Evaluating $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ using complex analysis Again, improper integrals involving $\ln(1+x^2)$
I am trying to get a result for the integral $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ - asked above link- using some complex analysis, however, I couldn't find an appropriate solution. (Of course, for $\alpha \in (0,3)$ as stated above link (o.w. it is divergent))
Any ideas?

Differentiation under the integral sign is another possibility. Let:
$$ I(\alpha,\beta) = \int_{0}^{+\infty}\frac{\log(1+\beta x^2)}{x^\alpha}\,dx. \tag{1}$$
Then assuming $1<\alpha<3$ we have:
$$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+\beta x^2}\,dx = \beta^{\frac{\alpha-3}{2}}\int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+x^2}\,dx\tag{2}$$
and by replacing $x$ with $\tan\theta$ we get, through the Euler beta function and the reflection formula for the $\Gamma$ function:
$$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \beta^{\frac{\alpha-3}{2}}\int_{0}^{\pi/2}\left(\tan\theta\right)^{2-\alpha}\,d\theta = -\beta^{\frac{\alpha-3}{2}}\frac{\pi}{2\cos\left(\frac{\pi \alpha}{2}\right)}\tag{3}$$
and by integrating $(3)$ with respect to $\beta$ we have:
$$ \int_{0}^{+\infty}\frac{\log(1+x^2)}{x^\alpha}\,dx = \frac{\pi}{1-\alpha}\,\sec\left(\frac{\pi\alpha}{2}\right).\tag{4}$$
