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I am considering the following term from the Riemann explicit formula (see here >>>):

$$\sum_{\rho(\Im>0)}{\mathrm{li}(x^\rho)}$$

with $\rho$ non-trivial zeros of $\zeta$-function.

I have a plot of this sum up to the first $100$ non-trivial zeros, see below. The oscilations are around a curve profile.

What is the function $f(x)$ for this curve profile, along which the above sum is oscillating? How can we get to this function analytically? Is there any reference?

enter image description here

APPEND: Meanwhile I found some approximation but can not find to the analytic relation. So may be the question could say: Why does the following function could be an approxiamtion for this profile curve?

$$f(x)=\frac{1}{2}\sum_{\rho(\Im>0)}{\mathrm{li}(x^{1/2})}$$

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  • $\begingroup$ Wouldn't this just be $\mathrm{li}(x^{1/2})$? $\endgroup$ – Peter Humphries Feb 12 '15 at 16:31
  • $\begingroup$ @PeterHumphries No. $\endgroup$ – al-Hwarizmi Feb 12 '15 at 16:37
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    $\begingroup$ Something seems wrong: I don't see why this function should always be positive - indeed I feel it should oscillate around an average of $0$. Are you including zeros $\rho$ below the real axis as well as above it? $\endgroup$ – Greg Martin Feb 13 '15 at 8:12
  • $\begingroup$ @GregMartin I only apply the zeros above the axis. You are right, I did not mention this. It is not wrong but consicous. And I only plot the $\Re$ part of the total sum. $\endgroup$ – al-Hwarizmi Feb 13 '15 at 8:17
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To understand the underlying "curve profile" you have to look at the integral that the machine is calculating. It turns out that most of the oscillation is due to irregular spacing of zeros.

For the integral $\int_2^{\rho_j}\frac{dt}{\log t}$ the computer is calculating the value of an integral from $2$ to each zero $1/2+it_j.$

The zeros are unevenly spaced but the $average$ spacing on $x=1/2$ is quite regular. Riemann estimated that the average number of zeros between $y = 0$ and $y = T$ was about

$$N = \frac{T}{2\pi}\log \frac{T}{2\pi} - \frac{T}{2\pi} .$$

If we plug the complex value of the $100th$ zero into this estimate (236.5) the estimate gives 98 zeros--quite close. We can use this to find the average distance $d$ between the first hundred zeros along the $y$ axis. It's about 2.36.

Now if instead of asking the machine to plot $\sum li(x^{\rho})$ we ask it to plot the real part of

$$\sum \text{li}(x^{1/2 + 2.36ki}), k = 1,2,3,... $$

for the first hundred "regularized zeros" we get a curve that in some sense underlies the curve you plotted. The curve you seek emerges as the oscillation due to uneven spacing of zeros diminishes.

Here is an image for 100 "regularized" zeros.

enter image description here

That the oscillation can be partly explained this way does not mean there is some ideal curve that the OP curve tracks or follows. The full expression for the number of zeros can be found here. The likelihood that the process above leads to a closed form is small and it would ignore complexities discussed in the link.

So the short answer is that we can find a calmer version of the curve in the OP and argue that this is what the eye sees behind the zig-zags, and perhaps use curve-fitting to find approximations for finite $x,\rho;$ but it is not really true that there is some simple curve about which the OP curve oscillates.

Hope this helps.

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  • $\begingroup$ appreciate your response. It is a beautiful description of the circumstances and how to view towards a "possible" solution. But your last line leaves my question essentially unanswered. The question was whether there is and if, what is the explicit function. The conclusion at the end of your answer "with more work..." does not solve the problem. The regularisation and smothening approach is comprehensive, but not taking the challenge to a mathematical final. $\endgroup$ – al-Hwarizmi Feb 13 '15 at 14:18
  • $\begingroup$ @al-Hwarizmi: added a couple of lines in response to your comment. Continuing the process in the answer does not yield a closed form so I don't think there is much to be done with this. It's not a simple function. $\endgroup$ – daniel Feb 16 '15 at 11:28
  • $\begingroup$ (It is supposedly much easier to look at the explicit formula for $\psi(x)$) $\endgroup$ – reuns Aug 9 '17 at 9:02
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Just Some Nerdy Teen made an interesting comment (in a now deleted answer) :
"You should put ExpIntegralEi(rho*log(x)) instead of LogIntegral(x^rho)".

