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Let $\zeta$ denote the Riemann zeta function. Using the Cauchy integral theorem, can you evaluate

$$I=\int_{\Re(s)=\frac{1}{2}} \frac{(2s-1)}{s^{2}(1-s)^2}\Bigg[\int \log((s-1) \zeta(s)) \mathrm{d}s\Bigg] \mathrm{d}s?$$

Note that $I$ converges since $\zeta(s)=O(|s|)$.

I have provided an answer below as an attempt.

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  • $\begingroup$ Assuming this isn't somehow related to Riemann Hypothesis (real part 1/2, and $\zeta$ itself are a bit concerning here)... perhaps if I were to look at it? But you need to show effort, context, etc. $\endgroup$ – Brevan Ellefsen Dec 11 '18 at 10:30
  • $\begingroup$ @BrevanEllefsen, see my attempted answer. $\endgroup$ – user507152 Dec 11 '18 at 11:28
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    $\begingroup$ "addendum" means something extra you added afterwards. I was saying your comments are very good, and you should add the material in them to your answer below. I did not edit your answer, but I recommend you do to add the material in your comments. $\endgroup$ – Brevan Ellefsen Dec 11 '18 at 13:09
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    $\begingroup$ What paper? I sincerely hope you don't mean you are submitting an attempted proof at the RH. Such an attempt would almost surely be ignored and never read if submitted out of nowhere, because so many crack proofs have been submitted. If this is what you mean, please run the paper by a professor first. Otherwise, I wish you the best of luck! $\endgroup$ – Brevan Ellefsen Dec 11 '18 at 14:37
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    $\begingroup$ There are a few journals that would look at such a paper if it's well written. As I said before, your thoughts look fine on a quick glance, but I can't verify their validity without very careful checking, which I lack time to do. If you intend to submit your argument as a proof you will have to make it super precise and clear to show it's valid. I wish you luck. $\endgroup$ – Brevan Ellefsen Dec 11 '18 at 14:53
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One trivial mistake that blows your argument away is that RZ is infinity at 1 so its log is not defined in your S; you need S to exclude the real line, so you have to work for t>T>0 as otherwise, you cannot apply the residue theorem since you would have a logarithmic singularity at 1

(edited later after tons of corrections by the original poster to address various objections for careless flaws that appeared in the original computation):

The main flaw of the computation lies in the misunderstanding of the complex-log, namely that while the complex-log behaves on branch points as noted in the paper by Bui et al referred in the comments from which the original poster copy pasted his argument - so it jumps by 2*Pii when crossing such-, its primitive doesn't as shown by the simple example of log(z) vs zlog(z)-z on the complex plane minus the negative axis.

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  • $\begingroup$ As you said ''trivial'' we can just ''trivially'' define $S$ so that it excludes $t=0$, in the same vein we excluded the segments $1/2 + i\gamma$ to $\beta + i\gamma$. I've edited the answer to include that. $\endgroup$ – user507152 Dec 12 '18 at 23:36
  • $\begingroup$ also your residue theorem application is just wrong as stated as you do not define what the path infinity - i*infinity means anyway $\endgroup$ – Conrad Dec 12 '18 at 23:43
  • $\begingroup$ basically you use a sleight of hand to make an integral zero (since the supposed path at infinity is technically empty), hence the sum of the terms is zero, hence no terms $\endgroup$ – Conrad Dec 12 '18 at 23:45
  • $\begingroup$ It seems you're not familiar with the concept of an indefinite shift of a contour. I suggest you familiarise yourself with that first, and also read the proofs of Theorems 1.2-1.3 of arxiv.org/abs/1306.0856 $\endgroup$ – user507152 Dec 12 '18 at 23:56
  • $\begingroup$ there the path is clear as they use a positive T as the imaginary part of the indefinite shift - your residue theorem application is wrong (the path used in the Bui et all paper is quite clear in both applications $\endgroup$ – Conrad Dec 13 '18 at 0:06

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