converges the fastest, so I suppose that's why s should be large. Thanks, I think I get it.2012-12-18

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