Using the definition, find the Laplace transform of \mathscr{L}\{1\}.

1. Apply the Laplace transform definition

The definition of Laplace transform of a function f(t) is:

\displaystyle \mathscr{L}\{f(t)\}=\int_{0}^{\infty}e^{-st}f(t)dt

So, for f(t)=1, we have:

\begin{aligned} \mathscr{L}\{1\} & = \int_{0}^{\infty}e^{-st}\cdot1dt \ & = \lim_{b\to\infty}\int_{0}^{b}e^{-st}dt \end{aligned}

2. Change of variables

To evaluate the integral above, let u=-st, then du=-sdt, which means:

dt=\displaystyle-\frac{du}{s}

Then:

\begin{aligned} \int_{0}^{b}e^{-st}dt & =\int e^{u}\cdot\left(-\frac{du}{s}\right) \\ & =-\frac{1}{s}\int e^{u}du \\ & =-\frac{1}{s}e^{u} \\ & =-\frac{1}{s}e^{-st} \end{aligned}

3. Integral boundaries evaluation

As we evaluate in the boundaries, we get:

\begin{aligned} \int_{0}^{b}e^{-st}dt & =-\frac{1}{s}e^{-sb}+\frac{1}{s}e^{-s\cdot0} \\ & =-\frac{1}{s}e^{-sb}+\frac{1}{s} \end{aligned}

4. Calculate the limit

So:

\mathscr{L}\{1\}=\lim_{b\to\infty}\left(-\frac{1}{s}e^{-sb}+\frac{1}{s}\right)=\lim_{b\to\infty}\left(-\frac{e^{-sb}}{s}\right)+\frac{1}{s}

For s<0, the limit above diverges and if s>0,

\displaystyle \lim_{b\to\infty}-\frac{e^{-sb}}{s}=0

5. Result

Hence:

\displaystyle\mathscr{L}\{1\}=\frac{1}{s}

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