The observed curve was indeed an artifact coming from the use of the logarithmic instead of the exponential integral!
The problem is that the logarithmic integral requires (at least) the evaluation of the logarithm of its parameter $x^\rho\,$ but $\,\log(x^\rho)\,$ will differ from $\,\rho\,\log(x)\,$ for $\rho\gg 1\,$ (CAS use simply the principal branch of the logarithm!).

The solution indicated by JSNT is to use the relation $\;\displaystyle\mathrm{li}(z)=\mathrm{Ei}(\ln z)\;$ with $\,\mathrm{Ei}\,$ the exponential integral and rewrite your sum for the $n$ first complex zeros (with positive imaginary part) as : $$\tag{1} S_n(x):=\sum_{i=1}^n{\mathrm{Ei}(\rho_i\ln x)}$$

(different 'traps' to avoid while evaluating the explicit formula(s) conclude this answer)

Let's compare the result using $\,\mathrm{Ei}(\ln x^{\rho_i})\,$ (your picture with $\,n=100$) : your fig.

with the actual result for $(1)$ : sum 1

Riemann's explicit formula restrained to the $\,(2)n\,$ 'smallest' non trivial zeros is (for $x>1$ and using $\,\mathrm{Ei}(\overline{\rho_i}\ln x)=\overline{\mathrm{Ei}(\rho_i\ln x)}$ for the $n$ first complex zeros with negative imaginary part): $$\tag{2}\Pi_{n}(x)=\mathrm{Ei}(\ln(x))-\sum_{k=1}^\infty \mathrm{Ei}(-2k\,\ln x)-2\,\Re\,S_n(x)$$ may then be used to visualize the Riemann Prime Counting Function $\Pi$.

For $n=100$ we get : PI0 that we may compare to the actual $\;\displaystyle\Pi(x)=\sum_{p^k<x}\frac 1k$ : real PI0


The artifact was thus eliminated but, out of curiosity, can we approximate it?

The actual error is of course given by the evaluation of (with $[x]$ the nearest integer) : \begin{align} \tag{3} \Delta_n(x)&:=\Re\sum_{i=1}^n \mathrm{Ei}(\ln x^{\rho_i})-\mathrm{Ei}(\rho_i\ln x)\\ \tag{4} &=\Re\sum_{i=1}^n \mathrm{Ei}\left(\frac{\ln x}2+\left(\frac{\Im\rho_i\ln x}{2\pi}-\left[\frac{\Im\rho_i\ln x}{2\pi}\right]\right)2\pi\,i\right)-\mathrm{Ei}(\rho_i\ln x)\\ \end{align} Which gives nearly your initial picture (observe the bounds on my second picture suggesting to neglect the right part :$\,\Re\sum_{i=1}^n\mathrm{Ei}(\rho_i\ln x)\,$). One could think at neglecting the imaginary part in the first term and write $\:\Delta_n(x)\approx\Re\sum_{i=1}^n \mathrm{Ei}\left(\frac{\ln x}2\right)=n\,\mathrm{li}(x^{1/2})$. This corresponds to your conjecture but only if divided by a value near of $2$.

A more precise result is obtained with the average value of $\;\mathrm{Ei}\left(\frac{\ln x}2+ti\right)$ for $t$ in $(-\pi,\pi)\,$ which can be done using integration by parts to get $\;\displaystyle\int \mathrm{Ei}(x) dx= x\mathrm{Ei}(x)-e^x\;$ and the better approximation : $$\tag{5}\Delta_n(x)\approx \left.-\,\frac n{2\pi}\,i\,\left(\frac{\ln x}2+ti\right)\mathrm{Ei}\left(\frac{\ln x}2+ti\right)\right|_{t=-\pi}^\pi $$

illustrated by following figure with $(5)$ in red and the OP's approximation $\;n\,\mathrm{li}(x^{1/2})/2$ at the bottom: approx 100

With $n=1000$ zeros (instead of $100$) and a larger range the approximation $(5)$ seems the only one to remain reliable : approx 1000

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