I have a simple Fourier transform problem, originating from mathematical physics (system of linear PDEs), which reduces to taking the integral
$$
I(\alpha)\equiv\int_{-\infty}^\infty e^{ikr} \cfrac{\alpha^2 + \beta k^2}{k(k^2+\alpha^2)}\, dk
$$
where $\alpha$ and $\beta$ are some free real-valued parameters. I need to compute $I(0)$. It turns out that if I simply set $\alpha=0$ in the integral above, I get an **absolutely different answer** than if I first compute the integral and set $\alpha\to 0$ in the final expression.

My question is as follows: **why do these two procedures lead to different answers**? From the physical viewpoint this means that a massless field behaves in a totally different way than a massive with infinitesimal mass, which seems unreasonable.

My attempt is as follows.

I lift the pole at $k=0$ to the upper half-plane: $$ I_\varepsilon(\alpha) \equiv \int_{-\infty}^\infty e^{ikr} \cfrac{\alpha^2 + \beta k^2}{(k-i\varepsilon)(k+i\alpha)(k-i\alpha)}\, dk \equiv \int_{-\infty}^\infty \cfrac{g(k)}{h(k)}\, dk $$ where $$ h(k) = (k-i\varepsilon)(k+i\alpha)(k-i\alpha)=k^3-i\varepsilon k^2+a^2k +i \varepsilon a^2, $$ $$ h'(k)=3k^2-2ik\varepsilon+\alpha^2 $$ I take the integral making use of the Jordan's lemma and Cauchy theorem: I choose a contour in the upper half-plane $\mathbb H$, so that the integral reduces to the sum of residues at $k=i\varepsilon$ and $k=i\alpha$: $$ I_1(\alpha)=2\pi i \lim_{\varepsilon \to 0}\left[\cfrac{g(i\varepsilon)}{h'(i\varepsilon)}+\cfrac{g(i\alpha)}{h'(i\alpha)}\right] $$ $$ =2\pi i\lim_{\varepsilon\to 0}\left[ \cfrac{\alpha^2 + \beta (i\varepsilon )^2}{3(i\varepsilon )^2-2i(i\varepsilon )\varepsilon +\alpha^2}\,e^{-\varepsilon r}+ \cfrac{\alpha^2 + \beta (i\alpha)^2}{3(i\alpha)^2-2i(i\alpha)\varepsilon +\alpha^2}\,e^{-\alpha r} \right] $$ $$ =2\pi i \left[ 1+ \cfrac{1 - \beta}{-3+1} \right]=2\pi i \cfrac{1+\beta}{2}=\pi i(1+\beta). $$ Thus, $I_1(\alpha) = \pi i(1+\beta)$. Clearly then, $\lim_{\alpha\to0}I_1(\alpha) = \pi i(1+\beta)$.

However, if I consider $$ I_2\equiv I(\alpha=0)= \int_{-\infty}^\infty e^{ikr} \cfrac{\beta}{k}\, dk = \lim_{\varepsilon \rightarrow 0}\int_{-\infty}^\infty e^{ikr} \cfrac{\beta}{k-i\varepsilon }\, dk = \lim_{\varepsilon \rightarrow 0}2\pi i \beta e^{-\varepsilon r} = 2\pi i \beta. $$ Hence, $I_2\neq \lim_{\alpha\to0}I_1$!!! Please, give a hint why this sort of thing happens. I clearly understand that in my reasoning there is a flaw -- but it escapes me.

Thank you for any help!