Let $g$ be in $C_c^\infty(\bf R)$, and $h$ its Fourier transform. There is the following equation:
\begin{align}\frac{1}{2\pi}\int^\infty_{-\infty}h(r)\frac{\Gamma'}{\Gamma}(1+ir)dr=&-\gamma g(0)+\int^\infty_0\log(u)g'(u)du\notag\\ &+\frac{1}{4}\int^\infty_{-\infty}g(u)du+\int^\infty_0\log\Big(\frac{\sinh(\frac{u}{2})}{\frac{u}{2}}\Big)g'(u)du.\notag \end{align}
I would like to know how to derive this so I can use it at other arguments of $\Gamma'/\Gamma$. (For which ones is it valid for?)
I know that for $z$ not a negative integer, $$\frac{\Gamma'}{\Gamma}(z)=-\gamma+\sum_{n=0}^\infty\Big(\frac{1}{n+1}-\frac{1}{n+z}\Big)$$ so the first term $\gamma g(0)$ follows immediately. The remaining ones I do not know how to deal with.