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Is there by any chance anything resembling a closed expression for, say, the integral $$I = \frac{1}{2 \pi} \int_{-\infty}^\infty \frac{dt}{|\zeta(1+i t)|^2 t^2} ?$$ It is easy to show (by Plancherel) that $$I = \int_0^\infty |m(x)|^2 \frac{d x}{x}$$ for $m(x):=\sum_{n\leq x} \mu(n)/n$, but I'd like to know whether there is something "more closed" than that.

H A Helfgott
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    Note that this has been crossposted to MO: https://mathoverflow.net/q/402657/14508. – davidlowryduda Aug 27 '21 at 16:11
  • you want it closed? $$\bbox[5px,border:2px solid]{\frac{1}{2 \pi} \int_{-\infty}^\infty \frac{dt}{|\zeta(1+i t)|^2 t^2}}$$ Now is closed. Problem solved! :) – Masacroso Aug 27 '21 at 19:15
  • Let the numbers 4 and 3 in the integral: Integrate[1/Sum[1/(E^(Round[Log[n]*4]/4))^(s), {n, 1, 3}],s] be (integer) variables that tend to infinity, then you might have a starting point for approximating the integral: Integrate[1/Sum[1/n^s, {n, 1, Infinity}], s] From
    Wolfram Alpha you can see that it gives a root sum, which if you included the analytic continuation in the integral would tend to a root sum over the Riemann zeta zeros.     – Mats Granvik Aug 30 '21 at 13:38
  • Related: https://mathematica.stackexchange.com/a/252896/328 And also: https://pastebin.com/EmvEAsyk Notice in the graph that Mathematica gives, that there is a weak tendency for the real part of the zeros to cluster around 1/2. See also: https://math.stackexchange.com/a/4216806/8530 – Mats Granvik Aug 30 '21 at 15:32
  • $$\displaystyle m(x)=\sum_{n\leq x} \frac{\mu(n)}{n^s} = \underbrace{1 - \sum_{2 \leq a \leq x} \frac{1}{a^{s}} + \underset{ab \leq x}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{1}{(ab)^{s}} - \underset{abc \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{1}{(abc)^{s}} + \underset{abcd \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{1}{(abcd)^{s}} - \cdots}_{\text{number of alternating sums} > \frac{\log(x)}{\log(2)}}$$ – Mats Granvik Feb 09 '22 at 17:22
  • $$\mu(n) = \underbrace{\underset{1 = n} 1 - \underset{a = n}{\sum_{a \geq 2}} 1 + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots}_{\text{#alternating sums}>\frac{\log(n)}{\log(2)}}$$ – Mats Granvik Feb 11 '22 at 15:34

1 Answers1

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The following Mathematica program:

(*Mathematica start*)
Clear[a, b, upper, lower, n, integral, t, nn];
nn = 10;
f[t_] := 1/(2*
      Pi)*1/(((Zeta[1 + I*t] + Zeta[1 - I*t])/
          2)^2 + ((Zeta[1 + I*t] - Zeta[1 - I*t])/I/2)^2)/t^2;
integral = Sum[(-1)^(n + 1)/n!*t^n*D[f[t], {t, n - 1}], {n, 1, nn}];
t = -1/2;
lower = %;
a = N[integral, 10];
t = 1/2;
upper = %;
b = N[integral, 10];
b - a
NIntegrate[f[t], {t, lower, upper}, WorkingPrecision -> 10]
(*end*)

solves the integral:

$$I = \frac{1}{2 \pi} \int_{-\frac{1}{2}}^\frac{1}{2} \frac{dt}{|\zeta(1+i t)|^2 t^2} $$

symbolically as b-a = 0.1567161649... by repeated integration by parts resulting in some kind of series expansion (probably Taylor expansion), and then at the end compares it to the numerical integral with NIntegrate[f[t], {t, lower, upper}, WorkingPrecision -> 10] = 0.1567161649...

For better precision I would increse nn = 10; but still the lower and upper integration limits are far from infinity.

Not maybe what you wanted, but it is something.

Mats Granvik
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  • $$\frac{\partial ^{n-1}}{\partial s^{n-1}}\frac{1}{\zeta (s)}=(-1)^{n-1} \left(\lim_{c\to 1} , \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k+1} \binom{n-1}{k-1}}{\zeta (c (k-1)-(k-1)+s)}\right)$$ – Mats Granvik Jun 03 '23 at 17:31
  • (*start*)Clear[n, k, c, h, s]; n = 5; aa = D[1/Zeta[s], {s, n - 1}]; s = 7; bb = N[aa, 40]; cc = (-1)^(n - 1)* Limit[Zeta[c]^(n - 1)* Sum[(-1)^(k + 1)* Binomial[n - 1, k - 1]/Zeta[-(k - 1) + (k - 1) c + s], {k, 1, n}], c -> 1]; dd = N[cc, 40]; bb - dd (*end*) – Mats Granvik Jun 03 '23 at 18:44