I am studying complex analysis but, because I'm an engineer, I have a lot of doubts. I'm going to present my doubts and it would be nice if someone helps me to see things clearly.
Let's start with the laplace transform of the heaviside function;
$$\int_0 ^\infty e^{-st} \ dt \ \ \ \ \ \ \ \ (1)$$
The way we did this in class was straightforwad: $\int_0 ^\infty e^{-st} \ dt = \frac{e^{-st}}{-s} \big|_0^\infty = \frac 1s$
I believe that it's not rigorous; what theorem did we just apply? It is not the fundamental theorem of calculus, because it deals with real functions.
It could be the fundamental theorem of calculus, complex version, that says that If $f$ is holomorphic in $U$ , then exists $F: F' = f$ and for every curve $\gamma: [a, b] \to \mathbb C$ we have $$\int_\gamma f(z) \ dz = F(\gamma(b)) - F(\gamma(a))$$
But our integrand in $(1)$ is a complex function of a real variable; it is not holomorphic, is it? I don't really know how to classify this type of functions.
So the only explanation is how we define complex integrals, that is $$\int f(t) = \int \text{Re } f(t) + i\int \text{Im } f(t)$$
Doing so, I have (setting $s = x + iy$)
$$\int_0^\infty e^{-st} \ dt = \int_0^\infty e^{-xt}\cos (yt) \ dt - i \int_0^\infty e^{-xt} \sin (yt) \ dt$$
They clearly converge ($x > 0$) but then I need to calculate them (how?) and show that their sum is indeed equal to $\displaystyle \frac 1{x +iy} = \frac 1s$
Can somebody answer a bit all the questions I have? I am a bit confused and it would really help me out a lot.
I have the matter explained by @PedroTamaroff in chat.
Basically $f(t)$ is a complex-valued function of a real variable, so I consider $f(z) = e^{-sz}$ which is its analytic continuation over $\mathbb C$ and consider the integral over the curve $[0, \infty]$; since $f(z)$ is holomorphic I can use the fundamental theorem of calculus and of course the resulting integral is the same I was set to evaluate.
(I hope I have understood it correctly)
Thanks again pedro!