Why are these integrals equivalent to the Riemann Hypothesis?

Question

Riemann Hypothesis is equivalent to the integral equation

$$\int_{-\infty}^{\infty} \frac{\log \mid \zeta (1/2+it)\mid }{1+4t^2} \ dt =0$$

Many other integral equations exist that are equivalent.

How to show that they are equivalent ?

They usually include absolute value of a function.

Why is that ?

I assume it came from a contour integral on the riemann sphere.

The growth rate of the zeta function on the critical line also probably relates to all those integrals , not ?

Another example is

Establishing the exact value $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$ is equivalent to the Riemann Hypothesis.

More examples :

Riemann's Hypothesis is true if and only if $$\frac{1}{\pi}\int_0^{\infty} \log\left|\frac{\zeta(\frac{1}{2}+it)}{\zeta(\frac{1}{2})}\right|\ \frac{dt}{t^2}=\frac{\pi}{8}+\frac{\gamma}{4}+\frac{\log 8\pi}{4}-2$$

Take $a\in R$ with $\frac{1}{2}\leq a<1$. Riemann's $\zeta$-function has no zeros in $\Re(s)>a$ if and only if $$\frac{1}{\pi}\int_0^{\infty} \log\left|\frac{\zeta(a+it)}{\zeta(a)}\right|\ \frac{dt}{t^2}=\frac{\zeta'(a)}{2\zeta(a)}-\frac{1}{1-a}$$

And many more exist.

I have no idea how to get to such conclusions or prove them.

Im not even sure how to prove integrals for "slightly easier" cases , meaning not famous open problems but zero's of other nontrivial functions that are not on a half-plane.

Answer 1

A long comment on the Riemann hypothesis equivalent by V.V.Volchkov found here:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt = \frac{\pi(3-\gamma)}{32} \label{1}\tag{1}$$

mentioned here.

I cannot give you the absolute correct answer but what I can give you is a plausibility argument in the form of a slightly incorrect calculation of the integral yielding the answers:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=-\frac{\pi (1-2 \gamma )}{32}\color{Blue}{-\frac{\pi}{32} \sum _{k=2}^{\infty } \frac{1}{k}} \label{2}\tag{2}$$

and

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-3\gamma)}{32} \label{2.1}\tag{2.1}$$

instead.

The starting points are \eqref{3} to \eqref{8} for results regarding the von Mangoldt function which have been proven by joriki (see here) and GH from MO (see here): assuming that $n>1$:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} \label{3}\tag{3}$$ $$\begin{align} a(n) &=-2\lim\limits_{s \rightarrow 0} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} \label{4}\tag{4} \\ &= \sum\limits_{d|n} d \cdot \mu(d) \label{5}\tag{5} \\ \end{align}$$

$$T = \left( \begin{array}{ccccccc} \color{Red}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{\cdots} \\ \color{Red}{+1}&-1&+1&-1&+1&-1&+1 \\ \color{Red}{+1}&+1&-2&+1&+1&-2&+1 \\ \color{Red}{+1}&-1&+1&-1&+1&-1&+1 \\ \color{Red}{+1}&+1&+1&+1&-4&+1&+1 \\ \color{Red}{+1}&-1&-2&-1&+1&+2&+1 \\ \color{Red}{+1}&+1&+1&+1&+1&+1&-6 \\ \color{Red}{\vdots}&&&&&&&\ddots \end{array} \right) \label{6}\tag{6}$$

