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.
When you calculate $I_2$ you should take the principal value of the integral, which is one half of what you are calculating when you shift the pole off the real axis, which is why you obtain $2\pi i\beta$ instead of $i\pi\beta$.