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
We prove that $$I=-\frac{\pi}{4}(\gamma+\log 4).$$ $$I=\int_0^\infty\frac{t\arg\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt.$$ $I$ is the imaginary part of the complex integral $$\int_0^\infty \frac{t\log\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt$$ using the usual branch of the logarithm of $\Gamma(s)$. Integrating by parts $$=\frac12\int_0^\infty \log\Gamma(\tfrac14+\tfrac{it}{2})\,d\Bigl\{-\frac{1}{\frac14+t^2}\Bigr\}= \Bigl.-\frac{1}{2(\frac14+t^2)}\log \Gamma(\tfrac14+\tfrac{it}{2})\Bigr|_{t=0}^\infty+\frac12\int_0^\infty \frac{1}{\frac14+t^2}\frac{i}{2}\frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\,dt$$ $$=2\log\Gamma(1/4)+\frac{i}{4}\int_0^\infty \frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\frac{dt}{\frac14+t^2}.$$ We have $$-\frac{\Gamma'(s)}{\Gamma(s)}=\gamma+\frac{1}{s}+\sum_{n=1}^\infty\Bigl(\frac{1}{s+n}-\frac{1}{n}\Bigr).$$ It is easy to justify the interchange here so that $$=2\log\Gamma(1/4)-\frac{i}{4}\Bigl\{\gamma\int_0^\infty \frac{dt}{\frac14+t^2}+ \int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}+\sum_{n=1}^\infty \int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2} \Bigr\}$$ With Mathematica we find $$\int_0^\infty \frac{dt}{\frac14+t^2}=\pi,\quad \int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}=2\pi-4i,$$ $$\int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2}=-\frac{\pi+i\log(1+4n)}{n(2n+1)}.$$ Taking the imaginary part we obtain $$I=-\frac14\Bigl\{\pi\gamma+2\pi-\sum_{n=1}^\infty \frac{\pi}{n(2n+1)}\Bigr\}$$ For the sum in $n$ we get with Mathematica $$\sum_{n=1}^\infty\frac{1}{n(2n+1)}=2-2\log 2.$$ It follows that $$I=-\frac14 \bigl\{\pi\gamma+2\pi-2\pi+2\pi\log2\bigr\}=-\frac{\pi}{4}(\gamma+\log 4).$$
The evaluations with Mathematica are not difficult to prove.