$$\displaystyle \begin{pmatrix} \color{Red}{\frac{T(1,1)}{1 \cdot 1}}&\color{Blue}{+\frac{T(1,2)}{1 \cdot 2}}&\color{Blue}{+\frac{T(1,3)}{1 \cdot 3}+}&\color{Blue}{\cdots}&\color{Blue}{+\frac{T(1,k)}{1 \cdot k}} \\ \color{Red}{\frac{T(2,1)}{2 \cdot 1}}&+\frac{T(2,2)}{2 \cdot 2}&+\frac{T(2,3)}{2 \cdot 3}+&\cdots&+\frac{T(2,k)}{2 \cdot k} \\ \color{Red}{\frac{T(3,1)}{3 \cdot 1}}&+\frac{T(3,2)}{3 \cdot 2}&+\frac{T(3,3)}{3 \cdot 3}+&\cdots&+\frac{T(3,k)}{3 \cdot k} \\ \color{Red}{\vdots}&\vdots&\vdots&\ddots&\vdots \\ \color{Red}{\frac{T(n,1)}{n \cdot 1}}&+\frac{T(n,2)}{n \cdot 2}&+\frac{T(n,3)}{n \cdot 3}+&\cdots&+\frac{T(n,k)}{n \cdot k} \end{pmatrix} = \begin{pmatrix} \color{Blue}{\frac{\infty}{1}} \\ +\frac{\Lambda(2)}{2} \\ +\frac{\Lambda(3)}{3} \\ \vdots \\ +\frac{\Lambda(n)}{n} \end{pmatrix} \label{7}\tag{7}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{pmatrix} \color{Red}{\frac{\infty}{1}}&+\frac{\Lambda(2)}{2}&+\frac{\Lambda(3)}{3}+&\cdots&+\frac{\Lambda(k)}{k} \end{pmatrix}\;\;\;\;\;\;\;\;\;\;\; \label{8}\tag{8}$$

The following limits and integrals can be done in Mathematica 14:

$$\begin{align} \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} &= \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \frac{a(\gcd(n,k))}{n^c \cdot k^s} \label{9}\tag{9} \\ &=\sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} \label{10}\tag{10}\\ &= \frac{\zeta(s) \zeta(c)}{\zeta(s + c - 1)} \label{11}\tag{11}\\ \end{align}$$

$$\begin{align} \frac{\zeta '(s)}{\zeta (s)}&=\lim_{c\to 1} \, \left( \color{Red}{\zeta (c)}-\frac{\zeta (s) \zeta (c)}{\zeta (s+c-1)}\right) \label{12}\tag{12}\\ &=\sum\limits_{n=1}^{\infty} \left( \color{Red}{\frac{1}{n^{1}}}-\frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^{1}}\right) \label{13}\tag{13} \\ &= \sum\limits_{n=1}^{\infty}\left(\color{Red}{\frac{1}{n^{1}}}-\sum\limits_{k=1}^{\infty} \frac{a(\gcd(n,k))}{n^{1} \cdot k^s}\right) \label{14}\tag{14} \\ &= \sum\limits_{n=1}^{\infty}\left(\sum\limits_{k=2}^{\infty} \frac{a(\gcd(n,k))}{n^{1} \cdot k^s}\right) \label{15}\tag{15} \\ &= \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}\frac{a(\gcd(n,k))}{n^{1} \cdot k^s} \label{16}\tag{16} \\ \end{align}$$

$$[k>1] = \text{Iverson bracket} = (\text{If k > 1 then 1 else 0}) \label{17}\tag{17} $$

$$\begin{align}\log | \zeta (\sigma+i t)| &=\log (| \zeta (\sigma+i t)| ) \label{18}\tag{18} \\ &=\frac{1}{2} \log (\zeta (\sigma+i t))+\frac{1}{2} \log (\zeta (\sigma-i t)) \label{19}\tag{19} \\ &=\int \frac{1}{2} \left(\frac{\zeta '(\sigma+i t)}{\zeta (\sigma+i t)}+\frac{\zeta '(\sigma-i t)}{\zeta (\sigma-i t)}\right) \, d\sigma \label{20}\tag{20} \\ &=\int \frac{1}{2} \left(\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}\frac{a(\gcd(n,k))}{n^{1} \cdot k^{\sigma+i t}}+\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}\frac{a(\gcd(n,k))}{n^{1} \cdot k^{\sigma-i t}}\right) \, d\sigma \label{21}\tag{21} \\ &=\int \frac{1}{2} \left(\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]} \left(\frac{a(\gcd (n,k))}{n^{1} \cdot k^{\sigma+i t}}+\frac{a(\gcd (n,k))}{n^{1} \cdot k^{\sigma-i t}}\right)\right) \, d\sigma \label{22}\tag{22}\\ &=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \int \frac{1}{2} \color{Red}{[k>1]} \left(\frac{a(\gcd (n,k))}{n^{1} \cdot k^{\sigma+i t}}+\frac{a(\gcd (n,k))}{n^{1} \cdot k^{\sigma-i t}}\right) \, d\sigma \label{23}\tag{23}\\ &=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}-\frac{\left(1+k^{2 i t}\right) k^{-\sigma-i t} a(\gcd (k,n))}{2 n \log (k)} \label{24}\tag{24} \\ \end{align}$$

