I calculated the Laplace Transform $\mathcal{L}\{sin(t)\}$ using the complexification of the integral and using the linearity property in combination with $\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia}$ and get different results.
The second approach is takeing from Wiki Proofs and is rather short:
Because: $\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$ and $\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia}$:
$$ \begin{align} \mathcal{L}\{e^{iat}\} &= \mathcal{L}\{\cos(t) + i\sin(t)\} = \mathcal{L}\{\cos(t)\} + i\mathcal{L}\{\sin(t)\} \\ &\Rightarrow \mathcal{L}\{\sin(t)\} = IM(\mathcal{L}\{e^{iat}\}) = IM(\frac{s+ia}{s^2 + a^2}) = \frac{a}{s^2 + a^2} \end{align} $$
Thus when $a = 1 \Rightarrow \mathcal{L}\{\sin(t)\} = \frac{1}{s^2 + 1}$.
However, when I calculate $\mathcal{L}\{\sin(t)\}$ using the complexification of the sine, I get the following result:
$$ \begin{align} \int e^{-st}\sin(t)dt &= IM(\int e^{-st}e^{it}dt) = IM(\int e^{(i - s)t} dt) \\ &= IM(\frac{e^{(i - s)t}}{i-s}) = IM(\frac{i+s}{(i-s)(i+s)}e^{-st}e^{it}) \\ &= \frac{1}{1 + s^2}e^{-st}IM((i+s)(\cos(t) + i\sin(t)) \\ &= \frac{1}{1 + s^2}e^{-st}IM(i(\cos(t) + i\sin(t)) + s(\cos(t) + i\sin(t))) \\ &= \frac{1}{1 + s^2}e^{-st}IM(i\cos(t) + i^2\sin(t)) + s\cos(t) + is\sin(t))) \\ &= \frac{1}{1 + s^2}e^{-st}IM(i(\cos(t) + s\sin(t)) + (s\cos(t)-\sin(t)))\\ &= \frac{1}{1 + s^2}e^{-st}(\cos(t) + s\sin(t))\\ \end{align} $$
Putting all together:
$$ \frac{1}{1 + s^2}e^{-st}(\cos(t) + s\sin(t)) = \frac{1}{s^2 + 1} \Rightarrow e^{-st}(\cos(t) + s\sin(t)) = 1 $$
is that correct? If not, where is my mistake?
EDIT
I followed these two sources for the complexification:
Complexifying the Integral (Arthur Mattuck, MIT) The dessert, complexifying the integral
where the latter calculates: $\int e^{-bx}sin(x)dx$ and the former $\int e^{-x}cos(x)dx$
If you get unexpected results, first be more strict with the symbolism you employ. Here it is the integration interval and the integration variable $t$ that is bound to it. \begin{align} {\cal L}\{\sin t\}(s)&=\int_0^\infty e^{-st}\sin t\,dt=\text{Im}\left(\int_0^\infty e^{(i-s)t}\,dt\right) \\ &=\text{Im}\left(\left[\frac{e^{(i-s)t}}{i-s}\right]_{t=0}^\infty\right) =\text{Im}\left(\frac{0-1}{i-s}\right) \end{align} for $s>0$, which then gives the correct result.