A formula found by Jacobi for $\psi(x)=\sum_{n=1}^{\infty} e^{-n^2\pi x}$ says $$\psi(\frac{1}{x})=\sqrt{x} \psi(x)+\frac{\sqrt{x}-1}{2}$$

If we define $$\phi(x)=\sum_p e^{-p^2\pi x}$$ where $p$ runs through all prime numbers. Now we want to know the behavior of this function near 0 and $+\infty$, then what can be told? Does there exist a similar relation like this $$\phi(\frac{1}{x})=\sqrt{x} \phi(x)+f(x)$$ where $f(x)$ is easy to handle?

  • 1
    $\begingroup$ I assume you mean $\phi(1/x) = \sqrt{x} \phi(x) + f(x)$? $\endgroup$ May 1 '17 at 14:21
  • $\begingroup$ Mellin transform of $\phi(x)$ is the so called Prime Zeta Function mathworld.wolfram.com/PrimeZetaFunction.html $\endgroup$
    – Negan
    May 1 '17 at 16:42
  • $\begingroup$ You can get a functional equation for $\sum_{n \ge 2} \Lambda(n) e^{-\pi n^2 x}$ and $\sum_{n \ge 2} \frac{\Lambda(n)}{\ln n} e^{-\pi n^2 x}$ $\endgroup$
    – reuns
    May 2 '17 at 3:15

There is no Jacobi-like functional equation for your $\phi(x)$. Nevertheless, with a bit of analysis, we can determine its asymptotic behavior as $x\to\infty$ and as $x\to 0+$.

When $x>0$ is large, the leading term $p=2$ dominates, so we have the asymptotics $$\phi(x)\sim e^{-4\pi x},\qquad x\to\infty.$$

When $x>0$ is small, we use the prime number theorem in the form $$\pi(x)\sim\mathrm{Li}(x):=\int_2^x\frac{dt}{\log t}.$$ Using this result and integration by parts, we get, as $x\to 0+$, \begin{align*} \phi(x)&=\int_{2-}^\infty e^{-t^2\pi x}\,d\pi(t) =\int_2^\infty \pi(t)\,d(-e^{-t^2\pi x})\\ &\sim \int_2^\infty \mathrm{Li}(t)\,d(-e^{-t^2\pi x}) =\int_2^\infty e^{-t^2\pi x}\,d\,\mathrm{Li}(t) =\int_2^\infty e^{-t^2\pi x}\,\frac{dt}{\log t}. \end{align*} Making the change of variable $t=u/\sqrt{x}$ in the last integral, we get $$\phi(x)\sim\frac{1}{\sqrt{x}}\int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})},\qquad x\to 0+.$$ Now the denominator $\log(u/\sqrt{x})$ always exceeds $2$, but for the bulk of the integral it is $\sim\log(1/\sqrt{x})$. More precisely, with the notation $r(x)=\sqrt{\log(1/x)}$, we can estimate the integral as \begin{align*} \int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})}&=\int_{2\sqrt{x}}^{\exp(-r(x))}+\int_{\exp(-r(x))}^{\exp(r(x))}+\int_{\exp(r(x))}^\infty\\[8pt] &=O(e^{-r(x)})+\frac{1+o(1)}{\log(1/\sqrt{x})}\int_{\exp(-r(x))}^{\exp(r(x))}e^{-u^2\pi}\,du+O(e^{-r(x)})\\[8pt] &=\frac{1/2+o(1)}{\log(1/\sqrt{x})}=\frac{1+o(1)}{\log(1/x)}. \end{align*} Hence we have proved that $$ \phi(x)\sim\frac{1}{\sqrt{x}\log(1/x)},\qquad x\to 0+.$$

  • $\begingroup$ I am really not seeing this decay rate at $0$ experimentally... $\endgroup$
    – Igor Rivin
    May 1 '17 at 18:38
  • $\begingroup$ @IgorRivin: Well, this is not a decay rate, but a blowup rate. Do you see experimentally the $1/(2\sqrt{x})$ blowup rate for $\psi(x)$ when $x\to 0+$? For $\phi(x)$ the result is natural in the light of this, since the density of primes around $n\approx 1/\sqrt{x}$ is approximately $\log(1/\sqrt{x})$. $\endgroup$
    – GH from MO
    May 1 '17 at 20:31
  • $\begingroup$ Sorry, blow-up rate. No, I am not seeing it (though this might be due to computational issues) - here are the values: $f(10^{-7}) = -0.0772, f(10^{-6}) = -0.092, f(10^{-5}) = -0.114, f(10^{-4}) = -0.146, f((10^{-3}) = -0.192, f(10^{-2} = -0.233, $ $\endgroup$
    – Igor Rivin
    May 1 '17 at 20:56
  • $\begingroup$ Notice that the blow-up seems sublinear in $\log x.$ $\endgroup$
    – Igor Rivin
    May 1 '17 at 20:57
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    $\begingroup$ @GH from MO, thanks for your wonderful analysis! $\endgroup$
    – Milin
    May 1 '17 at 23:52

Not an answer, just an experiment: If you define

$$f(x) = \varphi(1/x) - \sqrt{x} \varphi(x),$$

First, here is the graph of $f(x)$ between $0.00001$ and $0.1:$ enter image description here

And here is one between $0.01$ and $10:$enter image description here

Does not look like a nice function to me, but I am just a caveman (plus, computing this function is non-trivial)...

Secondly, $f(x)$ does seem to be approaching a limit as $x\to 0,$ the limit seems to be slightly negative (my guess is around $-0.05$)

  • $\begingroup$ I am typing an answer determining the asymptotics. Stay tuned. $\endgroup$
    – GH from MO
    May 1 '17 at 17:04
  • 2
    $\begingroup$ @GHfromMO Waiting with bated breath! $\endgroup$
    – Igor Rivin
    May 1 '17 at 17:05
  • $\begingroup$ I added my answer. I hope it is fine as I typed it fast. $\endgroup$
    – GH from MO
    May 1 '17 at 17:32
  • $\begingroup$ @Igor Rivin, good job! $\endgroup$
    – Milin
    May 1 '17 at 23:59

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