$$\begin{align} \int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma &=\int_{\frac{1}{2}}^{\infty }\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}-\frac{\left(1+k^{2 i t}\right) k^{-\sigma-i t} a(\gcd (k,n))}{2 n \log (k)} \, d\sigma \label{25}\tag{25} \\ &=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \int_{\frac{1}{2}}^{\infty } \color{Red}{[k>1]}-\frac{\left(1+k^{2 i t}\right) k^{-\sigma-i t} a(\gcd (k,n))}{2 n \log (k)} \, d\sigma \label{26}\tag{26} \\ &=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}-\frac{k^{-\frac{1}{2}-i t} \left(1+k^{2 i t}\right) a(\gcd (k,n))}{2 n \log ^2(k)} \label{27}\tag{27} \\ \end{align}$$

except for this integral which can not be done in Mathematica 14:

$$\begin{align} \int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt &=\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \color{Red}{[k>1]}-\frac{k^{-\frac{1}{2}-i t} \left(1+k^{2 i t}\right) a(\gcd (k,n))}{2 n \log ^2(k)} \, dt \label{28}\tag{28} \\ &=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\color{Red}{[k>1]}-\frac{k^{-\frac{1}{2}-i t} \left(1+k^{2 i t}\right) a(\gcd (k,n))}{2 n \log ^2(k)} \, dt \label{29}\tag{29} \\ \end{align}$$

Therefore we resort to computing instances of the integral above for integer values between $1$ and $6$ for both $n$ and $k$. That is: $$n=1,2,3,4,5,6 \label{30}\tag{30} $$ and $$k=1,2,3,4,5,6 \label{31}\tag{31} $$, with the following Mathematica 14 program:

Clear[sigma, t, a, n, k, B] (*takes several minutes to run*)
nn = 6;(* nn=3 for faster program but smaller matrix *)
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]]
TableForm[
 B = Integrate[(1 - 12*t^2)/(1 + 4*t^2)^3*
    Integrate[
     Integrate[
      Table[Table[
        If[k > 1, (a[GCD[n, k]]/n/k^(sigma + I*t) + 
            a[GCD[n, k]]/n/k^(sigma - I*t))/2, 0], {k, 1, nn}], {n, 1,
         nn}], sigma], {sigma, 1/2, Infinity}], {t, 0, Infinity}]]
TableForm[1/(-(Pi/32))*B]
TableForm[
 1/(-(Pi/32))*Table[Table[B[[n, k]]*n*k, {k, 1, nn}], {n, 1, nn}]]
-(Pi/32)*
 Limit[Zeta[s]*Zeta[s]/Zeta[s + s - 1] - Zeta[s] - (Zeta[s] - 1), 
  s -> 1]
N[%, 40]

The first output from the program is matrix $B$ that has the definition:
$$B(n,k)=\int_0^{\infty } \frac{\left(1-12 t^2\right) \int_{\frac{1}{2}}^{\infty } \left(\int \left[k>1\right] \frac{1}{2} \left(\frac{a(\gcd (n,k))}{n k^{\sigma+i t}}+\frac{a(\gcd (n,k))}{n k^{\sigma-i t}}\right) \, d\sigma\right) \, d\sigma}{\left(4 t^2+1\right)^3} \, dt \label{32}\tag{32} $$

and starts:

$$B=\left( \begin{array}{cccccc} \color{Red}{0} & \color{Blue}{-\frac{\pi }{64}} & \color{Blue}{-\frac{\pi }{96}} & \color{Blue}{-\frac{\pi }{128}} & \color{Blue}{-\frac{\pi }{160}} & \color{Blue}{-\frac{\pi }{192}} \label{33}\tag{33} \\ \color{Red}{0} & \frac{\pi }{128} & -\frac{\pi }{192} & \frac{\pi }{256} & -\frac{\pi }{320} & \frac{\pi }{384} \\ \color{Red}{0} & -\frac{\pi }{192} & \frac{\pi }{144} & -\frac{\pi }{384} & -\frac{\pi }{480} & \frac{\pi }{288} \\ \color{Red}{0} & \frac{\pi }{256} & -\frac{\pi }{384} & \frac{\pi }{512} & -\frac{\pi }{640} & \frac{\pi }{768} \\ \color{Red}{0} & -\frac{\pi }{320} & -\frac{\pi }{480} & -\frac{\pi }{640} & \frac{\pi }{200} & -\frac{\pi }{960} \\ \color{Red}{0} & \frac{\pi }{384} & \frac{\pi }{288} & \frac{\pi }{768} & -\frac{\pi }{960} & -\frac{\pi }{576} \\ \end{array} \right) $$

