Does there exist a closed-form for the integral

$$\int 8\frac { (\zeta(1/2+it))^2\pi}{-\Psi(1,1/4-i/2t) ( \zeta(1/2+it))^2 +\Psi(1,1/4+i/2t) (\zeta(1/2+it))^2 + 8\zeta(2,1/2+it) \zeta(1/2+it) -8(\zeta(1,1/2+it))^2}\,{\rm d}t$$

the latex expression is equivalent to enter image description here

A graph of the integrand is .. enter image description here and it has a pole at 5.5611757696135...

I posted a related question at

Does $z (s) = \int_0^s \zeta \left( \frac{1}{2} + i t \right) d t = s + \sum_{n = 2}^{\infty} \frac{i (n^{- i s} - 1)}{\ln (n) \sqrt{n}}$ converge?

  • 4
    $\begingroup$ I highly doubt this monstrosity has a closed form. Prove me wrong. $\endgroup$ – Frank W. Jul 7 '18 at 2:57
  • $\begingroup$ what about just the integral $\int \!\zeta \left( 1/2+it \right) \,{\rm d}t$ then? $\endgroup$ – crow Jul 7 '18 at 21:49
  • $\begingroup$ I asked a related question at math.stackexchange.com/questions/2845044/… $\endgroup$ – crow Jul 8 '18 at 22:30
  • $\begingroup$ I thought this might be useful because a root off the line would cause there to be a turning point that doesnt touch zero.. I think $\endgroup$ – crow Jan 13 at 2:06

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