parameter value in the solution of $\int_0^\pi\arctan\bigl(\frac{\ln\alpha\sin x}{x}\bigr)\mathrm{d}x$

Question

Evaluating $\int_0^\pi\arctan\bigl(\frac{\ln\sin x}{x}\bigr)\mathrm{d}x$

In the first two solutions of the above post,why is the parameter value $\alpha\in(0,1)$ or $\mu\in(0,1)$ . The solution does seem to work for $\alpha\geqslant 1$ or $\mu\geqslant 1$ also.

Answer 1

In the linked post, the first answer by @M.N.C.E. gave the answer.

Using your notation and his/her result $$I'(\alpha)=\frac {\pi^2}{2\alpha} \frac{1}{\log ^2\left(\frac{\alpha}{2}\right)+\frac{\pi ^2}{4}}$$ and $$\int_0^\pi \tan ^{-1}\left(\frac{\log (\alpha \sin (x))}{x}\right)\,dx=\pi \tan ^{-1}\left(\frac{2 }{\pi }\log \left(\frac{\alpha }{2}\right)\right)$$ which works $\forall \alpha >0$.