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
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. 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 138.
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. His statement on p. 340 that he used the Li function is highly misleading, as you found out.