Dividing this matrix $B$ with: $$-\frac{\pi}{32} \label{34}\tag{34}$$

$$\left( \begin{array}{cccccc} \color{Red}{0} & \color{Blue}{\frac{1}{2}} & \color{Blue}{\frac{1}{3}} & \color{Blue}{\frac{1}{4}} & \color{Blue}{\frac{1}{5}} & \color{Blue}{\frac{1}{6}} \\ \color{Red}{0} & -\frac{1}{4} & \frac{1}{6} & -\frac{1}{8} & \frac{1}{10} & -\frac{1}{12} \\ \color{Red}{0} & \frac{1}{6} & -\frac{2}{9} & \frac{1}{12} & \frac{1}{15} & -\frac{1}{9} \\ \color{Red}{0} & -\frac{1}{8} & \frac{1}{12} & -\frac{1}{16} & \frac{1}{20} & -\frac{1}{24} \\ \color{Red}{0} & \frac{1}{10} & \frac{1}{15} & \frac{1}{20} & -\frac{4}{25} & \frac{1}{30} \\ \color{Red}{0} & -\frac{1}{12} & -\frac{1}{9} & -\frac{1}{24} & \frac{1}{30} & \frac{1}{18} \\ \end{array}\right) \label{35}\tag{35}$$

Multiplying matrix $B$ with $$-\frac{n k B(n,k)}{\frac{\pi }{32}} \label{36}\tag{36}$$

$$\left( \begin{array}{cccccc} \color{Red}{0} & \color{Blue}{1} & \color{Blue}{1} & \color{Blue}{1} & \color{Blue}{1} & \color{Blue}{1} \\ \color{Red}{0} & -1 & 1 & -1 & 1 & -1 \\ \color{Red}{0} & 1 & -2 & 1 & 1 & -2 \\ \color{Red}{0} & -1 & 1 & -1 & 1 & -1 \\ \color{Red}{0} & 1 & 1 & 1 & -4 & 1 \\ \color{Red}{0} & -1 & -2 & -1 & 1 & 2 \\ \end{array} \right) \label{37}\tag{37}$$

Compare this modified matrix $B$ to matrix $T$:

$$T = \left( \begin{array}{ccccccc} \color{Red}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{+1}&\color{Blue}{\cdots} \\ \color{Red}{+1}&-1&+1&-1&+1&-1&+1 \\ \color{Red}{+1}&+1&-2&+1&+1&-2&+1 \\ \color{Red}{+1}&-1&+1&-1&+1&-1&+1 \\ \color{Red}{+1}&+1&+1&+1&-4&+1&+1 \\ \color{Red}{+1}&-1&-2&-1&+1&+2&+1 \\ \color{Red}{+1}&+1&+1&+1&+1&+1&-6 \\ \color{Red}{\vdots}&&&&&&&\ddots \end{array} \right) \label{38}\tag{38}$$

They look like they are the same.

Subtracting the numbers in red and blue and forming matrix $A$:

$$A=\left( \begin{array}{cccccc} \color{Red}{0} & \color{Blue}{0} & \color{Blue}{0} & \color{Blue}{0} & \color{Blue}{0} & \color{Blue}{0} \\ \color{Red}{0} & \frac{\pi }{128} & -\frac{\pi }{192} & \frac{\pi }{256} & -\frac{\pi }{320} & \frac{\pi }{384} \\ \color{Red}{0} & -\frac{\pi }{192} & \frac{\pi }{144} & -\frac{\pi }{384} & -\frac{\pi }{480} & \frac{\pi }{288} \\ \color{Red}{0} & \frac{\pi }{256} & -\frac{\pi }{384} & \frac{\pi }{512} & -\frac{\pi }{640} & \frac{\pi }{768} \\ \color{Red}{0} & -\frac{\pi }{320} & -\frac{\pi }{480} & -\frac{\pi }{640} & \frac{\pi }{200} & -\frac{\pi }{960} \\ \color{Red}{0} & \frac{\pi }{384} & \frac{\pi }{288} & \frac{\pi }{768} & -\frac{\pi }{960} & -\frac{\pi }{576} \\ \end{array} \right) \label{39}\tag{39}$$

