Proof of inverse Laplace transform Why is
$$f(t) = \frac{1}{2πj}\int_{\sigma-j\infty}^{\sigma+j\infty} F(s) e^{st} \, ds,$$
provided that
$$F(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt \ ?$$
I tried to find out myself, or searched online and found a term Bromwich integral, but I want to know how this expression is derived. (And I couldn't find any :()
Thank you.
 A: It is the Fourier inversion formula in disguise. In case you have never encountered this theorem before, let me prove the following version (which is obviously far from optimal).

Proposition. Let $F(s) = \int_{0}^{\infty} f(t)e^{-st} \, dt$ be the Laplace transform of $f : [0,\infty) \to \mathbb{R}$. Assume that the following technical conditions hold with some $g : [0,\infty) \to \mathbb{R}$ and $\sigma \in \mathbb{R}$:

*

*$f(t) = f(0) + \int_{0}^{t} g(u) \, du$. (In particular, $g$ is the 'derivative' of $f$.)

*Both $f(t)e^{-\sigma t}$ and $g(t)e^{-\sigma t}$ are Lebesgue-integrable on $[0, \infty)$.

Then for any $s > 0$, we have
$$ \lim_{R\to\infty} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz = f(s). $$

Proof. Define $S(x) = \frac{1}{2} + \frac{1}{\pi}\int_{0}^{x} \frac{\sin t}{t} \, dt$. Then $S(x)$ is bounded, and by Dirichlet integral, we have
$$ \lim_{R\to\infty} S(Rx) = H(x) := \begin{cases}
1, & x > 0 \\
\frac{1}{2}, & x = 0 \\
0, & x < 0
\end{cases} $$
(Obviously $H$ denotes the Heaviside step function.) Now we have
\begin{align*}
\frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz
&= \frac{1}{2\pi} \int_{-R}^{R} F(\sigma + i\xi)e^{s(\sigma+i\xi)} \, d\xi \\
&= \frac{1}{2\pi} \int_{-R}^{R} \left( \int_{0}^{\infty} f(t)e^{-(\sigma+i\xi)t} \, dt \right)e^{s(\sigma+i\xi)} \, d\xi.
\end{align*}
By Fubini's theorem, we can interchange the order of integral to obtain
\begin{align*}
\frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz
&= \int_{0}^{\infty} f(t)e^{-(t-s)\sigma} \left( \frac{1}{2\pi} \int_{-R}^{R} e^{(s-t)i\xi} \, d\xi \right) \, dt \\
&= \int_{0}^{\infty} f(t)e^{-(t-s)\sigma} \left( \frac{\sin R(t-s)}{\pi (t-s)} \right) \, dt
\end{align*}
By the assumption, both $f(t)e^{-\sigma t}$ and $(f(t)e^{-\sigma t})' = (f'(t) - \sigma f(t))e^{-\sigma t}$ are Lebesgue-integrable. In particular, this tells that $f(t)e^{-\sigma t}$ converges to $0$ as $t\to\infty$. So by integration by parts,
\begin{align*}
\frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz
&= - f(0)e^{s\sigma} S(-Rs)
- \int_{0}^{\infty} (f(t)e^{-(t-s)\sigma})' S(R(t-s)) \, dt.
\end{align*}
As $R \to \infty$, the right-hand side converges to
\begin{align*}
\lim_{R\to\infty} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz
&= - \int_{0}^{\infty} (f(t)e^{-(t-s)\sigma})' H(t-s) \, dt \\
&= - \left[ f(t)e^{-(t-s)\sigma} \right]_{t=s}^{t=\infty}
 = f(s).
\end{align*}
(Pushing the limit inside the integral is justified by the dominated convergence theorem.)
