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I try to show that $$ \int\limits_{R^3} \frac{e^{i\xi x} d\xi}{\xi^2 - k^2 - i0} = e^{ikx} \int\limits_{R^3} \frac{e^{i\xi x}d\xi}{\xi^2 + 2(k + i0\frac{k}{|k|})\xi}, \;\;\; k,x \in \mathbb R^3 $$ I tried to make the change $\xi \mapsto \xi + k$ in second integral. I obtained $$ \lim\limits_{\epsilon \to 0} \; \int\limits_{\mathbb R^d} \frac{e^{i\xi x} d\xi}{(\xi + i\epsilon \frac{k}{|k|})^2 - k^2 + \epsilon^2 - 2i|k|\epsilon} $$ So I don't know what to do...

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Why write out $i0$? Isn't that just zero? –  Thomas Andrews Nov 8 '11 at 19:59
    
$\xi^2-k^2 - i 0$ is a way of saying that you consider $\lim_{\epsilon \downarrow 0} \frac{e^{i\xi x} }{\xi^2 - k^2 - i \epsilon} \mathrm{d} \xi$. –  Sasha Nov 8 '11 at 20:20
    
foil the square in the denominator ;) –  N. S. Nov 8 '11 at 20:21
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2 Answers

Stuff does work out in spherical coordinates. Let $\kappa >0 $ be such that $\kappa^2 = k^2$. Align north pole of spherical $\xi$-coordinates along $x$ vector. Let $\vert x \vert = \rho$. Then $ \xi \cdot x = r \rho \cos(\theta)$, and $\mathrm{d} \xi = r^2 \sin \theta \mathrm{d} r \mathrm{d} \theta \mathrm{d} \phi$: $$ \begin{eqnarray} \lim_{\epsilon \downarrow 0}\int_{\mathbb{R}^3} \frac{\mathrm{e}^{i\xi x} }{\xi^2 - k^2 - i \epsilon} \mathrm{d} \xi &=& \lim_{\epsilon \downarrow 0} \int_0^{2 \pi} \mathrm{d} \phi \int_0^\infty r^2 \mathrm{d} r \int_0^\pi \mathrm{d} \theta \, \sin(\theta) \cdot \frac{\mathrm{e}^{i r \rho \cos \theta} }{r^2 - \kappa^2 - i \epsilon} \\ &=& \lim_{\epsilon \downarrow 0} \left( 2 \pi \int_0^\infty r^2 \mathrm{d} r \frac{1}{r^2 - \kappa^2 - i \epsilon} \cdot \frac{2 \sin(r \rho)}{r \rho} \right) \\ &=& \lim_{\epsilon \downarrow 0} \left( 2 \pi \int_0^\infty r^2 \mathrm{d} r \frac{1}{r^2 - \kappa^2 - i \epsilon} \cdot \frac{2 \sin(r \rho)}{r \rho} \right) \\ &=& \lim_{\epsilon \downarrow 0} \left( \frac{2 \pi^2}{\rho} \mathrm{e}^{i \rho \sqrt{\kappa^2 + i \epsilon}} \right) = \frac{2 \pi^2}{\rho} \mathrm{e}^{i \kappa \rho} \end{eqnarray} $$

The integral with respect to $r$ was carried out my method of residues.

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thank you, but integral on the left I can compute too) –  Nimza Nov 9 '11 at 12:40
    
@Nimza I am not understanding your comment. If this post not-helpful, or have you figured out a different way to compute the integral ? –  Sasha Nov 9 '11 at 13:10
    
Rather it is not helpful because I wanted to show that integrals on the left and on the right are equal. I don't need to count integral on the left –  Nimza Nov 9 '11 at 17:35
    
@Nimza Sorry, I missed your question, then. Notice that with formal substitution $\xi \to \xi + k + n \sqrt{\epsilon} \mathrm{e}^{i \pi/4}$, where $\vert n \vert =1$, the denominator on the l.h.s. becomes $ \xi^2 + 2 \xi \cdot \left( k + n \sqrt{\epsilon} \mathrm{e}^{i \pi/4} \right) + 2 k \cdot n \sqrt{\epsilon} \mathrm{e}^{i \pi/4}$. Now you should figure out why $n = \frac{k}{\vert k \vert}$ is a good choice. –  Sasha Nov 9 '11 at 18:28
    
But we leave $R^3$ making this substitution –  Nimza Nov 9 '11 at 19:52
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{{\mathbb R}^{3}}{\expo{i\vec{\xi}\cdot\vec{x}} \over \xi^{2} - k^{2} - \ic 0^{+}}\,d^{3}\vec{\xi}}$

Note that $\ds{\pars{\verts{k} + \ic 0^{+}}^{2} = k^{2} + \ic 0^{+}}$ such that \begin{align} &\color{#c00000}{% \int_{{\mathbb R}^{3}}{\expo{i\vec{\xi}\cdot\vec{x}} \over \xi^{2} - k^{2} - \ic 0^{+}}\,d^{3}\vec{\xi}} =\int_{0}^{\infty}\dd\xi\, {4\pi\xi^{2} \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}}\ \overbrace{\int\expo{i\vec{\xi}\cdot\vec{x}}\,{\dd\Omega_{\vec{\xi}} \over 4\pi}} ^{\ds{\sin\pars{\xi x} \over \xi x}} \\[3mm]&=2\pi\int_{-\infty}^{\infty} {\bracks{\xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} + \pars{\verts{k} + \ic 0^{+}}^{2}\over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi \\[3mm]&={2\pi^{2} \over x} + 2\pi k^{2}\color{#00f}{\int_{-\infty}^{\infty} {1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi}\tag{1} \end{align}

\begin{align}&\color{#00f}{% \int_{-\infty}^{\infty}{1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\sin\pars{x\xi} \over x\xi}\,\dd\xi} =\Im\,\pp\int_{-\infty}^{\infty}{1 \over \xi^{2} - \pars{\verts{k} + \ic 0^{+}}^{2}} \,{\expo{\ic x\xi} \over x\xi}\,\dd\xi \\[3mm]&=\Im\bracks{% 2\pi\ic\,{1 \over 2\pars{\verts{k} + \ic 0^{+}}}\, {\exp\pars{\ic x\pars{\verts{k} + \ic 0^{+}}} \over x\pars{\verts{k} + \ic 0^{+}}} -\int_{\pi}^{0}{1 \over -\pars{\verts{k} + \ic 0^{+}}^{2}}\,{\ic \over x}\,\dd\theta} \\[3mm]&={\pi \over x}\, \Re\pars{\expo{\ic\verts{k}x} \over \bracks{\verts{k} + \ic 0^{+}}^{2}} -{\pi \over x}\,\Re\pars{1 \over \bracks{\verts{k} + \ic 0^{+}}^{2}} =\color{#00f}{{\pi \over xk^{2}}\bracks{\cos\pars{\verts{k}x} - 1}} \end{align}

Now, we'll replace this result in expression $\pars{1}$: $$ \color{#00f}{\large% \int_{{\mathbb R}^{3}}{\expo{i\vec{\xi}\cdot\vec{x}} \over \xi^{2} - k^{2} - \ic 0^{+}}\,d^{3}\vec{\xi} =2\pi^{2}\,{\cos\pars{\verts{k}x} \over x}}\,,\qquad\quad x = \verts{\vec{x}} $$

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