Along the lines of thought given here, is it in general possible to substitute a summation over a function $f$ of primes like the following:
$$
\sum_{p\le x}f(p)=\int_2^x f(t) d(\pi(t))\tag{1}
$$
and further
$$
\int_2^x f(t) d(\pi(t))=f(t)\pi(t)\biggr|_2^{x}-\int_2^{x}f'(t)\pi(t)dt\tag{2}.
$$
If it's possible only for some cases, how can one specify them? answered in the comments
Let's continue from $(2)$ with an representation of the prime counting function:
$$
\pi(t) = \operatorname{R}(t^1) – \sum_{\rho}\operatorname{R}(t^{\rho}) \tag{3}
$$
with $ \operatorname{R}(u) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(u^{1/n})$ (the so-called Riemann's ${\rm R}$ Function, see e.g. $(11)$ here) and $\rho$ running over all the zeros (trivial and non-trivial) of $\zeta$ function. $\operatorname{li}(\cdot)$ is the logarithmic integral.
So we have
$$
\begin{eqnarray}
&=&f(t)\pi(t)\biggr|_2^{x}-\int_2^{x}f'(t)\pi(t)dt\\
&=&f(t)\left(\operatorname{R}(t^1) – \sum_{\rho}\operatorname{R}(t^{\rho})\right)\biggr|_2^{x}
-\int_2^{x}f'(t)\left(\operatorname{R}(t^1) – \sum_{\rho}\operatorname{R}(t^{\rho})\right)dt \phantom{somemorerspace}\\
&\phantom{AA}&\\
&=&\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\Big\{\left[f(t)\left( \sum_{z\in\{1,\rho\}} (-1)^{1-\delta_{1z}} \operatorname{li}(t^{z/n})\right)\right]_2^{x}\\
&&- \int_2^{x}f'(t)\left( \sum_{z\in\{1,\rho\}} (-1)^{1-\delta_{1z}} \operatorname{li}(t^{z/n})\right)dt\Big\}\\
&\phantom{AA}&\\
&=&\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}\left\{\left[f(t)\left( \operatorname{li}(t^{z/n})\right)\right]_2^{x} –
\int_2^{x}f'(t)\left( \operatorname{li}(t^{z/n})\right)dt\right\}\hskip0.9in(4)
\end{eqnarray}
$$
where I tried to combine the sum a little without introducing to much confusion by using
$$
(-1)^{1-\delta_{1z}}=
\cases{
+1&$ \text{if } z=1$\\
-1&$ \text{if } z=\rho$\\
}
$$
Now, what if we just take an approximation $\tilde{\pi}(t)$, where the sums over $n$ and $\rho$ are truncated. Is this approach still valid? I'm worried because $\tilde{\pi}(t)$ might not be monotone, which is a prerequisite of the Lebesgue-Stieltjes integration.
Let's work out the last integral, by parts:
We use
$$
\int_2^{x}f'(t) \operatorname{li}(t^{w})dt
=\left[ f(t)\operatorname{li}(t^{w}) \right]_2^x – \int_2^x \frac{f(t)wt^{w-1}}{\ln(t^w)}dt
\tag{5}
$$
which gives a nice result when $f(t)=t^{-s}$, see here:
$$
\int_2^{x}(-st^{-s-1}) \operatorname{li}(t^{w})dt
=\left[ t^{-s}\operatorname{li}(t^{w}) \right]_2^x – \int_2^x \frac{t^{-s}t^{w-1}}{\ln(t)}dt
=\left[ t^{-s}\operatorname{li}(t^{w}) \right]_2^x – \left[{\rm li}(t^{w-s})\right]^x_2.
$$
So overall we get
$$
\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}\left\{\left[f(t) \operatorname{li}(t^{z/n})\right]_2^{x}-
\left[ f(t)\operatorname{li}(t^{z/n}) \right]_2^x +\int_2^x \frac{zf(t)t^{z/n-1}}{n\ln(t^{z/n})}dt \right\}\\
=\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}\left\{\int_2^x \frac{f(t)t^{z/n-1}}{\ln(t)}dt \right\}\tag{6}\\
$$
and in the special case $f(t)=t^{-s}$ this simplifies to
$$
P_\color{red}x(\color{blue}s)=\sum_{p<\color{red}x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}
\left[
{\rm li}(t^{\frac zn-\color{blue}s})
\right]^{\color{red}x}_2
\tag{7}
$$
(for the interested reader: the story continues here…)
If anybody could confirm this, it would be ever so cool.
Thanks for your help and your time for reading all this,
Best Answer
If your function $f$ is smooth and compactly supported then the formula you are looking for already exists, and is called the "explicit formula". See for example Lemma 1 in http://arxiv.org/abs/math/0511092.
If you want to apply this lemma in the direction "primes to zeros" then you should "swap the hats" over $h$. Basically, once you have specified $\hat{h}$ to be a smooth compact function of your choice, the resulting function $h$ will be entire and you will be able to apply the lemma 1 to $h$, getting the desired formula a sum over the primes weighted by $\hat{h}$.