---
tag: E9Y670Y3QV
kind: statement
role: theorem
name: Differentiability of the Mellin transform
msc: [44A15, "30E20"]
uses: [SGZZ4Y0D36, N5PVGRJE1T, 0TGV80E6A4, 0NAHXR1FRH, X7GTF3D98F]
about: [N5PVGRJE1T]
sources:
  - url: https://en.wikipedia.org/wiki/Mellin_transform
---

## Statement

Let $f \colon (0, \infty) \to \C$ be integrable on every compact subinterval of $(0, \infty)$.
Let $a, b \in \R$ with $b < a$, and assume that $f(x) = O(x^{-a})$ as $x \to \infty$ and $f(x) = O(x^{-b})$ as $x \to 0^+$, in [[big O notation|0TGV80E6A4]].
Let $S = \set{s \in \C : b < \operatorname{Re} s < a}$.
Then:

1. For every $s \in S$, the function $x \mapsto (\log x) f(x)$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$.
2. The [[Mellin transform|N5PVGRJE1T]] $\mathcal{M}f$ is complex differentiable at every $s \in S$, with
$$(\mathcal{M}f)'(s) = \int_0^\infty x^{s-1} (\log x) f(x) \, dx = \mathcal{M}[(\log x) f(x)](s).$$

In particular $\mathcal{M}f$ is [[holomorphic|0NAHXR1FRH]] on the open strip $S$.

## Remarks

The derivative of $x^{s-1}$ in $s$ is $x^{s-1} \log x$.
Fix $s \in S$ and $\varepsilon > 0$ with $b < \operatorname{Re} s - 2\varepsilon$ and $\operatorname{Re} s + 2\varepsilon < a$.
For $\abs{z - s} \le \varepsilon$, the bound $\abs{\log x} \le \varepsilon^{-1} (x^{\varepsilon} + x^{-\varepsilon})$ dominates $\abs{x^{z-1} (\log x) f(x)}$ by a function that is integrable by the estimates of the fundamental strip.
So [[differentiation under the integral sign|0AR06W1E1B]] applies.
Iterating gives $(\mathcal{M}f)^{(n)}(s) = \mathcal{M}[(\log x)^n f(x)](s)$ for all $n \ge 0$.
Example: for $f(x) = e^{-x}$ the statement gives $\Gamma'(s) = \int_0^\infty x^{s-1} (\log x) e^{-x} \, dx$ for $\operatorname{Re} s > 0$, and at $s = 1$ this is $\Gamma'(1) = -\gamma$, with $\gamma$ the Euler–Mascheroni constant.
The open strip is needed: for $f(x) = 1/(1 + x)$, $\mathcal{M}f(s) = \pi / \sin(\pi s)$ on $0 < \operatorname{Re} s < 1$, and this function has poles at $s = 0$ and $s = 1$, the edges of the strip.

## Explanations

### L1 human

Under the usual bounds on $f$ at $0$ and at $\infty$, the Mellin transform is a smooth function of $s$ in the complex sense.
Its derivative is the Mellin transform of $f$ multiplied by $\log x$.

### L2 human

Holomorphy on the strip is what makes Mellin methods work: one can continue $\mathcal{M}f$ analytically beyond the strip, shift lines of integration, and pick up residues.
For instance the gamma function is first defined on $\operatorname{Re} s > 0$ as a Mellin transform, holomorphic there by this result, and then continued meromorphically to $\C$.
The same holds for the completed zeta function $\pi^{-s/2} \Gamma(s/2) \zeta(s)$, which is a Mellin transform of a theta series.