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Motivation:

I was looking at the approximation of the truncated Prime $\zeta$ function $$ P_x(s)=\sum_{p\leq x}p^{-s}= \mathrm{li}(x^{1-s}) + O \left(\cdot \right) $$ (to be found here with or without assuming the Riemann Hypothesis).

I wanted to know how the absolute value of $P_x(it)$ behaves ($\Re{s}=0$). So, I

  • neglected the $O(\cdot)$,
  • used $\mathrm{li}(x^{1-it})=\mathrm{Ei}\left((1-it)\ln(x)\right)$ as recommended here
  • and asked Wolfram, who told me that $$ \lim_{t\to \infty}|\mathrm{Ei}\left((1-it)\ln(x)\right)|=\pi.\tag{*} $$

Questions:

  1. How to prove $(*)$, without using a series expansion at $t=\infty$?
  2. What does the independance of $x$ mean for $P_x(it)$?
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1 Answer 1

up vote 3 down vote accepted

CORRECTED:
We may get the limit for $\operatorname{Ei}(\sigma+it)$ as $t\to \infty$ by using following asymptotic series : $$\tag{1} \operatorname{Ei}(z)\sim \operatorname{sgn}(\Im(z))i\pi+\frac {e^z}z\left(1+\frac {1!}{z}+\frac {2!}{z^2}+\frac {3!}{z^3}+\cdots\right)$$ (the terms at the right may be obtained by repetitive integration by parts of $\frac {e^z}{z^n}$ getting each time a greater power $n$ at the denominator, for more about this see $(12)$ in Pegoraro and Slusallek's paper of 2011 'On the Evaluation of the Complex-Valued Exponential Integral' where the Mathematica conventions are used)

If we apply this to $z=\sigma+it$ as $t\to \pm\infty$ we get $$\tag{2} \operatorname{Ei}(\sigma+it)\sim \operatorname{sgn}(t)i\pi+\frac {e^{\sigma+it}}{\sigma+it}$$ Since $\left|\frac {e^{\sigma+it}}{\sigma+it} \right|\sim \frac {e^{\sigma}}{|t|}$ as $t\to \pm\infty$ we get $\ \operatorname{Ei}(\sigma+it)\sim \operatorname{sgn}(t)i\pi\ $ and your result.


But you wanted a proof without series expansion at $\infty$ so let's try to use directly the integral definition of $\operatorname{Ei}$ and 'suppose' that : $$\tag{3} \overline{\operatorname{Ei}}(s):=\int_{-\infty}^s \frac {e^z}z\,dz$$

Let's study $\overline{\operatorname{Ei}}(it)$ as $t \to +\infty$ first (case $\sigma=0$) and use following contour at the top right of $0$ supposing that $R=t$ (for $t\to -\infty$ simply consider the symmetric contour!) :
contour

$\overline{\operatorname{Ei}}$ is of course analytic inside this contour so that any 'internal' path from $(-R)$ to $(it)$ should give the same integral.
Using the direct upper path we get :
$$\int_{-R}^{iR} \frac {e^z}z\,dz = \int_{\pi}^{\frac {\pi}2} \frac {e^{R e^{i\phi}}}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi=i\int_{\pi}^{\frac {\pi}2} e^{R e^{i\phi}}\,d\phi$$

Since $\cos(\phi)<0$ for $\phi \in (\frac {\pi}2,\pi)\ $ and $\ \left|\int_{-R}^{iR} \frac {e^z}z\,dz\right| \le \int_{\frac {\pi}2}^{\pi} e^{R \cos(\phi)}\,d\phi\ $ we conclude that the integral from $(-R)$ to $(iR)$ admits the limit $0$ as $R\to \infty$ (with an error of order $\frac 1R$).
Of course the other path should give the same result and :

$$\overline{\operatorname{Ei}}(it)-\overline{\operatorname{Ei}}(-R)=\int_{-R}^{-r} \frac {e^z}z\,dz + i\int_{\pi}^{\frac {\pi}2} e^{r e^{i\phi}}\,d\phi+\int_r^t \frac {e^{iu}}u\,du$$ (allowing us to compare the exponential integral of real and pure imaginary values but that's another subject...)

