Magnitude of the series over primes $p$ of exponential functions $\sum_{p} e^{-pt}$

Question

Is the magnitude of the series $\sum_{p} e^{-pt}$ summed over all primes $p$ with $t$ a positive real constant known?

Answer 1

Consider $$ f(t) = \sum_{p \text{ prime}} e^{-t p} $$ and let $g(z)$ be the related $$ g(z) = \sum_{p \text{ prime}} z^p $$ Clearly, we have the relation $g(e^{-t}) = f(t)$. One interesting note is that by the Fabry Gap Theorem, $g(z)$ has a natural boundary at $|z| = 1$, so $f(t)$ will have a natural boundary on the imaginary axis. Neither of these functions has a simple closed form in terms of any functions I have ever seen, though the Mellin transform of $f$ is relatively simple (see below).

For the asymptotics of $f(t)$, as $t$ goes to $\infty$, since $2$ is the smallest prime, we will certainly have $$ \lim_{t\rightarrow\infty} e^{2 t}f(t) = \lim_{t\rightarrow\infty} \sum_{p \text{ prime}} e^{-t (p-2)} = 1 + \lim_{t\rightarrow\infty} \sum_{p > 2 \text{ prime}} e^{-t (p-2)} = 1 $$ giving us the asymptotic equivalence $f(t) \sim e^{-2t}$. When $t$ approaches $0$, it's a lot more difficult. We can easily see that $$f(t) < \sum_{n\ge 2} e^{-t n} = \frac{e^{-2 t}}{1 - e^{-t}} \sim_{t\rightarrow 0^+} \frac1t$$ We can actually do a little better than this, as we will show that asymptotically as $t$ approaches $0$, we have $f(t) \sim \frac1{t\log t}$. To do this, observe that the Mellin transform of $f$ is: $$ \operatorname{M}[f](s) = \int_0^\infty t^{s-1} f(t) dt = \sum_p \int_0^\infty t^{s-1} e^{-t p} dt = \sum_p p^{-s} \Gamma(s) = P(s)\Gamma(s) $$ where $P(s)$ is the prime zeta function, which is known to have a logarithmic singularity at $s = 1$. Indeed, assuming RZH, $P(s) + \log(s - 1)$ can be analytically extended below $\Re{s} = 1$ (in fact, this follows from the weaker assertion that the zeta function does not have zeroes with real part arbitrarily close to 1). Therefore, we must have that $f(t)$ has a singularity of the appropriate type at $t=0$ to make this happen, and one can easily see that $\frac1{t\log t}$ is the correct function, because $$ \int_0^\alpha \frac{t^{s-1}}{t\log t} dt = \operatorname{Ei}((\log \alpha)(1-s)) = \log (1-s) + c - s + ... $$ (where $\alpha$ is an arbitrary constant in $(0,1)$, and $\operatorname{Ei}$ is the exponential integral function) Hence $f(t) - \frac1{t\log t}$ must have a singularity at $0$ such that $$ \int_0^\alpha t^{s-1} \left(f(t) - \frac1{t\log t}\right) dt $$ is analytic on some region $(1-\epsilon , \infty)$. In particular, $\int_0^\alpha f(t) - \frac1{t\log t} dt$ converges, whence we conclude $f(t) \sim \frac1{t\log t}$ when $t \rightarrow 0$.

Without the assumption that the Riemann zeta function does not have zeroes with real part approaching $1$, the picture becomes a lot more complicated, and I don't know exactly what sort of singularity we ought to expect, but it would probably be just slightly above $\frac1{t\log t}$.

Some other properties of f and g

We immediately have: $$ f^{(n)}(t) = \sum_{p \text{ prime}} (-p)^n e^{-tp} $$ Interestingly, $g$ relates to the Goldbach conjecture: $$ g(z)^2 = \sum_{n\ge 1} g_n z^n $$ where $g_n$ is the number of ways of writing $n$ as a sum of two primes. According to the Goldbach conjecture, $g_{2n} > 0$ for $2n \ge 4$. This function has probably been studied at some point, but I couldn't find anything in Google.

Answer 2

Let $F$ be your original sum.

Let $F_N$ be the $N$th partial sum.

If one accepts that $$p_n\approx n\ln n$$

then $$F-F_N\approx\sum_{n\ge N}n^{-tn}$$ which may have some linkage with the sophomore dream.