---
tag: 9C8SS4522J
kind: statement
role: proposition
name: Cahen–Mellin integral
msc: [44A15, 33B15, "30E20"]
uses: [0H5SYBSED9, CHESR1W1CY, 0RNGDMHT7M]
about: [0H5SYBSED9, CHESR1W1CY]
sources:
  - url: https://en.wikipedia.org/wiki/Mellin_transform
---

## Statement

Let $c > 0$ and let $y \in \C$ with $\operatorname{Re} y > 0$.
Let $y^{-s} = e^{-s \log y}$, with the principal branch of $\log y$.
Then
$$e^{-y} = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \Gamma(s) \, y^{-s} \, ds,$$
where $\Gamma$ is the [[gamma function|0H5SYBSED9]] and the integral converges absolutely.

## Remarks

For real $y > 0$ this is the [[inverse Mellin transform|CHESR1W1CY]] of $\Gamma = \mathcal{M}[e^{-x}]$, so it follows from the Mellin inversion theorem.
For complex $y$, put $\abs{\arg y} = \theta < \pi/2$; then $\abs{y^{-c - it}} = \abs{y}^{-c} e^{t \arg y} \le \abs{y}^{-c} e^{\theta \abs{t}}$, while $\abs{\Gamma(c + it)}$ decays like $\abs{t}^{c - 1/2} e^{-\pi \abs{t}/2}$ by Stirling's formula.
So the integral converges absolutely and locally uniformly, it is holomorphic in $y$, and the identity extends from $y > 0$ by the identity theorem.
A direct proof moves the line of integration to the left: by the [[residue theorem|032HKYS329]], the poles of $\Gamma$ at $s = -k$ with residues $(-1)^k / k!$ give $\sum_{k \ge 0} (-1)^k y^k / k! = e^{-y}$.
The hypothesis $c > 0$ is needed: for $-1 < c < 0$ the line passes to the left of the pole at $0$, and the integral equals $e^{-y} - 1$.

## Explanations

### L1 human

The exponential function $e^{-y}$ can be written as an integral of powers $y^{-s}$ against the gamma function along a vertical line.
It is the inverse of the fact that the gamma function is the Mellin transform of $e^{-x}$.

### L2 human

This integral is the prototype of [[Mellin–Barnes integrals|1PDGBGEHM2]] and of the method of shifting contours.
In analytic number theory it is applied to $y = n x$ and summed over $n$: the sum $\sum_n a(n) e^{-n x}$ then becomes $\frac{1}{2\pi i} \int \Gamma(s) F(s) x^{-s} \, ds$ with $F$ the Dirichlet series of the $a(n)$.
Shifting the line past the poles of $\Gamma(s) F(s)$ gives asymptotic expansions of the exponential sum as $x \to 0^+$.
It is the smooth counterpart of Perron's formula, which uses the kernel $x^{s}/s$ instead of $\Gamma(s) x^{-s}$ and has worse convergence.