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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.

Care should be taken because of the pole at $k=0$, let me first take the principal value of the integral. I note that $I(\alpha,-r)=\bar{I}(\alpha,r)$ (complex conjugate), for convenience I will restrict myself to $r>0$. 

The principal value integral 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. The same result would have been obtained if we would have set $\alpha=0$ before carrying out the integral, because the principal value integral $\int dk e^{ikr}k^{-1}=i\pi$ for $r>0$. 

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.

We have recoved the result $I_1$, where the limit $\epsilon\downarrow 0$ is taken before the limit $\alpha\rightarrow 0$. These two limits do not commute, which is why the result $I_2$ in the OP differs from $I_1$.

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 $I_2$.

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.

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 $i\pi\beta$.

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 $I_2$.