I tried to recreate the proof of the inverse Laplace transform formula based on the blurry memory from an electrical engineering book I've read and I would like a real mathematician to look at it and tell me if I've omitted some little detail that electrical engineers are not told about. It goes as follows:
Let
$$ f(t) = \left\{\begin{split}&g(t),~t \in [0, +\infty),\\&0, t \in (-\infty, 0).\end{split}\right. $$
The Laplace transform of $f(t)$ can be written as
$$ \mathcal{L}\{f(t)\} = \int_{0}^{+\infty}{f(t)e^{-st}}dt = \int_{-\infty}^{+\infty}{f(t)e^{-st}}dt = \int_{-\infty}^{+\infty}{f(t)e^{-\sigma t}e^{-i\omega t}}dt = \mathcal{F}\{f(t)e^{-\sigma t}\}. $$
Thus,
$$ f(t)e^{-\sigma t} = \mathcal{F}^{-1}\{\mathcal{L}\{f(t)\}\} = \frac{1}{2\pi}\int_{-\infty}^{+\infty}{\mathcal{L}\{f(t)\}e^{i\omega t}}d\omega, $$
and
$$ f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}{\mathcal{L}\{f(t)\}e^{(\sigma + i\omega)t}}d\omega = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty}{\mathcal{L}\{f(t)\}e^{st}}dt. $$
The thing which seems the most precarious to me is the extension of the integration range at the beginning because it relies on the fact that $$ \lim_{t \rightarrow -\infty}(0 \cdot e^{-\sigma t}) = 0, $$
which should be fine in my mind because $0$ is a constant, and not a function which approaches zero at $-\infty$.
If something is missing, I would be very grateful for any pointers on how to make this derivation more robust.
The development looks good. And yes, since $f(t)\equiv 0$ for $t< 0$, we can assert that
$$\int_0^\infty f(t)e^{-st}\,dt=\int_{-\infty}^\infty f(t)e^{-st}\,dt$$
where we assume that $f(t)e^{-\sigma t}H(t)$ is in $L^1$. That is to say , $\int_{0}^\infty |f(t)|e^{-\sigma t}\,dt<\infty$.
Then, let $F$ repersent the Laplace Transform of $f$ as given by
$$F(s)=\mathscr{L}\{f\}(s)=\int_{-\infty}^\infty f(t)e^{-st}\,dt$$
Denoting the real and imaginary parts of $s$ be denote $\sigma$ and $\omega$, respectively, we see that
$$\begin{align} F(s)&=F(\sigma+i\omega)\\\\ &=G_\sigma (\omega)\\\\ &=\int_{-\infty}^\infty f(t) e^{-\sigma t}e^{-i\omega t}\,dt\\\\ &=\mathscr{F}\{g_\sigma\}(\omega) \end{align}$$
where $G_\sigma(\omega)=F(\sigma+i\omega)$, and $g_\sigma(t)=f(t)e^{-\sigma t}$.
Next, we assume that $G_\sigma\in L^1$ and $g_\sigma$ is continuous at the point $t$.. Then, applying the Fourier Inversion Theorem, we find that
$$\begin{align} f(t)e^{-\sigma t}&=g_\sigma(t)\\\\ &=\mathscr{F^{-1}}\{G_\sigma\}(t)\\\\ &=\frac1{2\pi}\int_{-\infty}^\infty G_\sigma(\omega)e^{i\omega t}\,d\omega\\\\ &=\frac1{2\pi}\int_{-\infty}^\infty F(\sigma+i\omega)e^{i\omega t}\,d\omega \end{align}$$
Finally, enforcing the substitution $s=\sigma+i\omega$ such that $d\omega =\frac1i \,ds$, we arrive at the coveted Inverse Laplace Transform
$$f(t)=\frac1{2\pi i }\int_{\sigma - i\infty}^{\sigma+i\infty} F(s)e^{st}\,ds$$