Asymptotics of an improper integral I have to show that if $x \to \infty$, then
$$
 \int\limits_{\mathbb{R}^d} \frac{e^{i\xi x}}{\xi^2 + 2k\xi}d\xi = O\left(|x|^{-\frac{d-1}{2}} \right)  \;\;\; \; d\geqslant2, \;\;\; k\in \mathbb{C}^d
$$
where $k^2 = 0$ and $k\neq0$ if $d = 2$. I made a variable change $\xi = |x| \eta$ but it didn't help. What I have to do?
 A: How about using $\int_0^\infty \exp(-t u) \mathrm{d} t = \frac{1}{u}$ for $u>0$. Now let $u = \xi^2 + 2 k \xi = (\xi + k)^2 -k^2 = (\xi+k)^2$ and first solve:
$$ \begin{multline}
  \int_{\mathbb{R}^d} \exp\left( i \xi x - t(\xi +k)^2\right) \mathrm{d} \xi =
  \int_{\mathbb{R}^d} \exp\left( -t ( \xi + k - \frac{i}{2t} x)^2 - \frac{x^2}{4 t} - i k x \right) \mathrm{d} \xi = \\
  \int_{\mathbb{R}^d} \exp\left( -t \zeta^2 - \frac{x^2}{4 t}  - i k x\right) \mathrm{d} \zeta =
   \left( \frac{\pi}{t} \right)^{d/2}  \exp\left(- \frac{x^2}{4 t} - i k x\right) 
 \end{multline}
$$
Now the answer your original integral $\mathcal{I}$ is reduced to univariate:
$$
  \mathcal{I} = \int_0^\infty \left( \frac{\pi}{t} \right)^{d/2}  \exp\left(- \frac{x^2}{4 t} - i k x \right)  \, \mathrm{d} t \, \stackrel{t = u^{-1}}=  \,
 \pi^{d/2} \int_0^\infty u^{d/2 - 2}  \exp\left(- \frac{x^2}{4} u - i k x \right) \, \mathrm{d} u
$$
The latter integral is the defining integral for Euler's $\Gamma$-function and after suitable rescaling yields:
$$
 \mathcal{I} = (\pi)^{d/2} \mathrm{e}^{-i k x} \left(\frac{x^2}{4}\right)^{1-d/2} \Gamma\left(\frac{d}{2} -1\right) =  (\pi)^{d/2} \mathrm{e}^{-i k x} \left(\frac{\vert x \vert}{2}\right)^{2-d} \Gamma\left(\frac{d}{2} -1\right)
$$
Notice that $d > 2$ is required for convergence.
