Bromwich integral

Remarks

Example: let $F(s) = 1/s$ on $\operatorname{Re} s > 0$, and $\gamma > 0$. For $t > 0$ one closes the segment by a large arc to the left, whose contribution tends to $0$. The residue of $e^{s t}/s$ at $0$ is $1$, so the integral is $1$. For $t < 0$ one closes to the right, where there are no poles, and gets $0$; for $t = 0$ the symmetric limit is $\frac{1}{2 \pi i} \lim_{T \to \infty} \left( \log(\gamma + i T) - \log(\gamma - i T) \right) = \tfrac{1}{2}$. Non-example: for $F(s) = 1$ and $t \ne 0$, the integral over the segment equals $e^{\gamma t} \sin(T t) / (\pi t)$, which has no limit as $T \to \infty$; so the Bromwich integral does not exist. The limit is a symmetric one; it can exist even when $e^{s t} F(s)$ is not absolutely integrable on the line. The integral is also called the Fourier–Mellin integral. When $F$ is the Laplace transform of a suitable function $f$, the integral equals $f(t)$ at each $t > 0$ where $f$ is continuous; this is Mellin's inverse formula.