Find the limit of $\overline{\operatorname{Ei}}(\sigma+it)$ as $t\to \infty$ is not more difficult since : $$\overline{\operatorname{Ei}}(\sigma+it)-\overline{\operatorname{Ei}}(it)=\int_0^{\sigma} \frac {e^{u+it}}{u+it}\,du$$

So that, for $\sigma > 0$, the absolute value of the difference will be bounded by $\frac{\sigma e^{\sigma}}{|t|}$ and go to $0$ as $t\to \infty$ (consider a rectangle with a larger value of $t$ say $T$ that we bring to $\infty$ would be better but this should be enough to get the idea).

This is all nice and well and seems to prove that $\overline{\operatorname{Ei}}(\sigma+it)$ goes to $0$ as $t\to \infty$ but where is the $\pi$ if not in the sky ?? :-)

The $\pi$ is merely an artifact of the cut chosen for $\operatorname{Ei}$ (on the negative real axis) instead of the cut on the positive real axis implied by the integral expression I used in $\overline{\operatorname{Ei}}$ (that's the reason of the over-line everywhere). To go from the cut on the positive axis ($\overline{\operatorname{Ei}}$) to the cut at the negative axis ($\operatorname{Ei}$) is not really difficult : add $\pi i$ to the result if the imaginary part of the parameter is positive and subtract $\pi i$ if the imaginary part is negative (I updated my answer to this older thread with graphics to make things clearer ; $\overline{\operatorname{Ei}}(z)$ corresponds there to the definition given by Lebedev and is equal to $-\operatorname{E_1}(-z)$ ).

This study showed that the limit of $\overline{\operatorname{Ei}}(z)$ was $0$ when $z$ went to $\infty$ in the left half-plane or along vertical lines in the right half-plane (it should go to $\infty$ in the remaining cases).


Concerning your second question I still don't know what to think about it (or even if this specific approach is useful at all as $t\to \infty$). Isn't $\operatorname{li}\left(x^{1-s}\right)$ decreasing while the error term is increasing ?

By the way let's observe that a more precise bound was given by Schoenfeld if you suppose the R.H. and $x\ge 2657$ : $\displaystyle |\pi(x)-\operatorname{li}(x)| < \frac{\sqrt{x}\ln(x)}{8\pi}$

I tried some numerical experiments and found that for $x$ fixed and growing values of $t$ $P_x(t)$ was oscillating 'randomly' without noticeable modification. In fact it seemed scale invariant relatively to $t$ and this for small or large values of $\sigma$ (so that it doesn't behave as $\operatorname{li}$ nor as the error term!).


It could perhaps be interesting to know the algorithm used by Wolfram Alpha to compute the 'prime zeta function' ($P_x(s)$ as $x \to \infty$) in the complex plane since they could evaluate a finite sum combined with an extrapolation (see Neat examples for the case $\sigma =\frac 12$). For real parts of order $0.4$ or more if not on the imaginary line I got this plot of PrimeZetaP[0.4+t*i] :

re=0.4

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Maybe we are talking about different things: (i) looking at $Ei(1-it)$ doesn't look like that is growing without bounds, but rather converges to $-i\pi$, as does e.g. $Ei((1-it)*1000)$. (ii) I should have stated that my PrimeZeta is truncated to primes smaller than $x$. –  draks ... Jun 22 '12 at 10:01
    
@draks: I noticed the part (ii) but fear I was wrong on the part (i). Sorry... I missed the limit because of the large values for moderate values for example. The asymptotic formula I gave becomes indeed $\operatorname{Ei}(u+iv)\sim \frac {e^{u}}{|v|}$ that is $0$ when $v\to \infty$. And only remains the $\pi$ term from the $\log$ term of $\operatorname{Ei}$. I'll try to find a more appropriate answer this evening or remove it... –  Raymond Manzoni Jun 22 '12 at 12:14
    
+1 Thanks so far. –  draks ... Jun 23 '12 at 20:53
    
In fact it seemed scale invariant relatively to t and this for small or large values of σ (so that it doesn't behave as li nor as the error term!). That's exactly what I'm after. Why is that and what does that mean? –  draks ... Jun 28 '12 at 13:04
    
@draks: (sorry for the delay...) We are adding $e^{-\sigma\log(p)}e^{-i t\log(p)}$ terms. The first is slowly decreasing (for $\sigma \in (0,1)$) while the second is nearly random for large values of $t$ so that all contributions will compensate giving this $t$ scale invariant picture. In fact you should get nearly the same picture if you replace $p$ by an approximation like $i\ln i$ (rounded or not!) even if the phase is replaced by a random number in $(0,2\pi)$. –  Raymond Manzoni Jun 30 '12 at 9:09

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