There is nothing blocking us from making the $ a $ in the previous post complex, so we get this Laplace transform pair for free!
$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{at}e^{-st}dt} &=& \frac{1}{s-a} \end{eqnarray*} $
This, of course, imply $ |a - s| < 0 $. Interesting things happen when we use the Euler's formula. Let $ a = r + i \theta $, we have $ e^{at} = e^{rt}(\cos \theta t + i \sin \theta t) $. Plugging this into the formula above we have
$ \begin{eqnarray*} \mathcal{L}(e^{rt}(\cos \theta t + i \sin \theta t)) &=& \frac{1}{s-a} \\ \mathcal{L}(e^{rt}(\cos \theta t)) &=& \operatorname{Re}(\frac{1}{s-a}) \\ &=& \operatorname{Re}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Re}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Re}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{s-r}{(s-r)^2 + \theta^2} \\ \mathcal{L}(e^{rt}(\sin \theta t)) &=& \operatorname{Im}(\frac{1}{s-a}) \\ &=& \operatorname{Im}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Im}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Im}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{\theta}{(s-r)^2 + \theta^2} \\ \end{eqnarray*} $
Of course we can further specialize so that $ r = 0 $, and in that case we will have
$ \begin{eqnarray*} \mathcal{L}(\cos \theta t) &=& \frac{s}{s^2 + \theta^2} \\ \mathcal{L}(\sin \theta t) &=& \frac{\theta}{s^2 + \theta^2} \end{eqnarray*} $
See now little we need to come up with so many results!
$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{at}e^{-st}dt} &=& \frac{1}{s-a} \end{eqnarray*} $
This, of course, imply $ |a - s| < 0 $. Interesting things happen when we use the Euler's formula. Let $ a = r + i \theta $, we have $ e^{at} = e^{rt}(\cos \theta t + i \sin \theta t) $. Plugging this into the formula above we have
$ \begin{eqnarray*} \mathcal{L}(e^{rt}(\cos \theta t + i \sin \theta t)) &=& \frac{1}{s-a} \\ \mathcal{L}(e^{rt}(\cos \theta t)) &=& \operatorname{Re}(\frac{1}{s-a}) \\ &=& \operatorname{Re}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Re}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Re}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{s-r}{(s-r)^2 + \theta^2} \\ \mathcal{L}(e^{rt}(\sin \theta t)) &=& \operatorname{Im}(\frac{1}{s-a}) \\ &=& \operatorname{Im}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Im}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Im}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{\theta}{(s-r)^2 + \theta^2} \\ \end{eqnarray*} $
Of course we can further specialize so that $ r = 0 $, and in that case we will have
$ \begin{eqnarray*} \mathcal{L}(\cos \theta t) &=& \frac{s}{s^2 + \theta^2} \\ \mathcal{L}(\sin \theta t) &=& \frac{\theta}{s^2 + \theta^2} \end{eqnarray*} $
See now little we need to come up with so many results!
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