For a function $\sin(at)$, I get the exponential order $\gamma = 0$ because the function is limited through $|\cdot|\leq 1.$ I try to calculate the following: $$\mathcal{L}\{\sin(at)\}(s)$$ where $s$ is my complex frequency in the newly calculated transform. My Idea is the following: $$\mathcal{L}\{\sin(at)\}(s)=\int_0^{\infty}e^{-st}\sin(at)\;\mathrm{d}t=\int_0^{\infty}e^{-st}\Im\left ( e^{iat} \right )\;\mathrm{d}t.$$
$\Im$ is the imaginary part of my term and $\Re$ would return the real part. But because I have to assume that $s$ has the form $$s=\sigma + i\omega,$$ I cannot combine those integrals as easy as I thought. How to solve this particular problem and handle the $\Im(\cdot)$ or $\Re(\cdot)$ in such cases in general?
There is a much simpler way to do the same.
We have, $$L[e^{bt}] =\frac1{s-b} $$ Set, $b=ia$, it gives: $$L[e^{iat}] =\frac1{s-ia} =\frac{s+ia} {s^2+a^2} = L[\cos at +i\sin at] $$
Equating the real and imaginary parts on both sides, we have: $$L[\sin at] =\frac{a} {s^2+a^2}$$ and $$L[\cos at] =\frac{s} {s^2+a^2}$$