We can define the prime number function as $$\pi(x) = \int_{-\infty}^x \sum_{p}\delta(p-x).dx$$ That is, we include each prime p as a delta function $\delta_p(x) = \delta(p-x)$, set $P(x) = \sum_{p}\delta_p(x)$, and then integrate $P(x)$ from $-\infty$ to $x$ to calculate $\pi(x)$.
In this form, $\pi(x)$ is a step-wise function, discontinuous at each prime p, and constant between consecutive primes.
I wondered whether it might be possible to get a smooth approximation to $\pi(x)$ by smoothly approximating $P(x)$. For example, consider $$q_p(x) = (1/\sqrt(2\pi))\exp(-(x-p)^2/2),$$ and then let $Q(x) = \sum_{p}q_p(x)$. So $Q(x)$ is a function on the real line that has normal distribution "bumps" at each prime, and is close to zero between primes.
I will admit to not having gone through the details, but it feels intuitively clear that $Q(x)$ is a well-defined function, that is, $\sum_{p}q_p(x)$ converges for all $x$ because $\exp(-(x-p)^2/2) < (x-p)^{-2}$.
Similarly, I am pretty sure that $q(x) = \int_{-\infty}^x Q(x).dx$ is well-defined for all $x$. But for clarity, I'd like to understand the following questions:
- Is $Q(x)$ well-defined?
- Is $Q(x)$ continuous?
- Is $Q(x)$ smooth?
- Is $q(x) = \int_{-\infty}^x Q(x).dx$ well-defined?
- Is $\pi(x) \sim q(x)$?
Apologies if these are dumb or naive questions, but I wonder if functions like $q(x)$ provide a handle on $\pi(x)$ that allow for a different angle of attack on $\pi(x)$.
