Example 4.
Compute \(\lap{ e^{7t}} \text{.}\)
Solution. Solution
We begin by applying the definition of the Laplace transform:
\begin{align*}
\lap{ e^{7t} }
=\amp\ \int_0^{\infty} e^{-st} \cdot e^{7t}\ dt\\
=\amp\ \lim_{b \to \infty}\int_0^b e^{(-s+7)t}\ dt
\end{align*}
For this improper integral to converge, the exponent \(-s+7\) must be negative, meaning:
\begin{equation*}
-s+7 \lt 0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} -s \lt -7 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} s \gt 7\text{.}
\end{equation*}
Thus, we proceed under the assumption that \(s \gt 7\text{.}\)
\begin{align*}
\lap{ e^{7t} }
=\amp\ \lim_{b \to \infty}\int_0^b e^{(-s+7)t}\, dt\\
=\amp\ \lim_{b \to \infty} \frac{1}{7-s} e^{(-s+7)t} \Bigg|_0^b\\
=\amp\ \frac{1}{-s+7} \left[ \lim_{b \to \infty} \left( e^{(-s+7)b} - e^{0} \right) \right]\\
=\amp\ \frac{1}{-s+7} \left[ \lim_{b \to \infty} e^{(-s+7)b} - 1 \right]\\
=\amp\ \frac{1}{-s+7} \left[ 0 - 1 \right] \qquad \Big(e^{(\text{negative})b} \to 0\Big)\\
=\amp\ -\frac{1}{-s+7} = \frac{1}{s-7}. \quad \text{for } s \gt 7
\end{align*}
Thus, the Laplace transform of \(e^{7t}\) is:
\begin{equation*}
\ds \lap{e^{7t}} = \frac{1}{s - 7}, \quad s \gt 7.
\end{equation*}
