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Reading on control theory and the Laplace transform of the unit step function, I came upon the following in my textbook.

The Laplace transform defined as: $$Y(s)=\int_{0}^{\infty}y(t)e^{-st}dt$$

s being a complex number $s=\mu+j\omega$

Given the unit step function $\sigma(t)=\begin{cases} 1 & t\ge0\\ 0 & t<0 \end{cases} $

The Laplace transform of $\sigma$ is

$$ Y(s)=\int_{0}^{\infty}1e^{-st}dt $$

$$ =-\left.\frac{e^{-st}}{s}\right|_{0}^{\infty}=\frac{1}{s} $$

My question is about the last part and how to calculate it step by step. I tried and did the following:

$$ -\left.\frac{e^{-st}}{s}\right|_{0}^{\infty}=\lim_{t\rightarrow\infty}-\frac{e^{-s\cdot t}}{s}-(-\frac{e^{-s\cdot0}}{s}) $$

$$ =-\lim_{t\rightarrow\infty}\frac{1}{s\cdot e^{s\cdot t}}+\frac{e^{0}}{s} $$

$$ =-\lim_{t\rightarrow\infty}\frac{1}{s\cdot e^{s\cdot t}}+\frac{1}{s} $$

Since the right term is $\frac{1}{s}$, which is the end result, I conclude the left term must become 0.

If $s$ was a real number I would have no trouble with this, but I wasn't certain how to do this when $s$ is a complex number. Trying with Wolfram Alpha, it says this expression has value zero , but doesn't specify if using real or complex numbers.

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    $\begingroup$ When you see $\infty$ there, you cannot just plug in $\infty$. You have to do a limit: $$ \lim_{t\to\infty}\frac{e^{-st}}{s}$$ $\endgroup$
    – GEdgar
    Mar 24, 2015 at 11:58
  • $\begingroup$ Thanks, I did the limit with wolfram alpha as linked. Also found this post which relates math.stackexchange.com/questions/585766/… $\endgroup$
    – user985366
    Mar 24, 2015 at 11:59
  • $\begingroup$ You get limit zero here only if $s$ has positive real part. $\endgroup$
    – GEdgar
    Mar 24, 2015 at 12:01
  • $\begingroup$ How can we assume anything about s having a positive real part? $\endgroup$
    – user985366
    Mar 24, 2015 at 12:06
  • $\begingroup$ The integral definition of $Y(s)$ diverges when $s=-1$. Try it. So you can assume $\mathrm{Re}\;s>0$ or you can try something other than that integral. $\endgroup$
    – GEdgar
    Mar 24, 2015 at 12:10

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