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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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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.

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    I think that after variable change $\xi \to \zeta$ the integral is no more over the $R^d$.2011-10-07
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    It would be over $\mathbb{R}^d$ translated in $\mathbb{C}^d$ by a complex vector. Because the integrand is holomorphic, the integral is the same as if we integrated over $\mathbb{R}^d$.2011-10-07
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    Thanks, i didn't hear about this property... Tell me please, where can I read about it (demostration of this property)?2011-10-07
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    @Nimza Consider the 1D problem first, as multivariate one reduces to this. Let $I = \int_\mathbb{R} \exp(-(x+i y)^2) \mathrm{d} x = \int_{-\infty}^\infty \exp( -x^2 + y^2 - 2 i x y) \mathrm{d} x = 2 \mathrm{e}^{y^2} \int_0^\infty \exp(-x^2) \cos(2 x y) \mathrm{d} x$. You can show that $I = \sqrt{\pi} = \int_\mathbb{R} \exp(-x^2) \mathrm{d} x$. If you are unsure on how to show it, please ask a separate question.2011-10-07
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    And what if $Re(\xi + k)^2 < 0$? I think we should consider 2 integrals: one over $B = \{ \xi :|Re(k) + \xi| \leq |Im(k)|\}$ and other over $R^d \setminus B$2011-10-08