Why does the Newman/Zagier proof of the PNT invoke complex analysis? I was thinking, specifically, of this paper, in which Zagier offers a proof of the PNT, inspired by a paper of Newman's, the Cliffs Notes version of which would be that, first, it's fairly easy to show that
$$\left|\int_1^{\infty}\frac{\vartheta(x) - x}{x^2}dx\right| < \infty \implies \vartheta(x) \sim x$$
As well as that
$$\lim_{s\to1^+}\int_1^{\infty}\frac{\vartheta(x) - x}{x^{1+s}}dx = \lim_{s\to1^+}\left(\sum\frac{((1-s)p^s - 1)\log p}{sp^s(p^s - 1)} - \frac{d}{ds}\log((s-1)\zeta(s))\right)$$
which converges.  What Zagier does from here to prove the convergence of the integral itself is that he uses a particular Tauberian theorem for a slightly rewritten integral, whose preconditions take a fair amount of complex analysis to justify.
What I was wondering was, from this step, since the absolute value of the integrand will at every point be increasing as $s$ decreases, couldn't you prove its convergence more easily with the monotone convergence theorem?  Since Zagier is a world-famous mathematician and I collect academic suspensions like expensive pogs, I'm guessing the answer is "no," but I'd like to know why.
 A: The monotone convergence theorem requires monotone - the integrand should, as a function of $s$, be monotonic in the same direction for all $x$ - convergence, it is not sufficient that the absolute value is monotonically increasing at each point. Consider
$$\int_0^{\infty} e^{-sx}\cos x\,dx = \frac{s}{1+s^2}$$
for $s > 0$. Clearly the absolute value of the integrand is increasing at each point for $s \to 0$, and the limit of the integrals also exists (and is $0$), but of course
$$\lim_{a \to \infty} \int_0^a \cos x\,dx$$
does not exist.
The existence of
$$\lim_{s \searrow 0} \int_0^{\infty} e^{-sx} f(x)\,dx$$
is the integration analogue of Abel summability for series, and just like not all Abel-summable series are convergent, not all Abel-integrable functions are improperly Riemann-integrable.
Newman's theorem gives a - relatively mild - sufficient condition for the existence of the improper Riemann integral. Conversely, if $f$ is a bounded function such that the improper Riemann integral
$$\int_0^{\infty} f(x)\,dx$$
exists, then it is not too hard (but on the other hand neither is it obvious) to show that the holomorphic function defined for $\operatorname{Re} s > 0$ by
$$F(s) = \int_0^{\infty} e^{-sx}f(x)\,dx \tag{$\ast$}$$
cannot have a meromorphic continuation with a pole on the imaginary axis. Thus Newman's theorem does in some sense not require "much" more than necessary.
To see that $F$ as defined in $(\ast)$ cannot have poles on the imaginary axis if the improper integral of $f$ exists, one shows that for every fixed $t \in \mathbb{R}$ one has $\lim\limits_{\sigma \searrow 0} \sigma\cdot F(\sigma + it) = 0$. For $y \geqslant 0$ define
$$R(y) := \int_y^{\infty} f(x)\,dx$$
and $S(y) = \sup \:\{ \lvert R(x)\rvert : x \geqslant y\}$, and integrate by parts in $(\ast)$:
\begin{align}
F(s) &= \int_0^{\infty} e^{-sx}\bigl(-R'(x)\bigr)\,dx \\
&= \bigl(-e^{-sx}R(x)\bigr)\biggr\rvert_0^{\infty} - s\int_0^{\infty} e^{-sx}R(x)\,dx \\
&= R(0) - s\int_0^{\infty} e^{-sx}R(x)\,dx\,.
\end{align}
(Although $R$ is in general not differentiable everywhere, so replacing $f(x)$ by $-R'(x)$ is not valid everywhere, the integration by parts is correct.) Then for fixed $t \in \mathbb{R}$, all $\sigma > 0$ and arbitrary $y \in (0,\infty)$ one has
\begin{align}
\lvert \sigma F(\sigma + it)\rvert
&\leqslant \sigma \lvert R(0)\rvert + \lvert\sigma + it\rvert\cdot \sigma \int_0^{\infty} e^{- \sigma x} \lvert R(x)\rvert\,dx \\
&\leqslant \sigma \lvert R(0)\rvert + \sigma\lvert \sigma + it\rvert \int_0^y \lvert R(x)\rvert\,dx + S(y)\lvert \sigma + it\rvert \underbrace{\int_y^{\infty} \sigma e^{-\sigma x}\,dx}_{{} = e^{-\sigma y}}
\end{align}
and therefore
$$\limsup_{\sigma \searrow 0} \lvert \sigma F(\sigma + it)\rvert \leqslant S(y)\lvert t\rvert.$$
Since $\lim_{y \to \infty} S(y) = 0$, the assertion follows.
