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I need to solve $$\int_{\mathbb{R}^3}d^3k\frac{e^{i\vec{k}\cdot\vec{\rho}}}{|\theta|+k^2}$$ I have a feeling that I should use contour integration, but in three variables I do not know how to employ it. Moreover, Mathematica cannot solve it for me.

I should say that I have some notes that say that I should pass from a form like $-i4\pi\int_{-1}^{1}d\cos\theta\int_0^\infty\frac{dk\sin(kr)}{|\theta|+k^2}$ but I do not know why.

I'd appreciate a help in solving that.

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Since you are a new user, you should explains what did you try. In this way, it would be easier for any user to help you. –  Felix Marin Jan 23 at 15:42
    
In the integral, do you mean $\rho^2=k^2$ ? –  V. Rossetto Jan 23 at 15:52
    
V.Rossetto, you are right. Corrected –  Rimon Jan 23 at 16:45

1 Answer 1

up vote 1 down vote accepted

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{\mathbb{R}^{3}} {\expo{\ic\vec{k}\cdot\vec{\rho}} \over \verts{\theta} + k^{2}}\,\dd^{3}\vec{k} &=\int_{0}^{\infty}\dd k\, {4\pi k^{2}\, \over \verts{\theta} + k^{2}}\, \overbrace{\int\expo{\ic\vec{k}\cdot\vec{\rho}}\,{\dd\Omega_{\vec{k}} \over 4\pi}} ^{\ds{=\ {\sin\pars{k\rho} \over k\rho}}} \\[3mm]&= {4\pi \over \rho^{3}}\int_{0}^{\infty}\rho\dd k\, {\pars{k\rho}^{2} \over \verts{\theta} +\pars{k\rho}^{2}/\rho^{2}} \,{\sin\pars{k\rho} \over k\rho} \\[3mm]&= {4\pi \over \rho} \int_{0}^{\infty}{x\sin\pars{x} \over k^{2} + \mu^{2}}\,\dd x\quad \mbox{where}\quad\mu \equiv \rho\verts{\theta}^{1/2}\,,\quad \rho \equiv \verts{\vec{\rho}} \end{align}

\begin{align} \int_{\mathbb{R}^{3}} {\expo{\ic\vec{k}\cdot\vec{\rho}} \over \verts{\theta} + k^{2}}\,\dd^{3}\vec{k} &={2\pi \over \rho}\int_{-\infty}^{\infty}{x\sin\pars{x} \over k^{2} + \mu^{2}}\,\dd x = {2\pi \over \rho}\, \Im\int_{-\infty}^{\infty}{x\expo{\ic x} \over k^{2} + \mu^{2}}\,\dd x \\[3mm]&={2\pi \over \rho}\, \Im\bracks{2\pi\ic\,{\ic\mu\expo{\ic\pars{\ic\mu}} \over \ic\mu + \ic\mu}} ={2\pi^{2} \over \rho}\expo{-\mu} \end{align}

$$\color{#00f}{\large% \int_{\mathbb{R}^{3}} {\expo{\ic\vec{k}\cdot\vec{\rho}} \over \verts{\theta} + k^{2}}\,\dd^{3}\vec{k} = {2\pi^{2} \over \rho}\expo{-\rho\root{\verts{\theta}}}} $$

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Oh, wow, thanks! You spared me much time, I have never seen that substitution of the sine –  Rimon Jan 23 at 16:50
    
@Rimon It's a usual one since $\large{\rm e}^{{\rm i}\theta} = \cos\left(\theta\right) + {\rm i}\sin\left(\theta\right)$. Thanks. –  Felix Marin Jan 23 at 16:55
    
Actually I was referring to the first substitution (in polar coordinates?) –  Rimon Jan 23 at 22:26

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