Calculate the Laplace transform of \mathscr{L}\{e^{2t}\} using the definition.

1. Apply the Laplace transform definition

By definition, the Laplace transform F(s) of f(t)=e^{2t} is:

\begin{aligned} F(s) & = \int_{0}^{\infty}e^{-st}f(t)dt \\ & = \int_{0}^{\infty}e^{-st}\cdot e^{2t}\\ & = \int_{0}^{\infty}e^{-(s-2)t}dt \end{aligned}

and since is a improper integral, we rewrite as follows:

\displaystyle F(s)=\lim_{b\to\infty}\int_{0}^{b}e^{-(s-2)t}dt

2. Change of variables

Now, let u=-(s-2)t which implies:

du=-(s-2)dt

i.e.,

\displaystyle dt=-\frac{du}{s-2}

The above limit becomes:

\begin{aligned} \int_{0}^{b}e^{-(s-2)t}dt & =\int e^{u}\cdot\left(-\frac{du}{s-2}\right) \\ & =-\frac{1}{s-2}\int e^{u}du \\& =-\frac{1}{s-2}e^{u} \\ & =-\frac{1}{s-2}e^{-(s-2)t} \end{aligned}

3. Integral boundaries evaluation

At t=b and t=0, we get:

\begin{aligned}\int_{0}^{b}e^{-(s-2)t}dt & =-\frac{e^{-(s-2)b}}{s-2}+\frac{e^{-(s-2)\cdot0}}{s-2} \\& =-\frac{e^{-(s-2)b}}{s-2}+\frac{1}{s-2} \end{aligned}

4. Calculate the limit

As we apply \displaystyle\lim_{b\to\infty}, we have for s>2:

\displaystyle \lim_{b\to\infty}\left(-\frac{e^{-(s-2)b}}{s-2}+\frac{1}{s-2}\right)=\frac{1}{s-2}

Otherwise, the limit diverges.

5. Result

The Laplace transform of the exponential function is given by:

\displaystyle F(s)=\frac{1}{s-2}

We reached the end of this post, for the general case \mathscr{L}\{ e^{at}\}, check this post. And for more exercises with detailed solutions, check this page!