---
tag: X2H35BP974
kind: statement
role: theorem
name: Mellin's inverse formula
msc: [44A10]
uses: [05XT5AVJBK, 9WRE8KEJJX, F34S6P61SW, T4Z8YQZKZV]
about: [07WBTX1MHA, 5KNAXNHRY6]
sources:
  - url: https://en.wikipedia.org/wiki/Inverse_Laplace_transform
---

## Statement

Let $f \colon [0, \infty) \to \C$ be integrable on each interval $[0, T]$, and let $\gamma$ be a real number with $\int_0^\infty \abs{f(t)} e^{-\gamma t} \, dt < \infty$.
Let $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.
Let $t > 0$, and assume that $f$ has [[bounded variation|F34S6P61SW]] on some interval around $t$.
Then the [[Bromwich integral|9WRE8KEJJX]] of $F$ along the line $\operatorname{Re} s = \gamma$, that is, the symmetric limit below, exists at $t$, and
$$\frac{1}{2 \pi i} \lim_{T \to \infty} \int_{\gamma - i T}^{\gamma + i T} e^{s t} F(s) \, ds = \frac{f(t^+) + f(t^-)}{2}.$$
In particular the integral equals $f(t)$ at each $t > 0$ where $f$ is continuous.
For every $t < 0$ the same integral exists and equals $0$.

## 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|T4Z8YQZKZV]] 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.

## Explanations

### L1 human

If you know the Laplace transform $F$ of a function, this formula gives the function back.
You integrate $e^{s t} F(s)$ along a vertical line far enough to the right in the complex plane.
At a jump, the formula returns the average of the values on the two sides.

### L2 human

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|E14W2V09HM]].
The same formula, after the change of variables $x = e^{-t}$, gives the inversion theorem for the Mellin transform.