On a chapter on Dirichlet series, there is this lemma:
Lemma: if $f \in C^\infty$ and $\forall n, k, |f^{(n)}(t)| =o(t^k)$ when $t \to +\infty $ then
$M(f, s) = \frac{1}{\Gamma(s)} \int_0^\infty f(t)t^s \frac{dt}{t} $, defined for $Re(s) > 0$ has an holomorphic prolongation on $\mathbb{C}$
Demonstration (steps only without gory details):
Let $\phi \in C^\infty$ on $\mathbb{R}^+$ such that $\phi = 1$ on $[0, 1]$ and $\phi = 0$ on $[2, +\infty [$
$M((1-\phi)f, s)$ is holomorphic on $\mathbb{C}$, that's the "easy" part.
Also, $M(\phi f, s) = (-1)^n M((\phi f)^{(n)}, s+n)$ by an integration by parts. So, $M(\phi f, s)$ has an holomorphic prolongation on $Re(s) > -n$ and thus on $\mathbb{C}$.
my question: Why bother with $\phi$ and not say directly that $M(f, s) = (-1)^n M(f^{(n)}, s+n)$? What am I missing?
There is IMO a more readable proof :
If $\forall k$, as $t \to \infty $ : $f(t) = \mathcal{O}(t^{-k})$ then $F(s) = \int_1^\infty f(t) t^{s-1}dt$ converges absolutely and is analytic for $\Re(s) < k$ and hence it is entire.
Now it doesn't tell us anything of $G(s) = \int_0^1 f(t) t^{s-1}dt$ and this is what we are looking at.
Since $f$ is $C^\infty$ at $0$ we have as $t \to 0$ : $f(t) - \sum_{n=0}^k \frac{f^{(n)}(0)}{n!}t^k = \mathcal{O}(t^{k+1})$ so that $$G_k(s)=\int_0^1 (f(t) - \sum_{n=0}^k \frac{f^{(n)}(0)}{n!}t^k)t^{s-1}dt= G(s)-\sum_{n=0}^k \frac{f^{(n)}(0)}{n!} \frac{1}{s+k}$$ converges absolutely and is analytic for $\Re(s) > -k-1$.
Since this is true for every $k$, $G(s)$ is analytic on $\mathbb{C} \setminus -\mathbb{N}$ and $\frac{G(s)}{\Gamma(s)}$ is entire, as well as $\frac{F(s)+G(s)}{\Gamma(s)}$.