regarding the integral form of the inverse Laplace transform The integral form of the inverse Laplace transform is given as
$$f(t)=\frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{st}F(s)ds$$
where $s'$ is larger than the real parts of all the possible singularities of $F(s)$ on the complex plain.
I am quite curious why it can be given in this way, even though it works pretty well in some concrete cases. I tried to plug the Laplace transform $F(s)$ of $f(t)$ into the above expression, hoping that I can recover the original function $f(t)$, but it didn't go anywhere simply because it's really hard to precede without knowing the explicit form of the  functions involved. However, I still believe there should be a general proof under certain conditions. Any help is appreciated.
 A: Recall the definition of the Laplace Transform:
$$
F(s)=\int_{0}^{\infty}e^{-st}f(t)\ dt.
$$
Let's plug this into the Inverse Laplace Transform:
$$\begin{align*}
\frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{st}F(s)\ ds&=\frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{st}\int_{0}^{\infty}e^{-su}f(u)\ du\ ds\\
\text{(Fubini)}\qquad&=\int_{0}^{\infty}f(u)\left(\frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{s(t-u)}\ ds\right)\ du\\
\text{(Complex Analysis)}\qquad&=\int_{0}^{\infty}f(u)\delta(t-u)\ du\\
&=f(t).
\end{align*}$$
Okay so I clearly glossed over the two named steps, because they are applications of analysis. The first one, Fubini, says that when our functions decay sufficiently rapidly, we are allowed to interchange the order of integration. This will typically hold for all the functions that you will try to plug in to the Laplace Transform.
The second step is even more subtle. It relies both on the theory of residues, and on the concept of a distribution, which is a sort of generalized function. But an intuitive explanation is that $\delta(t-u)$ is an infinite spike centered at $u=t$, such that its total mass is 1. A little playing around will show that the inner integral behaves roughly like this (although to fully explain where the $\frac{1}{2\pi i}$ is coming from, you'll need to do some residue calculus.)
A: Write
$$F(s) = \int_0^{\infty} dt' \, f(t') e^{-s t'} $$
Then, reversing the order of integration, we have
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c + i \infty} ds \, F(s) e^{s t} &= \frac1{i 2 \pi} \int_0^{\infty} dt' f(t') \, \int_{c-i \infty}^{c+i \infty} ds \, e^{s (t-t')}\\ &= \frac1{2 \pi}  \int_0^{\infty} dt' f(t') e^{c (t-t')} \int_{-\infty}^{\infty} dy \, e^{i (t-t') y} \\ &= \int_0^{\infty} dt' f(t') e^{c (t-t')} \delta(t-t') \\ &= f(t) \theta(t) \end{align} $$
where $\theta$ is the Heaviside step function.
I should mention that this was a little fast and non rigorous.  I really should have had the ILT as a limit os some $T \to \infty$, and the delta function is a result of that limit.  But I only wanted a simple, heuristic outline, so here it is.
