The first integral reads:
$$ f(\vec{r})=\iiint_{-\infty}^\infty \mathrm{d}k_x \, \mathrm{d}k_y \, \mathrm{d}k_z \, \frac{e^{i\vec{k}\cdot\vec{r}}}{(k_0-\alpha k_z)^2-k^2+i(k_0-\alpha k_z)\eta} $$
the second integral is very similar, except the infinitesimal parts: $$ f(\vec{r})=\iiint_{-\infty}^\infty \mathrm{d}k_x \, \mathrm{d}k_y \, \mathrm{d}k_z \, \frac{e^{i\vec{k}\cdot\vec{r}}}{(k_0-\alpha k_z)^2-k^2+i\eta} $$
where $\alpha>1$, $k_0>0$ is finite positive number, $\eta = 0^+$ is the infinitesimal positive number. $\vec{k}=(k_x,k_y,k_z), \vec{r}=(x,y,z), k=\sqrt{k_x^2+k_y^2+k_z^2}$.
As a hint, the integral for $\alpha=0$ can be done as following:
\begin{align} &\int\mathrm{d}^3k \frac{e^{i\vec{k}\cdot\vec{r}}}{k_0^2-k^2+i\eta} \quad\text{--- making $\vec{r}$ as $z$ axis for $\vec{k}$ } \\ =& 2\pi\int_{-1}^1\mathrm{d}\xi\int_0^\infty \mathrm{d}k k^2\frac{e^{ikr\xi}}{k_0^2-k^2+i\eta}\quad\text{--- integral over $\phi_k$, and let $\xi=\cos\theta_k$}\\ =&-\frac{2\pi}{i r} \int_0^\infty \frac{k\,\mathrm{d}k}{k^2-(k_0^2+i\eta)}(e^{ikr}-e^{-ikr})\\ =&-\frac{2\pi}{i r}\int_{-\infty}^\infty \frac{k\,\mathrm{d}k}{k^2-(k_0^2+i\eta)}e^{ikr} \quad\text{--- use the upper semicircle as contour.} \\ =&-(2\pi)^2\frac{e^{ik_0r}}{2r} \end{align}
It seems that the first integral is zero as pointed by @Fabian. As for the second integral, @Felix Martin makes a variable substitution and integrate out one dimension, it seems that it's hard to proceed and get an analytical form.
Therefore, is it possible to get the asymptotic form of the second integral when $r\to\infty$? (i.e. keep only $1/r$ part, ignore $1/r^2$ etc.)
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\mrm{f}\pars{\vec{r}}} & = \iiint\limits_{-\infty}^{\infty}\dd k_{x}\,\dd k_{y}\,\dd k_{z}\,{\expo{\ic\vec{k}\cdot\vec{r}} \over \pars{k_{0} - \alpha k_{x}}^{\, 2} - k^{2} + \ic\eta} \\[5mm] & = \iiint\limits_{-\infty}^{\infty}{\expo{\ic\vec{k}\cdot\vec{r}} \over \alpha^{2}\pars{k_{0}/\alpha - k_{x}}^{\, 2} - k^{2} + \ic\eta} \,\dd k_{x}\,\dd k_{y}\,\dd k_{z} \label{1}\tag{1} \end{align} Now, I make the substitution $\ds{\pars{k_{x} - {k_{0} \over \alpha}} \mapsto k_{x}}$ in \eqref{1}: \begin{align} \color{#f00}{\mrm{f}\pars{\vec{r}}} & = \iiint\limits_{-\infty}^{\infty} {\expo{\ic\pars{k_{x}\ +\ k_{0}\,/\alpha}x}\ \expo{\ic k_{y}\,y\ +\ \ic k_{z}\,z} \over \alpha^{2}k_{x}^{\, 2} - \pars{k_{x} + k_{0}/\alpha}^{2} - k_{y}^{2} - k_{z}^{2} + \ic\eta}\,\dd k_{x}\,\dd k_{y}\,\dd k_{z} \\[5mm] & = \exp\pars{{k_{0} \over \alpha}\,x\,\ic}\iiint\limits_{-\infty}^{\infty}{\expo{\ic\vec{k}\cdot\vec{r}} \over \pars{\alpha^{2} - 1}k_{x}^{2} -2k_{0}k_{x}/\alpha- k_{y}^{2} - k_{z}^{2} - k_{0}^{2}/\alpha^{2} + \ic\eta} \,\dd k_{x}\,\dd k_{y}\,\dd k_{z} \end{align}
