Polynomials
Linear Functions
Given $f(t) = t$
\[\begin{align} \mathcal{L}\{ t \} = F(s) &= \int_{0}^{\infty} t \ e^{-st} \; dt \qquad \text{(using integration by parts)} \\[10pt] &= \frac{1}{s} f(0) - \frac{1}{s} \lim_{t \rightarrow \infty} f(t)e^{-st} + \frac{1}{s} \int_{0}^{\infty} f'(t)e^{-st} \ dt \\[10pt] &= \frac{1}{s} \lim_{t \rightarrow \infty} te^{-st} + \frac{1}{s} \int_{0}^{\infty} e^{-st} \ dt \\[10pt] &\text{assume } \lvert s \rvert > 0 \text{ as part of the condition on } \mathcal{L}\{ 1 \}\\[10pt] &= \frac{1}{s} \cdot 0 + \frac{1}{s} \cdot \mathcal{L}\{ 1 \} \\[10pt] &= \frac{1}{s^2} \end{align}\]To complete the proof, we will quickly show the evaluation of the limit. It is an easy application of L’Hôpital’s rule.
\[\begin{align} \lim_{t \rightarrow \infty} te^{-st} &= \lim_{t \rightarrow \infty} \frac{t}{e^{st}} \\[10pt] &= \lim_{t \rightarrow \infty} \frac{1}{se^{st}} \\[10pt] &= 0 \qquad (\text{assuming } \lvert s \rvert > 0) \end{align}\]Monomials
Given $f(t) = t^n$ for any integer $n > 0$
\[\begin{align} \mathcal{L}\{ t^n \} = F(s) &= \int_{0}^{\infty} t^n \ e^{-st} \; dt \qquad \text{(using integration by parts)} \\[10pt] &= \frac{1}{s} f(0) - \frac{1}{s} \lim_{t \rightarrow \infty} f(t)e^{-st} + \frac{1}{s} \int_{0}^{\infty} f'(t)e^{-st} \ dt \\[10pt] &= \frac{1}{s} \lim_{t \rightarrow \infty} t^ne^{-st} + \frac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \ dt \\[10pt] &= \frac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \ dt \\[10pt] &\text{Now, we just recurrively apply this result} \\[10pt] &= \frac{n!}{s^n} \int_{0}^{\infty} e^{-st} \ dt \\[10pt] &\text{assume } \lvert s \rvert > 0 \text{ as part of the condition on } \mathcal{L}\{ 1 \}\\[10pt] &= \frac{n!}{s^n} \mathcal{L}\{ 1 \} \\[10pt] &= \frac{n!}{s^{n+1}} \end{align}\]Again, to complete the proof, we will quickly show the evaluation of the limit. It’s just successive applications of L’Hôpital’s rule.
\[\begin{align} \lim_{t \rightarrow \infty} t^n e^{-st} &= \lim_{t \rightarrow \infty} \frac{t^n}{e^{st}} \\[10pt] &= \lim_{t \rightarrow \infty} \frac{1}{s^n e^{st}} \\[10pt] &= 0 \qquad (\text{assuming } \lvert s \rvert > 0) \end{align}\]Polynomials
Given $\displaystyle f(t) = \sum_{k=0}^n a_k t^k = a_0 + a_1 t + a_2 t^2 + \ldots + a_n t^n$ for any integer $n \geq 0$
\[\begin{align} \mathcal{L} \left \{ \sum_{k=0}^n a_k t^k \right \} = F(s) &= \int_{0}^{\infty} \sum_{k=0}^n a_k t^k \ e^{-st} \; dt \qquad \text{(using linearity of integrals)} \\[10pt] &= \sum_{k=0}^n a_k \int_{0}^{\infty} t^k \ e^{-st} \; dt \\[10pt] &\text{assume } \lvert s \rvert > 0 \text{ as part of the condition on } \mathcal{L}\{ t^k \}\\[10pt] &= \sum_{k=0}^n a_k \mathcal{L}\{ t^k \} \\[10pt] &= \sum_{k=0}^n \frac{k! \ a_k}{s^{k+1}} \end{align}\]Scaling and Translation
First, we can apply our scaling property.
\[\mathcal{L}\{ (bt)^n \} = \frac{1}{b^n} \frac{n!}{(s/b^n)^{n+1}} = \frac{b^{n^2} n!}{s^{n+1}}\]For translation, notice we can multiply the terms of $(bt+c)^n$ in order to get a polynomial in standard form. However, we can also notice that in the proof for Monomials, adding a translation term will not effect the derivatives. It will only effect when we evaluate $f(0)$, which will now be $c^n$. Thus, the result should be (although I did not double check this so please let me know if it’s wrong)
\[\mathcal{L}\{ (bt+c)^n \} = \mathcal{L}\{ (bt)^n \} + \frac{1}{s} \sum_{i=0}^n c^i\]