Lemma 1.1.1. on page 4 of these notes sates that for a $\mathcal{C^{\infty}}$-smooth function $f:\mathbb{R}_{+}\rightarrow\mathbb{C}$ such that $\lim_{x\rightarrow\infty}x^{n}f\left(x\right)=0$ for all integers $n\geq0$, then the Mellin transform: $$L\left(f,s\right)=\frac{1}{\Gamma\left(s\right)}\int_{0}^{\infty}x^{s-1}f\left(x\right)dx,\textrm{ Re}\left(s\right)>0$$ has a continuation to an analytic function of $s\in\mathbb{C}$.
However, a bit of scrutiny reveals that something is amiss. Let's take, for example (as done in the notes) $$f\left(x\right)=\frac{x}{e^{x}-1}$$ then, as is well-known: $$\zeta\left(s\right)=\frac{1}{\Gamma\left(s\right)}\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$$ and so: $$L\left(\frac{x}{e^{x}-1},s\right)=\frac{1}{\Gamma\left(s\right)}\int_{0}^{\infty}\frac{x^{s}}{e^{x}-1}dx=s\zeta\left(s+1\right)$$
Now, let $\mathbb{P}$ be the set of prime numbers, and let: $$P\left(x\right)=x\sum_{p\in\mathbb{P}}e^{-px}$$ Then, note that for all $x>0$: $$P\left(x\right)\leq x\sum_{n=1}^{\infty}e^{-nx}=x\frac{e^{-x}}{1-e^{-x}}=\frac{x}{e^{x}-1}$$ Thus, the smoothness of $\frac{x}{e^{x}-1}$ for all $x\in[0,\infty)$ then guarantees the same for $P\left(x\right)$ (the series being uniformly convergent on the positive real numbers). Moreover, for all integers $n\geq0$, L'Hôpital's rule gives: $$\lim_{x\rightarrow\infty}x^{n}P\left(x\right)\leq\lim_{x\rightarrow\infty}\frac{x^{n+1}}{e^{x}-1}=\left(n+1\right)\lim_{x\rightarrow\infty}x^{n}e^{-x}=0$$ Thus, $P\left(x\right)$ satisfies all the hypotheses of the above Lemma.
However: $$L\left(P,s\right)=\frac{1}{\Gamma\left(s\right)}\sum_{p\in\mathbb{P}}\int_{0}^{\infty}x^{s}e^{-px}dx
=\frac{1}{\Gamma\left(s\right)}\sum_{p\in\mathbb{P}}\frac{\Gamma\left(s+1\right)}{p^{s+1}}
=s\zeta_{\mathbb{P}}\left(s+1\right)$$
where: $$\zeta_{\mathbb{P}}\left(s\right)=\sum_{p\in\mathbb{P}}\frac{1}{p^{s}}$$ is the Prime Zeta Function. According to the Lemma, $s\zeta_{\mathbb{P}}\left(s+1\right)$ should then be an analytic function for all $s\in\mathbb{C}$. However, it is known that the Prime Zeta Function has a natural boundary at the line $\textrm{Re}\left(s\right)=0$; $\zeta_{\mathbb{P}}\left(s\right)$ cannot be analytically continued to any domain in $\mathbb{C}$ larger than the open right half-plane bounded by this line. This contradicts the Lemma.
What's going on here? Is my reasoning somehow flawed, or is the Lemma to blame? Moreover, if it's the Lemma that's at fault, how might the Lemma be modified in order to produce a reliable criterion on f to guarantee the existence of an analytic continuation of $L\left(f,s\right)$?
Best Answer
As you see $x\sum_p e^{-px}$ has a complicated singularity at $0$, which prevents from an easy analytic continuation of $\sum_p p^{-s}$.
If $f_m(x) =x^m \sum_{n\ge 1} a_n e^{-nx}$ is smooth at $0$ with $F(s)=\sum_{n \ge 1} a_n n^{-s}$ then $$\Gamma(s+m)F(s+m)-\sum_{k=0}^K \frac{f_m^{(k)}}{k! (s+k)} = \int_0^\infty x^{s-1} (f_m(x)-1_{x < 1}\sum_{k=0}^K \frac{f_m^{(k)}}{k!} x^k)dx$$ is analytic for $\Re(s) > -K-1$
The converse is a Tauberian theorem : if the analytic continuation of $\Gamma(s+m)F(s+m)-\sum_{k=0}^K \frac{c_k}{k! (s+k)}$ exists and is analytic for $\Re(s)> -K-1$ and $F(s)$ has at most polynomial growth on vertical lines then $f_m(x)=\sum_{k=0}^K \frac{f_m^{(k)}}{k!} x^k + O(x^{K+1-\epsilon})$ as $x \to 0^+$. For L-functions the growth on vertical lines follows from the functional equation plus if needed Phragmén–Lindelöf for the critical strip.