I was looking at this beautiful answer, but I got stuck almost at the end of the demonstration, I've been thinking about how to solve it for days but I can't, could you help me understand it? The function is defined $F(x) = \displaystyle\int_1^x \frac{ \log t}{(t+a)(t-1)} dt,\quad x\ge 0$. I copy just the part that interests me (red is what he couldn't see): For $z=-1+it, t>0$, the second term can be estimated as \begin{align*} |1+z|\left|\int_0^\infty \frac{F(x)}{(x+z)^2} dx\right| \le& t \int_0^\infty \frac{|F(x)|}{|x-1+it|^2} dx \\ \le& \int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx\\ &\xrightarrow{t\to 0} \pi \color{red}{|F(1)|} = 0. \end{align*}
For a moment I thought I could overbound $|F(x)|$ with some $C$ such that \begin{align*} \int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx&\le C \int_0^\infty \frac{t}{(x-1)^2 + t^2}dx\\ &=C\cdot \arctan(\frac{x-1}{t})\Big|_0^{+\infty}\\ &=C\cdot (\frac{\pi}{2}+\arctan \frac{1}{t}) \end{align*} If $t\to 0^+$ then we would have $\displaystyle\int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx\le \pi C$. But that would suggest that the upper bound $C$ is $|F(1)|=0$. Which I don't understand why. Maybe my ideas are wrong, could someone throw me a lifeline with this?
Simply let $x-1 = tu$:
$$\int_0^\infty \frac{t}{(x-1)^2+t^2}|F(x)|dx = \int_{-\frac{1}{t}}^\infty \frac{t^2}{t^2u^2+t^2}|F(tu+1)|du$$
$$\longrightarrow \int_{-\infty}^\infty \frac{1}{1+u^2}|F(1)|du = \pi|F(1)|$$
as $t\to0^+$ by dominated convergence.