[proofplan]
We derive the Bromwich integral from the [Fourier inversion theorem](/theorems/364). The key substitution is $g(t) = f(t) e^{-ct}$, which is $L^1(\mathbb{R})$ for $c > \sigma_0$ (the exponential order). The Fourier transform of $g$ is $\hat{f}(c + i\omega)$, and Fourier inversion recovers $g(t)$. The change of variable $p = c + i\omega$ transforms the Fourier integral into the Bromwich contour integral along $\operatorname{Re} p = c$.
[/proofplan]
[step:Reduce the Laplace transform to a Fourier transform by exponential damping]
Define
\begin{align*}
g: \mathbb{R} &\to \mathbb{C} \\
t &\mapsto f(t) e^{-ct},
\end{align*}
where $c > \sigma_0$. Since $f(t) = 0$ for $t < 0$ and $|f(t)| \leq Me^{\sigma_0 t}$ for $t \geq 0$, we have $|g(t)| \leq Me^{(\sigma_0 - c)t}$ for $t \geq 0$ and $g(t) = 0$ for $t < 0$. Since $\sigma_0 - c < 0$, $g$ decays exponentially and $g \in L^1(\mathbb{R})$.
The Fourier transform of $g$ is
\begin{align*}
\hat{g}(\omega) = \int_{-\infty}^\infty g(t) e^{-i\omega t} \, dt = \int_0^\infty f(t) e^{-(c + i\omega)t} \, dt = \hat{f}(c + i\omega),
\end{align*}
where the last equality holds because $c + i\omega$ has real part $c > \sigma_0$, placing it in the domain of convergence of the Laplace transform.
[guided]
The idea is to exploit the relationship between the Laplace and Fourier transforms.
The Laplace transform $\hat{f}(p) = \int_0^\infty f(t) e^{-pt} \, dt$ evaluated at $p = c + i\omega$ becomes $\int_0^\infty f(t) e^{-ct} e^{-i\omega t} \, dt$, which is exactly the Fourier transform of the damped function $g(t) = f(t)e^{-ct}$ (extended by zero for $t < 0$).
Why do we need $c > \sigma_0$? The exponential order condition $|f(t)| \leq Me^{\sigma_0 t}$ means $f$ may grow exponentially as $t \to \infty$.
The damping factor $e^{-ct}$ with $c > \sigma_0$ ensures $|g(t)| \leq Me^{(\sigma_0 - c)t}$ for $t \geq 0$, and since $\sigma_0 - c < 0$, this decays exponentially:
$\int_0^\infty |g(t)| \, dt \leq M \int_0^\infty e^{(\sigma_0 - c)t} \, dt = M/(c - \sigma_0) < \infty$.
Hence $g \in L^1(\mathbb{R})$.
Without sufficient damping ($c \leq \sigma_0$), the function $g$ would not be integrable and its Fourier transform would not exist in the classical sense.
The parameter $c$ controls the position of the Bromwich contour: larger $c$ provides stronger damping but moves the contour further to the right in the complex plane.
[/guided]
[/step]
[step:Apply the Fourier inversion theorem to recover $g(t)$]
By the [Fourier inversion theorem](/theorems/364) (applicable since $g \in L^1(\mathbb{R})$), at every point of continuity of $g$ (equivalently, of $f$):
\begin{align*}
g(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{g}(\omega) e^{i\omega t} \, d\omega = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(c + i\omega) e^{i\omega t} \, d\omega.
\end{align*}
[/step]
[step:Substitute $p = c + i\omega$ to obtain the Bromwich contour integral]
Perform the change of variable $p = c + i\omega$, so $dp = i \, d\omega$ (hence $d\omega = dp / i$), $e^{i\omega t} = e^{i(p-c)t/i \cdot i} = e^{(p-c)t}$, and the integration path $\omega \in (-\infty, \infty)$ becomes the vertical line $\operatorname{Re} p = c$ traversed from $c - i\infty$ to $c + i\infty$:
\begin{align*}
g(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(c + i\omega) e^{i\omega t} \, d\omega = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \hat{f}(p) e^{(p - c)t} \, dp.
\end{align*}
Multiplying both sides by $e^{ct}$ and using $f(t) = g(t) e^{ct}$:
\begin{align*}
f(t) = e^{ct} g(t) = \frac{e^{ct}}{2\pi i} \int_{c - i\infty}^{c + i\infty} \hat{f}(p) e^{(p-c)t} \, dp = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \hat{f}(p) e^{pt} \, dp.
\end{align*}
[guided]
The substitution $p = c + i\omega$ is the bridge between Fourier analysis on the real line and contour integration in the complex plane. Under this substitution: $\omega = (p - c)/i$, so $d\omega = dp/i$, and $e^{i\omega t} = e^{i \cdot \frac{p-c}{i} \cdot t}$... more carefully: $\omega = \operatorname{Im}(p)$, $p = c + i\omega$, $dp = i \, d\omega$, hence $d\omega = dp/i$.
The Fourier inversion integral $g(t) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(c + i\omega) e^{i\omega t} \, d\omega$ becomes, after substituting $d\omega = dp/i$:
\begin{align*}
g(t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \hat{f}(p) e^{(p-c)t} \, dp.
\end{align*}
The integration path changes from $\omega \in (-\infty, \infty)$ to the vertical line $\operatorname{Re} p = c$, traversed from $c - i\infty$ to $c + i\infty$. Multiplying both sides by $e^{ct}$ and using $f(t) = g(t)e^{ct}$: $f(t) = \frac{e^{ct}}{2\pi i}\int_{c-i\infty}^{c+i\infty} \hat{f}(p) e^{(p-c)t} \, dp = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \hat{f}(p) e^{pt} \, dp$, since $e^{ct} \cdot e^{(p-c)t} = e^{pt}$.
The requirement $c > \sigma_0$ ensures the vertical contour lies in the half-plane $\operatorname{Re} p > \sigma_0$ where $\hat{f}(p)$ converges absolutely, and to the right of all singularities of $\hat{f}$.
[/guided]
[/step]
