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. Alternatively, you could shift the pole off the real axis, still taking $r>0$ the answer then becomes

$$I(\alpha,r)=\lim_{\epsilon\downarrow 0}\int_{-\infty}^\infty dk\, e^{ikr} \cfrac{\alpha^2 + \beta k^2}{(k-i\epsilon)(k^2+\alpha^2)}=2i\pi+i\pi(\beta-1)e^{-|\alpha|r}.$$
So now $I(0,r)=i\pi(\beta+1)$, still continuous and with a discontinuous derivative.