Evaluate $$I=\int_{0}^{\infty} \frac{t\arg \Gamma(\frac{1}{4}+\frac{it}{2})}{(\frac{1}{4}+t^2)^2}\mathrm{d}t$$ where $\Gamma(s)=\int_{0}^{\infty}e^{-x}x^{s-1}\mathrm{d}x.$

Note that $I$ converges since $\Gamma(s)\sim s\log s$. I tried Wolfram Alpha, but it hasn't given me an answer after almost 90 minutes of computation, hence perhaps will never do.

PS: Migrated from https://math.stackexchange.com/q/3045147

ADDENDUM: Let $\zeta$ be the Riemann zeta function. Since it can be shown by the Riemann functional equation (see, for example https://math.stackexchange.com/q/3045666/627517) that $2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big)$, note that

$$I_2=\int_{0}^{\infty} \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x.$$

Thus by invoking the fact that $\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x=\frac{1}{2}\pi \log \pi$ and @Juan's answer below, we conclude that $I_2 > \frac{1}{2}\pi\log \pi$. Note that this disproves the Riemann Hypothesis since Volchkov proved that the RH is equivalent to the statement that $I=\pi(\gamma-3)$, where $\gamma=0.577\cdots$ is the Euler constant.