Since $I(\alpha,-r)=\bar{I}(\alpha,r)$ (complex conjugate), we may restrict ourselves to $r>0$. The integral then evaluates to
$$I(\alpha,r)=\int_{-\infty}^\infty dk\, e^{ikr} \cfrac{\alpha^2 + \beta k^2}{k(k^2+\alpha^2)}=i\pi+i\pi(\beta-1)e^{-|\alpha|r}.$$
So for $\alpha=0$ the result is $I(0,r)=i\pi\beta$. There is no discontinuity at $\alpha=0$, but there is a discontinuous derivative.

Care should be taken because of the pole at $k=0$, here I am taking the principal value of the integral. In the question the pole is shifted off the real axis in a way that leads to an inconsistency in the calculation of $I_1$ and $I_2$.