Does logint function work well? (for PNT)

On the 'Prime Obsession' book, 20 to the power 1/2 + 14.134725i is −0.302303 − 4.46191i. Take the logarithmic integral—the Li function—of that to get the answer −0.105384 + 3.14749i.

I tried as belows, but failed.

>> a=20^(1/2+14.134725i)
a =
  -0.3023 - 4.4619i % OK
>> logint(a)-logint(2)
ans =
   0.9528 - 3.9138i % Wrong

[Notae]

Li(x)= integral(1/log(t)) from x to 2, so Li(x)=li(x)-li(2) => It's a minor thing. Anyhow the result on the book cannot be explained by my trials.

[KALYAN ACHARJYA]

Explain what are you looking for?

[the cyclist]

According to this MATLAB documentation page, Li(x) is defined as

Li(x) = integral(1/log(t)) from 0 to x

not from x to 2 as you state. I don't have the symbolic math toolbox to test.

[Notae]

You are right. logint integral from 0 to x. Li(x) = int(1/ln(t)dt from 0 to x or from 2 to x => It's a minor problems. The result of The Book(page 340) and mine is very different. I used logint(x^a); a is first roots of zeta function, and is about 1/2+14.134725i). logint( ) works well. I tested as integral(@(x)1./log(x), 2, x)+logint(2) == logint(x). But the book cannot be wrong, the author did many trials with the Mathematica(I don't have and use).

[the cyclist]

If you go to WolframAlpha and evaluate the expression

Li(20^(1/2+14.134725i)) - Li(2)

you get the answer

0.952805 - 3.91384i

which is what you say that MATLAB also gives. Therefore, I think the problem is not with MATLAB.

Maybe you could post more of what it says in the book you mention. Maybe there is some other part you are missing.

[Notae]

Thank you for your efforts. I have tired to roughly understand the prime number theory and Riemann zeta function(and hypothesis). For the final error complement, all sum of Li(x^zeros of zeta function) is essential. But the number(s) is too big. With psi function, I did well, all sum of x^p/p is necessary(p is rho, zeros of zeta function)

Answer 1

[David Goodmanson]

Hi Notae,

The reason for this is something that Derbyshire relegated to a footnote several pages before (p. 335, 128). If you look at the logarithmic integral, its primary definition is

Li(z) = Ei(log(z))

(Abramowitz & Stegun 5.1.3) where Ei is a particular exponential integral which Matlab does have, although that fact is not so obvious and you need the symbolic toolbox. [there is a workaround for Ei, see below]. However, there is a 2pi ambiguity when taking the log, since for any integer n, log(z) and log(z)+2*pi*i*n are equally valid answers. That means that Ei(logz)) depends on n and you have to pin down the actual angle.

If z = Ae^(i*b) where b is restricted to the range -pi<b<=pi in the first place, then Li(z) and Ei(log(z)) give the same result.

b = logint(2+i)
c = ei(log(2+i))
b = 1.4113 + 1.2247i
c = 1.4113 + 1.2247i

In this case, though,

a = 20^(.5+14.134725i)

and you have to keep track of the number of times you go around the circle so as to remove ambiguity.

%  a = 20^(.5+14.134725i)
loga = log(20)*(.5+14.134725i)
  loga = 1.4979 +42.3439i       % large angle, no 2pi ambiguity
format long
ei(loga)
  ans = -0.105384042414102 + 3.147487521958689i

as advertised. That's what he did in footnote 128.

If you don't have the symbolic toolbox, since expint is available in basic Matlab you can use

function y = ei_alt(z)
  y = -expint(-z) + (log(z)-log(1/z))/2 - log(-z); 
end

Ei(logz)) can distinguish how many times you go around the circle, but Li(z) as defined on p.114 (and generalized so that the path of integration is a straight line from 0 to complex number z) cannot. To say as he did on p. 340 that he used the Li function is highly misleading, as you found out.

[Notae]

Very thankful for your answer. I tried as below without reading your answer. >> c=log(20^a) c = 1.4979 - 1.6384i >> ei(c) ans = 1.9980 - 3.9138i

I saw it on the 'http://demonstrations.wolfram.com/HowTheZerosOfTheZetaFunctionPredictTheDistributionOfPrimes/' They say 'Ei(p*log(x), where p(rho) is the complex zero of the zeta function. In this formula, (Mathematica's built-in function ExpIntegralEi[z]) is the generalization of the logarithmic integral to complex numbers.' And I read logint(z)=ei(log(z)) by MATLAB doc. So I tried as abovee. But failed again. Anyway logint(z) works well if a input arg. is complex number. So ei(log(z)) doesn't neccessory.

[Notae]

I skipped your good answer and just insisted my biased, one-way thought that I had tried all as I can including your answer. But now I read your answer one word by one, and I understand what you mean. Anyway thank you really. (But the plot of pi(x) and R(x)-R(x^p) appeared bad. The x and psi(x) (here Li(sum of x^p/p) exist) showed very good looks which was the same as many papers.)