Taking the limit out in integral

Question

Let $x\in\mathbb{R}$, let $\alpha\in\mathbb{R}_{>0}$, define $\alpha^+=\lim_{t\downarrow0}(\alpha+it)$, and consider the integral \begin{equation} I(x)=\int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-\alpha^+}. \end{equation} For any $t\in\mathbb{R}_{>0}$, the integral \begin{equation} \int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-(\alpha+it)} \end{equation} is readily done using complex analysis. However, I'm a bit confused as to how the limit behaves with respect to the integral. I want to say that \begin{equation} \int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-\alpha^+}=\int_{-\infty}^\infty dy\lim_{t\downarrow0}\frac{\exp(ixy)}{y^2-(\alpha+it)}, \end{equation} and consequently that \begin{equation} \int_{-\infty}^\infty dy\lim_{t\downarrow0}\frac{\exp(ixy)}{y^2-(\alpha+it)}=\lim_{t\downarrow0}\int_{-\infty}^\infty dy\frac{\exp(ixy)}{y^2-(\alpha+it)}, \end{equation} but I'm not sure if this is how it works and right now I can't think of any theorems or restrictions which play a role here. I think that the first equality holds because the integrand is continuous for all $y\neq\pm\sqrt{\alpha}$ and $\{\pm\sqrt{\alpha}\}$ is a set of measure zero, but I'm not sure. Any thoughts?