# A computation in Conrey's paper on Riemann zeta function

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: \begin{align}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},\end{align} 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.

P.S. I have posted the question on Mathematics Stack Exchange few days ago. But I did not get an answer. I am sorry if this question does not appear to be about research level mathematics, and I will delete my question if it is the case.

• It looks fine to me: take $u = e^{2Ry}/2$, $v = (1+\lambda y)^2$. Then $\mathrm{d}u = R2^{2Ry}\, \mathrm{d}y$, $\mathrm{d}v = \mathrm{d}(1+\lambda y)^2$ and the third line is $\theta (\int_0^1 v\, \mathrm{d}u + \int_0^1 u\, \mathrm{d}v)$ which by integration by parts is $\theta [uv]_0^1$. – Mark Wildon Dec 2 '18 at 13:17
• The m.se posting is math.stackexchange.com/questions/3015359/… – Gerry Myerson Dec 2 '18 at 22:07