How to do this two complex integrals? 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.)
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\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}
A: Your integral is zero for $\alpha>1$. For that you only have to look at the integral over $k_z$. As the integral decays fast enough, it is amenable to be treated by the method of residues. Let us take a look at the denominator ($\rho^2 = k_x^2 + k_y^2$)
$$D = (k_0 - \alpha k_z)^2 - \rho^2 -k_z^2 + i (k_0 -\alpha k_z) \eta.$$
The roots of this quadratic polynomial (for $k_z$) are in the complex plane situated at ($\beta =\alpha^2 -1 > 1$)
$$ k_{3\pm} = \frac{1}{\beta} \left( k_0 \alpha \pm \sqrt{k_0^2 +\beta \rho^2}\right) + i \tilde\eta_\pm.$$
The imaginary parts $\tilde \eta$ have to be only evaluated to first order in $\eta$, we obtain
$$ \tilde \eta_\pm = \frac{\eta}{2 \beta} \left( \alpha \pm \frac{1}{\sqrt{1 + \beta \rho^2 /k_0^2}} \right). \tag{1}$$
Now observe that $\sqrt{1 + \beta \rho^2 /k_0^2}>1$ such that the term in the bracket of (1) is always positive ($\alpha>1$). Thus, we have that $\tilde \eta_\pm>0$ (both poles are above the real axis). Closing the contour below the real axis, we immediately obtain that your integral vanishes.
