Talk:Laplace Transform of Cosine

I saw this proof on many places online, and also in my lecture notes, but seeing as $\mathcal Lf\left({t}\right)$ is a transform on the real variable t, perhaps it's not justifiable? --GFauxPas (talk) 10:25, 7 May 2014 (UTC)

To be strictly rigorous, the L.T. requires for a start that $t \in \R: t > 0$.

Further, note that $F(s)$ is a complex function. A L.T. converts a real function $f(t)$ to a complex function $F(s)$. Hence you cannot talk about $s - a$, but you can (in this context, and that of the L.T. of an exponential) use $\operatorname{Re} (s) > a$, and this is what is usually done. In the practical engineering applications, you only get $\operatorname{Re} (s) > a$ anyway, as the other way about leads to negative damping, i.e. divergent solutions, which, er, makes things break. --prime mover (talk) 11:24, 7 May 2014 (UTC)

But further to this, in all this work, it is essential to establish the domains of your functions, whether real or complex. If at this stage your source works are unclear on this, then I advise to lay them aside and either work on something else or get some better (i.e. more advanced) source works. I believe the one you'e working from is elementary and cuts a lot of corners in order to establish the simple results needed to design electronic circuitry, etc. --prime mover (talk) 11:27, 7 May 2014 (UTC)