Mellin's inverse formula

Remarks

Put $g(\tau) = f(\tau) e^{-\gamma \tau}$ for $\tau \ge 0$ and $g(\tau) = 0$ for $\tau < 0$; then $g$ is integrable on $\R$. On the line $s = \gamma + i y$, one has $F(\gamma + i y) = \hat{g}(y) = \int_\R g(\tau) e^{-i y \tau} \, d\tau$, and the Bromwich integral is $e^{\gamma t} \cdot \frac{1}{2\pi} \lim_{T \to \infty} \int_{-T}^{T} \hat{g}(y) e^{i y t} \, dy$. So the statement is the Fourier inversion theorem in its Dirichlet–Jordan form, applied to $g$. Example: for $f(t) = e^{b t}$ with $b$ real and $\gamma > b$, $F(s) = 1/(s - b)$, and the residue of $e^{s t}/(s - b)$ at $b$ gives back $e^{b t}$ for $t > 0$. At a jump of $f$ the integral gives the midpoint of the jump, not either one-sided value.

Going deeper

For a student

The formula shows that the Laplace transform is a Fourier transform in disguise: along a vertical line it is the Fourier transform of $f(t) e^{-\gamma t}$. In practice one rarely integrates along the line directly; one deforms the contour and uses residues, as in residue calculus for Laplace inversion. The same formula, after the change of variables $x = e^{-t}$, gives the inversion theorem for the Mellin transform.