Why is Newman's Analytic Theorem neccessary In a proof of the prime number theorem along the lines of Newman's, we establish that $-\frac {\zeta'(s)}{\zeta(s)}-\frac 1{s-1}$ possesses an analytic continuation to $\Re(s)\ge 1$ and that $\psi(x)=O(x)$ and then use an auxiliary Tauberian theorem, which Newman calls the Analytic Theorem, to show that the integral $$\int_1^\infty \frac{\psi(x)-x}{x^2}\; dx=\int_0^\infty e^{-t}\psi(e^t)\; dt$$ converges. My question is this: why is the Analytic Theorem necessary? Can't we simply say that since $$-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^\infty \frac{\psi(x)}{x^{s+1}}\; dx$$ by Abel's summation formula (this formula is used in Newman's proof) then we have $$-\frac{\zeta'(s)}{\zeta(s)}-\frac 1{s-1}-1=s\int_1^\infty \frac{\psi(x)-x}{x^{s+1}}\; dx.$$ Now, the function $-\frac{\zeta'(s)}{\zeta(s)}-\frac 1{s-1}$ without analytic continuation cannot be evaluated directly at $s=1$; however, the limit from above does exist and is bounded. Therefore the singularity at $s=1$ is a removable singularity, and as such we may define the value of $-\frac{\zeta'(s)}{\zeta(s)}-\frac 1{s-1}$ at $s=1$ by this limit. In the above formula, it is clear that either $\psi(x)\sim x$ or the integral on the right-hand side diverges to $\pm\infty$ at $s=1$ (by the bound $\psi(x)=O(x)$). However, we know that the left-hand side is bounded at $s=1$ by the above argument. We conclude that $\psi(x)\sim x$.
What is the problem with this reasoning that makes it necessary to use Newman's Analytic Theorem?
 A: I see two problems. First, from the existence of
$$\lim_{u\to 0} \int_0^\infty f(t) e^{-ut}\,dt,$$
where $u$ approaches $0$ from the right half-plane, it does not follow that
$$\int_0^\infty f(t)\,dt$$
exists. Consider $f(t) = \sin t$. We have
$$\int_0^\infty e^{-ut}\sin t\,dt = \frac{1}{1+u^2},$$
so the limit exists, but $\int_0^\infty \sin t\,dt$ does not exist.
And second, from $\psi \in O(x)$, it does not follow that either $\psi(x) \sim x$ or
$$\int_1^\infty \frac{\psi(x) - x}{x^{s+1}}\,dx$$
diverges to $+\infty$ or $-\infty$ at $s = 1$. If we take for example
$$g(x) = x + \frac{x}{4}\sin (\log x),$$
then $g$ is monotonically increasing, $g\in O(x)$ but $g(x) \nsim x$, and
$$\int_1^\infty \frac{g(x)-x}{x^2}\,dx = \frac{1}{4}\int_1^\infty \frac{\sin (\log x)}{x}\,dx$$
does not diverge to either $+\infty$ or $-\infty$, it oscillates between $0$ and $2$. But, see above, the limit
$$\lim_{s\searrow 1} \int_1^\infty \frac{g(x)-x}{x^{s+1}}\,dx$$
exists. To conclude $\psi(x)\sim x$, we need something stronger. The "analytic theorem" in Newman's proof is that "something stronger" that gives the desired conclusion.