Since the conjectured formula for matrix $B$ is:

$$B(n,k) = -\frac{\pi a(\gcd (n,k))}{32 n k} \label{40}\tag{40}$$

and given that the matrix in display \eqref{37} is essentially matrix $T$ in displays \eqref{38} and \eqref{6}, then matrix $A$ is known to have the generating function and limit:

$$\begin{align} \sum _{n=1}^{\infty } \left(\sum _{k=1}^{\infty } \frac{A(n,k)}{k^s n^s}\right) &=-\frac{\pi}{32} \underset{s\to 1}{\text{lim}}\left(\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}-\color{Red}{\zeta (s)}-\color{Blue}{(\zeta (s)-1)}\right) \label{41}\tag{41} \\ &=-\frac{\pi (1-2 \gamma )}{32} \label{42}\tag{42} \\ \end{align}$$

Since: $$-\frac{\pi}{32} \underset{s\to 1}{\text{lim}}\left(\color{Blue}{(\zeta (s)-1)}\right) = \color{Blue}{-\frac{\pi}{32} \sum _{k=2}^{\infty } \frac{1}{k}} \label{43}\tag{43}$$

we get the wrong answer:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=-\frac{\pi (1-2 \gamma )}{32}\color{Blue}{-\frac{\pi}{32} \sum _{k=2}^{\infty } \frac{1}{k}} \label{44}\tag{44}$$

compared to V.V.Volchov:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt = \frac{\pi(3-\gamma)}{32} \label{45}\tag{45}$$

It is hard for me to tell exactly what I did wrong, but I am pretty sure it is a matter of analytic continuation that I don't know how to deal with. Unless also V.V.Volchov was slightly wrong regarding the resulting pole in blue, but I don't know about that because I have not read his paper.

---

Edit 9.11.2024:

Mathematica 14 for the comment below:

Clear[s, sigma, t]
int = Integrate[1/7^(s - 1)/(s - 1), s]
s = sigma + I*t
v = int
Integrate[(ExpIntegralEi[-((-1 + sigma + I  t)  Log[7])] + 
    ExpIntegralEi[-((-1 + sigma - I  t)  Log[7])])/2, {sigma, 1/2, 
  Infinity}]
Integrate[
 1/4  ((1 - 2  I  t)  ExpIntegralEi[
      1/2  (1 - 2  I  t)  Log[7]] + (1 + 2  I  t)  ExpIntegralEi[
      1/2  (1 + 2  I  t)  Log[7]] - (2 7^(1/2 - I t) (1 + 7^(2 I t)))/
    Log[7])(1 - 12t^2)/(1 + 4*t^2)^3, {t, 0, Infinity}]
Clear[k, m]
Limit[1/32   [Pi]   (1 + Log[m]) - Pi/32*Sum[1/k, {k, 2, m}], m -> Infinity]

where seven plays the role of $m$

It is possible to get rid of the sum

$$\color{Blue}{-\frac{\pi}{32} \sum _{k=2}^{\infty } \frac{1}{k}} \label{46}\tag{46}$$

in blue, if one is allowed the following:

Continuing from display \eqref{43}: The indefinite integral of the analytic continuation of the sum in blue is:

$$\color{Blue}{\int \zeta (s)-1 \, ds=\int \underset{m\to \infty }{\text{lim}}\left(\sum _{k=2}^m \frac{1}{k^s}+\frac{1}{(s-1) m^{s-1}}\right) \, ds} \label{47}\tag{47}$$

$$\color{Blue}{=-\sum _{k=2}^m \frac{k^{-s}}{\log (k)}+\text{Ei}(-((s-1) \log (m)))} \label{48}\tag{48}$$

Substituting:

$$s=\sigma+i t \label{49}\tag{49}$$

and integrating the real part of $\color{Blue}{\text{Ei}(-((s-1) \log (m)))}$: $$\color{Blue}{\int_{\frac{1}{2}}^{\infty } \frac{1}{2} (\text{Ei}(-((\sigma+i t-1) \log (m)))+\text{Ei}(-((\sigma-i t-1) \log (m)))) \, d\sigma=\frac{1}{4} \left((1+2 i t) \text{Ei}\left(\frac{1}{2} (2 i t+1) \log (m)\right)+(1-2 i t) \text{Ei}\left(\frac{1}{2} (1-2 i t) \log (m)\right)-\frac{2 \left(1+m^{2 i t}\right) m^{\frac{1}{2}-i t}}{\log (m)}\right)} \label{50}\tag{50}$$

