# help with fourier transform integral

I need to know how to calculate

$$\int^\infty_{-\infty} \, d^2q \frac{e^{i\mathbf{q}\cdot\mathbf{x}}}{a +bq^2}$$

basically a 2-dimensional fourier transform of that function.

I think the answer is some sort of exponentially decaying function but I don’t know how to proceed. Possibly using polar coordinates or complex integration or something is in order Please help

#### Solutions Collecting From Web of "help with fourier transform integral"

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

Whenever $\displaystyle{\quad{a \over b} > 0}$:

\begin{align}
&\color{#f00}{\int_{{\bf R}^{2}}{\rm d}^{2\,}\vec{q}\,
{{\rm e}^{{\rm i}\,\vec{q}\cdot\vec{r}} \over a + bq^{2}}}
=
\int_{0}^{\infty}{\rm d}q\,{q \over a + bq^{2}}
\int_{0}^{2\pi}{\rm d}\theta\,{\rm e}^{{\rm i}\,qr\cos\left(\theta\right)}
\\[5mm]&=
2\pi\int_{0}^{\infty}{\rm d}q\,{q \over a + bq^{2}}
\overbrace{\quad{1\over \pi}\int_{0}^{\pi}{\rm d}\theta\,{\rm e}^{{\rm i}\,qr\cos\left(\theta\right)}\quad}^{\displaystyle{{\rm J}_{0}\left(qr\right)}}
=
2\pi\int_{0}^{\infty}{\rm d}q\,{q\,{\rm J}_{0}\left(qr\right) \over a + bq^{2}}
\\[5mm]&=
{2\pi \over b}\int_{0}^{\infty}{\rm d}q\,{q\,{\rm J}_{0}\left(qr\right)
\over
q^{2} + \left(\sqrt{a/b}\right)^{2}}
=\color{#f00}{%
{2\pi \over b}\ {\rm K}_{0}\left(\sqrt{a \over b\,}\ r\right)}
\end{align}

where $\ds{\rm J_{0}}$ and $\ds{\rm K_{0}}$ are
Bessel Functions.

Switch to plane polar coordinates to get
$$\frac{\pi}{b}K_0(r/\sqrt{b})$$