1
$\begingroup$

I am reading the Conrey's paper "More than two fifths of the zeros of the Riemann zeta function are on the critical line" (see here). I have question/doubt in a particular step: In P.10, it claimed that $B=0$. I wonder why it is true, because, in my opinion, there should be an extra term coming from the integration by parts: $$B=\theta\int_0^1 w(y)\overline{w}'(y)dy =\theta\int_0^1 e^{2Ry}[R(1+\lambda y)^2+\lambda(1+\lambda y)]dy\\ =\theta\left(\int_0^1 e^{2Ry}R(1+\lambda y)^2dy+\frac{1}{2}\int_0^1e^{2Ry}d((1+\lambda y)^2)\right)\\ \\ =\theta \Bigg[\frac{1}{2}e^{2Ry}(1+\lambda y)^2\Bigg]_{y=0}^{y=1},$$ where the last equality follows from doing integration by parts. I think this paper have been read and checked by many people. I would appreciate if I can get help from some of you who are familiar with this paper. Thank you very much.

$\endgroup$
  • $\begingroup$ I cannot access the paper. The only thing I could say is that your integration is perfect. Now, the stupid question : is there any relation between $R$ and $\lambda$ ? $\endgroup$ – Claude Leibovici Nov 27 '18 at 5:37
  • $\begingroup$ @ClaudeLeibovici Thanks for the comment. But it seems that $R$ and $\lambda$ are not related. In P.10 of the paper, it was chosen such that $R=1.2$ and $\lambda=-1.02$, and it does not make the above term vanishes. I am not sure if I miss something. $\endgroup$ – Tong Nov 27 '18 at 11:51
  • $\begingroup$ Now posted to MO, mathoverflow.net/questions/316712/… $\endgroup$ – Gerry Myerson Dec 2 '18 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.