Integrating again: $$\color{Blue}{\int_0^{\infty } \frac{\left(1-12 t^2\right) \left((1+2 i t) \text{Ei}\left(\frac{1}{2} (2 i t+1) \log (m)\right)+(1-2 i t) \text{Ei}\left(\frac{1}{2} (1-2 i t) \log (m)\right)-\frac{2 \left(1+m^{2 i t}\right) m^{\frac{1}{2}-i t}}{\log (m)}\right)}{4 \left(4 t^2+1\right)^3} \, dt} \label{51}\tag{51}$$

we get the answer:

$$\color{Blue}{=\frac{1}{32} \pi (\log (m)+1)} \label{52}\tag{52}$$

plugging \eqref{52} into \eqref{46}:

$$\color{Blue}{-\frac{\pi}{32} \sum _{k=2}^{\infty } \frac{1}{k}} \label{53}\tag{53}$$

$$\color{Blue}{\underset{m\to \infty }{\text{lim}}\left(\frac{1}{32} \pi (\log (m)+1)-\frac{1}{32} \pi \sum _{k=2}^m \frac{1}{k}\right)=-\frac{1}{32} (\gamma -2) \pi} \label{54}\tag{54}$$

Taking the black part of the right hand side of \eqref{2} and substituting the blue part of \eqref{2} with the right hand side of \eqref{54}:

$$\frac{1}{32} \pi (1-2 \gamma )\color{Blue}{-\frac{1}{32} (\gamma -2) \pi} =\frac{\pi(3-3\gamma)}{32} \label{55}\tag{55}$$

we get the less incorrect answer:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-3\gamma)}{32} \label{56}\tag{56}$$

compared to V.V.Volchov:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt = \frac{\pi(3-\gamma)}{32} \label{57}\tag{57}$$

---

Edit 10.11.2024:

Mathematica 14:

nn = 10^3;
NIntegrate[(1 - 12*t^2)/(1 + 4*t^2)^3*
  NIntegrate[Log[Abs[Zeta[sigma + I*t]]], {sigma, 1/2, 9*nn}], {t, 0, nn}]

gives:

0.237853...

while:

N[Pi*(3 - EulerGamma)/32]

gives:

0.237856...

which numerically supports V.V.Volchov's number $$\frac{\pi(3-\gamma)}{32}=0.237856...\text{=probably right}$$ in displays \eqref{1}, \eqref{45} and \eqref{57} being probably right, while my number $$\frac{\pi(3-3\gamma)}{32}=0.12452...\text{=probably wrong}$$ in displays \eqref{2.1} and \eqref{56} is probably wrong.

Answer 2

The right hand side of display $(54)$ in the other answer:

$$\color{Blue}{\underset{m\to \infty }{\text{lim}}\left(\frac{1}{32} \pi (\log (m)+1)-\frac{1}{32} \pi \sum _{k=2}^m \frac{1}{k}\right)=-\frac{1}{32} (\gamma -2) \pi}$$

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\Re\left(\left.\int(\zeta(s)-1)~ds\right|_{s=\sigma+it}\right)~d\sigma ~dt = \color{Blue}{\frac{\pi(2-\gamma)}{32}}$$

is not far from display $(1)$:

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\Re\left(\left.\int\frac{\zeta'(s)}{\zeta(s)}~ds\right|_{s=\sigma+it}\right)~d\sigma ~dt = \frac{\pi(3-\gamma)}{32}$$

Mathematica 14:

Clear[m, s, d, sigma, t]
m = 7;
d = 11;
int1 = Integrate[1/m^s, s];
int2 = Integrate[1/d^(s - 1)/(s - 1), s];
s = sigma + I*t;
v1 = int1;
w1 = int2;
s = sigma - I*t;
v2 = int1;
w2 = int2;
inte1 = Integrate[(v1 + v2)/2, {sigma, 1/2, Infinity}];
inte2 = Integrate[(w1 + w2)/2, {sigma, 1/2, Infinity}];
Integrate[inte1(1 - 12t^2)/(1 + 4*t^2)^3, {t, 0, Infinity}]
Integrate[inte2(1 - 12t^2)/(1 + 4*t^2)^3, {t, 0, Infinity}]