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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.

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$$\int_0^\infty\underbrace{\frac{log(1+x^2)}{x^\alpha}}_{x^2\rightarrow t}dx=\frac{1}{2}\int_0^\infty log(1+t)t^{\frac{1-\alpha}{2}-1}dt=\frac{\pi \csc\left(\frac{\pi-\pi\alpha}{2}\right)}{2\left(\frac{1-\alpha}{2}\right)}=\frac{\pi\sec\left(\frac{\pi\alpha}{2}\right)}{1-\alpha}$$

$$-1<\mathfrak{R}\left(\frac{1-\alpha}{2}\right)<0$$ $$\boxed{1<\mathfrak{R}\left(a\right)<3}$$

The result was derived using the Mellin Transform of $log(1+t)$.

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