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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?

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    If $\xi$ ranges over $\mathbb R^d$, then what does, say, $e^{i\xi x}$ even mean?2011-10-07
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    @Henning: the usual thing, $\xi\cdot x$, where $\xi,x\in\mathbb{R}^d$. What is more puzzling is the $k\xi$ term. $\xi^2$ I expect to mean $|\xi|^2$. But I cannot see how to form a scalar from $k$, which is defined to be in $\mathbb{C}$, and $\xi$. Considering the comment $k^2 = 0$ but $k\neq 0$, I am guessing the OP may mean that $k\in\mathbb{C}^d$?2011-10-07

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