Zagier's proof of the prime number theorem. In Zagier's paper, "Newman's Short Proof of the Prime Number Theorem", (link below) his theorem ${\bf (V) }$ states that,
$$ \int_{1}^{\infty} \frac{\vartheta(x) - x}{x^2} dx \text{ is a convergent integral.} $$
Note: $\vartheta(x) = \sum_{p \le x} \log(p)$, where $p$ is a prime.
Zagier proceeds to say that, for $\Re(s) > 1$ we have
$$\sum_{p} \frac{\log p}{p^s} = \int_{1}^{\infty} \frac{ d \vartheta(x)}{x^s} = s \int_{1}^{\infty} \frac{ \vartheta(x)}{x^{s+1}} dx = s \int_{0}^{\infty} e^{-st} \vartheta(e^{t})dt. $$
My question is how the 2nd equality holds. Using integration by parts, it's easily verified that
$$ \int_{1}^{\infty} \frac{d \vartheta(x)}{x^s} = x^{-s} \vartheta(x) |_{x=1}^{x=\infty} + s \int_{1}^{\infty} \frac{\vartheta(x)}{x^{s+1}}dx. $$
As this theorem is used to show that $\vartheta(x) \sim x$, I do not understand how we can claim that for $\Re(s) > 1,$
$$\lim_{x \rightarrow \infty} x^{-s} \vartheta(x) = 0.$$
http://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf

In step $\mathbf{III}$, it was shown that $\vartheta(x) \leqslant C\cdot x$ for some constant $C$. That is enough to ensure
$$\lim_{x\to\infty} x^{-s}\vartheta(x) = 0$$
for $\Re s > 1$.
