## 1. Apply Laplace transform definition

Using the definition of the Laplace transform, We get:

\mathscr{L}\{a\}=\int_{0}^{\infty}ae^{-st} dt=a\lim_{b\to\infty}\int_{0}^{b}e^{-st}dt

## 2. Change integral variables

In the integral above, if u=-st then:

du=-sdt\Rightarrow dt=-\frac{du}{s}

So, changing the variables:

\begin{align} \mathscr{L}\{a\} & = a\lim_{b\to\infty}\int _{0}^{b} e^{u}\cdot\left(-\frac{du}{s}\right) \\ & = a\lim_{b\to\infty}\int _{0}^{b} -\frac{e^{u}}{s}du \end{align}

## 3. Calculate the integral

As \displaystyle-\frac{1}{s} is constant due to the fact we integrate in respect to u, then

\begin{align*} \mathscr{L}\{a\} & = a\lim_{b\to\infty}-\frac{1}{s}\int_{0}^{b} e^{u}du \\ & = a\lim_{b\to\infty}-\frac{1}{s}e^{u} \\ & = a\lim_{b\to\infty}\left.-\frac{1}{s}e^{-st}\right|_{t=0}^{t=b} \end{align*}

because u=-st. Now, evaluating in the boundaries:

\begin{align*} \mathscr{L}\{a\} & = a\lim_{b\to\infty}-\frac{1}{s}(e^{-s\cdot b}-\overbrace{e^{{-s\cdot0}}}^{=1}) \\ & = a\lim_{b\to\infty}-\frac{1}{s}(e^{-sb}-1) \\ & = \frac{a}{s}\lim_{b\to\infty}(-e^{-sb}+1) \\ & = \frac{a}{s}(\lim_{b\to\infty}-e^{-sb}+\lim_{b\to\infty}1) \end{align*}

## 4. Simplify the result

Since the limit:

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

for s>0 and

\lim_{b\to\infty}1=1

then, the Laplace Transform of a constant f(t)=a is the function:

F(s)=\frac{a}{s}

which is defined for every s in (0,+\infty).

That’s all!