Then, \begin{align} \color{#f00}{\mrm{f}\pars{\vec{r}}} & = \exp\pars{{k_{0} \over \alpha}\,x\,\ic} \int_{-\infty}^{\infty}\int_{0}^{\infty}\rho\int_{0}^{2\pi}{\expo{\ic k_{x}\,x} {\expo{\ic\rho\bracks{y\cos\pars{\theta} + z\sin\pars{\theta}}}} \over \pars{\alpha^{2} - 1}k_{x}^{2} -2k_{0}k_{x}/\alpha - \rho^{2} - k_{0}^{2}/\alpha^{2} + \ic\eta}\,\,\,\,\,\dd\theta\,\dd\rho\,\dd k_{x} \end{align} The $\ds{\theta}$-integral can be expressed in terms of the $\ds{\,\mathrm{J}_{0}}$ Bessel Function: \begin{align} \color{#f00}{\mrm{f}\pars{\vec{r}}} & = 2\pi\exp\pars{{k_{0} \over \alpha}\,x\,\ic} \int_{0}^{\infty}\rho\,\mrm{J}_{0}\pars{\root{y^{2} + z^{2}}\rho}\times \\& \underbrace{\int_{-\infty}^{\infty}{\expo{\ic k_{x}\,x} \over \pars{\alpha^{2} - 1}k_{x}^{2} -2k_{0}k_{x}/\alpha - \rho^{2} - k_{0}^{2}/\alpha^{2} + \ic\eta}\,\,\,\,\,\dd k_{x}} _{\ds{\mc{F}_{1}\pars{\rho} + \mc{F}_{2}\pars{\rho}}}\,\dd\rho \end{align}
I assume $\ds{\alpha > 0}$ and $\ds{\eta = 0^{+}}$ ( for the purpose of the example ): \begin{align} \mc{F}_{1}\pars{\rho} & = \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic k_{x}\,x} \over \pars{\alpha^{2} - 1}k_{x}^{2} -2k_{0}k_{x}/\alpha - \rho^{2} - k_{0}^{2}/\alpha^{2}}\,\dd k_{x} \\[5mm] \mc{F}_{2}\pars{\rho} & = -\ic\pi \int_{-\infty}^{\infty}\expo{\ic k_{x}\,x} \delta\pars{\bracks{\alpha^{2} - 1}k_{x}^{2} -2k_{0}k_{x}/\alpha - \rho^{2} - k_{0}^{2}/\alpha^{2}}\,\dd k_{x} \end{align} The $\ds{\delta}$-argument vanishes out when $\ds{k_{x}}$ takes the values $\ds{k_{x\pm}\pars{\rho} = {k_{0} \pm \alpha\root{\pars{\alpha^{2} - 1}\rho^{2} + k_{0}^{2}} \over \alpha\pars{\alpha^{2} - 1}}}$. \begin{align} \mc{F}_{2}\pars{\rho} & = -\ic\,{\pi \over \alpha^{2} - 1} \int_{-\infty}^{\infty}\expo{\ic k_{x}x} \delta\pars{\bracks{k_{x} - k_{x+}\pars{\rho}} \bracks{k_{x} - k_{x-}\pars{\rho}}}\,\dd k_{x} \\[5mm] & = -\ic\,{\pi \over \alpha^{2} - 1} \int_{-\infty}^{\infty}\expo{\ic k_{x}x}\, {\delta\pars{k_{x} - k_{x+}\pars{\rho}} + {\delta\pars{k_{x} - k_{x-}\pars{\rho}}} \over \verts{2k_{x} - k_{x+}\pars{\rho} - k_{x-}\pars{\rho}}}\,\dd k_{x} \\[5mm] & = -\ic\,{\pi \over \root{\pars{\alpha^{2} - 1}\rho^{2} + k_{0}^{2}}} \bracks{\exp\pars{\ic k_{x+}\pars{\rho}x} + \exp\pars{\ic k_{x-}\pars{\rho}x}} \\[5mm] & = -\ic\,{2\pi \over \root{\pars{\alpha^{2} - 1}\rho^{2} + k_{0}^{2}}}\, \exp\pars{{k_{0} \over \alpha\bracks{\alpha^{2} - 1}}\,x\ic} \cos\pars{{\root{\bracks{\alpha^{2} - 1}\rho^{2} + k_{0}^{2}} \over \alpha^{2} - 1}\,x\ic} \end{align}