# Showing for which parameter values an improper integral of $\ln(1+x^2)/x^\alpha$ converges

I need to show for which values of the parameter $$\alpha$$ does the integral $$\int_0^\infty \frac {\ln(1+x^2)} {x^\alpha}dx$$ converge. I only managed to deduce that the integral can't converge for $$\alpha\leq0$$ since the function diverges at infinity, that for $$\alpha\geq2$$ the function diverges at $$0$$ and for $$0<\alpha<2$$ the function approaches $$0$$ at $$0$$. I can't find an explicit formula for the antiderivative and I can't find any comparison test that will help me, so I'm not sure how to go about this.

Thanks to all the helpers.

$$\int_0^\infty\underbrace{\frac{log(1+x^2)}{x^\alpha}}_{x^2\rightarrow t}dx=\frac{1}{2}\int_0^\infty log(1+t)t^{\frac{1-\alpha}{2}-1}dt=\frac{\pi \csc\left(\frac{\pi-\pi\alpha}{2}\right)}{2\left(\frac{1-\alpha}{2}\right)}=\frac{\pi\sec\left(\frac{\pi\alpha}{2}\right)}{1-\alpha}$$
$$-1<\mathfrak{R}\left(\frac{1-\alpha}{2}\right)<0$$ $$\boxed{1<\mathfrak{R}\left(a\right)<3}$$
The result was derived using the Mellin Transform of $$log(1+t)$$.