Differentiability of the Mellin transform

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 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.

Going deeper

For a student

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.