Fourier transform as Laplace transform on the imaginary axis

Remarks

In general $\mathcal{B}f(\sigma + 2\pi i \xi) = \widehat{f_\sigma}(\xi)$ with $f_\sigma(x) = e^{-\sigma x} f(x)$, by writing $e^{-(\sigma + 2\pi i \xi) x} = e^{-\sigma x} e^{-2\pi i x \xi}$. Example: $f(x) = e^{-\abs{x}}$ is integrable, and $\mathcal{B}f(s) = \frac{1}{1 + s} + \frac{1}{1 - s} = \frac{2}{1 - s^2}$ for $\abs{\operatorname{Re} s} < 1$. At $s = 2\pi i \xi$ this gives $\hat f(\xi) = 2/(1 + 4\pi^2 \xi^2)$, the known Fourier transform of $e^{-\abs{x}}$. Non-example: $f(x) = e^{x}$ for $x \ge 0$ and $f(x) = 0$ for $x < 0$ is not integrable, so the imaginary axis is not in its region of convergence. Here $\mathcal{B}f(s) = 1/(s - 1)$ for $\operatorname{Re} s > 1$, and this function is not the Fourier transform of $f$, which does not exist as an integral. With the convention $\int f(x) e^{-i \omega x} \, dx$, the same relation reads $\hat f(\omega) = \mathcal{B}f(i\omega)$.