Constants
Zero
Given $f(t) = 0$
\[\begin{align} \mathcal{L}\{ 0 \} = F(s) &= \int_{0}^{\infty} 0 \cdot e^{-st} \; dt \\[10pt] &= \int_{0}^{\infty} 0 \; dt \\[10pt] &= 0 \end{align}\]Condition: $s \in \mathbb{C}$
One
Given $f(t) = 1$
\[\begin{align} \mathcal{L}\{ 1 \} = F(s) &= \int_{0}^{\infty} 1 \cdot e^{-st} \; dt \\[10pt] &= \int_{0}^{\infty} e^{-st} \; dt \\[10pt] &= -\frac{1}{s} e^{-st} \biggr\rvert_{0}^{\infty} \\[10pt] &= \left ( \lim_{t \rightarrow \infty} -\frac{1}{s} e^{-st} \right ) - \left ( - \frac{1}{s} e^{0} \right ) \\[10pt] &\text{We must assume } \lvert s \rvert > 0 \text{ otherwise the limit goes to } -\infty \\[10pt] &= -\frac{1}{s} \cdot 0 + \frac{1}{s} \cdot 1 \\[10pt] &= \frac{1}{s} \end{align}\]Any Constant
Given $f(t) = \alpha$ for any constant $\alpha \in \mathbb{R}$
\[\begin{align} \mathcal{L}\{ \alpha \} = F(s) &= \int_{0}^{\infty} \alpha \ e^{-st} \; dt \\[10pt] &= \alpha \ \int_{0}^{\infty} e^{-st} \; dt \\[10pt] &\text{assume } \lvert s \rvert > 0 \text{ as part of the condition on } \mathcal{L}\{ 1 \}\\[10pt] &= \alpha \cdot \mathcal{L}\{ 1 \} \\[10pt] &= \frac{\alpha}{s} \end{align}\]