On pages 35-36 here, we have that the integral
$$\frac{1}{2i\sqrt{y}}\int_{1/2-i\infty}^{1/2+i\infty}\phi(s-1/2)\phi(1/2-s)(s-1)\Gamma(1+s/2)\pi^{-s/2}\zeta(s)y^sds$$
equals for $\phi(s)=1$ to: $$\frac{1}{i\sqrt{y}}\sum_{n=1}^\infty \int_{2-i\infty}^{2+i\infty} (\Gamma(2+s/2)-\frac{3}{2}\Gamma(1+s/2))\left(\frac{y}{n\sqrt{\pi}}\right)^sds$$
I don't see how to derive this? I assume it has to do with the previous section, but I took a long pause from it before returning to this section, so if someone can explain this to me, that would be excellent.
We move the contour to the line $\operatorname{Re}(s)=2$ then a residue at $s=1$ appears, but this residue is zero. Replacing $\phi(s)=1$
$$ \begin{align*} I&:=\frac{1}{2i\sqrt{y}}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\pi^{-s/2}\zeta(s)y^sds\\ &=\frac{1}{2i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\pi^{-s/2}\frac{1}{n^s}y^sds\\ &=\frac{1}{2i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\left(\frac{y}{n\sqrt{\pi}}\right)^s ds\\ &=\frac{1}{i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}\left(\frac{s}{2}\Gamma(1+s/2)-\frac{1}{2}\Gamma(1+s/2)\right)\left(\frac{y}{n\sqrt{\pi}}\right)^s ds\\ \end{align*} $$ Finally, see that $$ \frac{s}{2}\Gamma(1+s/2)=\left(\frac{s}{2}+1-1\right)\Gamma(1+s/2)=(1+s/2)\Gamma(1+s/2)-\Gamma(1+s/2). $$ Then $(s/2)\Gamma(1+s/2)=\Gamma(2+s/2)-\Gamma(1+s/2)$. Replacing this we are done.