integration of $\int \frac{e^{-i\mathbf q\cdot r}}{r} r\,dr\sin \theta d\theta$

Question

I would like to prove that:

\begin{equation} V_q = \int \frac{e^{-i\mathbf q\cdot\mathbf r}}{r} d^2r =\frac{2\pi}{q} \end{equation} throughout the 2D space where $\mathbf q$ is a vector that can be arbitrarily oriented.

In polar coordinates, we obtain : \begin{equation} V_q = \int_0^{\infty} dr\int_0^{2\pi}e^{-i\mathbf q\cdot\mathbf r}d\theta \end{equation}

choosing q aligned with the $x$ axis such that $\mathbf q\cdot\mathbf r=qr\cos\theta$. I unsuccessfully tried the change of variables $x=\cos \theta$.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} V_{\vec{q}} & \equiv \iint_{\large\mathbb{R}^{2}} {\expo{-\ic\vec{q}\cdot\vec{r}} \over r}\,\dd^{2}\vec{r} = \int_{0}^{\infty}\int_{0}^{2\pi}\expo{-\ic qr\cos\pars{\theta}}\dd\theta\,\dd r = 2\pi\int_{0}^{\infty}\mrm{J}_{0}\pars{qr}\dd r = \bbx{2\pi \over q} \end{align}

where $\ds{\quad q \equiv \verts{\vec{q}}\quad\mbox{and}\quad\mrm{J}_{\nu}}$ is a Bessel Function.

Answer 2

We are integrating over all of $p=(x,y) \in \mathbb{R}^2$, we can choose the axis along $q$ such that $$(x,y) \cdot q = r \,|q|\cos\phi$$ with $(r,\phi)$ the polar coordinates of the point $p$.

So, we immediately obtain the result that the integral does only depend on $|q|$ and not on the vector $q$ itself.

Finally, we obtain $$V_q =\int_{[0,\infty]\times[0,2\pi]} \frac{r}{r} e^{-ir |q| \cos \phi} dr \,d\phi =\int_{0}^{2\pi} \frac{e^{-i r |q| \cos\phi}}{-i |q| \cos\phi}\Bigg|_{r=0}^{\infty} \,d\phi \,.$$

You observe that the integral is not convergent. In fact, in order to obtain a finite result, you have to regularize the integral (or treat it in the proper distributive sense). A typical choice would be to multiply the integrand by a factor $e^{-\eta r}$ and let $\eta \to0$ at the end.

With this regularization, the upper limit vanishes, and we obtain the result $$ V_q = \int_0^{2\pi}\frac{1}{i |q| \cos \phi+\eta}d\phi.$$

The integral lends itself to be treated with complex analysis. We subsitute $z=e^{i\phi}$ and obtain $$V_q = \oint_{|z|=1} \frac{1}{iz} \frac{1}{i |q|(z+z^{-1})/2 +\eta} =-\oint_{|z|=1} \frac{2}{|q|(1+z^2) -2 i z \eta} .$$

The last integral can be solve with the help of the residue theorem. For small $\eta$ (we let $\eta\to0$ in the end), we have poles at $$z_\pm = \pm i + \frac{i}{|q|}.$$ In particular, only $z_-$ is within the unit circle and thus, we obtain $$V_q = - 2\pi i \operatorname{Res}_{z=z_-}\frac{2}{|q|(z-z_-)(z-z_+)} = \frac{4\pi i}{|q|(z_+-z_-)} =\frac{2\pi}{|q|}. $$