I read a paper where the 2D spatial Fourier transform of this function is needed
\begin{equation*}
F(r,\theta) = \delta(r-a)\cos(\theta)
\end{equation*}
where $(r,\theta)$ are the polar coordinates and $a$ is a positive real.
The 2D spatial Fourier transform in cartesian coordinates is defined as
\begin{equation*}
\bar{F}(\xi_1,\xi_2) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} F(x_1,x_2)e^{i(\xi_1x_1+\xi_2x_2)} dx_1 dx_2
\end{equation*}
where $(x_1,x_2)$ and $(\xi_1,\xi_2)$ are respectively the cartesian coordinates and the cartesian wavenumbers. Then we switch to polar coordinates
\begin{align*}
\bar{F}(\xi_1,\xi_2) &= \int_{0}^{+\infty}\int_{0}^{2\pi} F(r,\theta)e^{ir(\xi_1\cos(\theta)+\xi_2\sin(\theta))} rdrd\theta \\
\bar{F}(\xi_1,\xi_2) &= \int_{0}^{+\infty}\int_{0}^{2\pi} \delta(r-a)\cos(\theta)e^{ir(\xi_1\cos(\theta)+\xi_2\sin(\theta))} rdrd\theta
\end{align*}
We first need to compute the integral relative to $\theta$.
\begin{equation*}
I = \int_{0}^{2\pi} \cos(\theta)e^{ir(\xi_1\cos(\theta)+\xi_2\sin(\theta))} d\theta
\end{equation*}
We define the polar wavenumbers as
\begin{align*}
\xi_1 &= \xi \cos(\psi) \\
\xi_2 &= \xi \sin(\psi)
\end{align*}
Thus the integral becomes
\begin{equation*}
I = \int_{0}^{2\pi} \cos(\theta)e^{ir\xi \cos(\theta-\psi)} d\theta
\end{equation*}
We make the following change of variable
\begin{equation*}
\alpha = \theta - \psi
\end{equation*}
The integral becomes
\begin{equation*}
I = \int_{0-\psi}^{2\pi-\psi} \cos(\alpha+\psi)e^{ir\xi \cos(\alpha)} d\alpha
\end{equation*}
Since $\cos(\alpha+\psi) = \cos(\alpha)\cos(\psi) - \sin(\alpha)\sin(\psi)$ and since $\sin$ is an odd function
\begin{equation*}
I = \cos(\psi)\int_{0}^{2\pi} \cos(\alpha)e^{ir\xi \cos(\alpha)} d\alpha
\end{equation*}
\begin{equation*}
I = 2\cos(\psi)\int_{0}^{\pi} \cos(\alpha)e^{ir\xi \cos(\alpha)} d\alpha
\end{equation*}
We know that the Bessel function of first kind is defined as
\begin{equation*}
J_1(z) = \frac{1}{i\pi}\int_{0}^{\pi} \cos(\alpha)e^{iz\cos(\alpha)} d\alpha
\end{equation*}
Then the expression of the integral is
\begin{align*}
I &= 2\cos(\psi)i\pi J_1(\xi r) \\
I &= 2\frac{\xi_1}{\xi}i\pi J_1(\xi r)
\end{align*}
The 2D spatial Fourier transform can now be easily calculated
\begin{align*}
\bar{F}(\xi_1,\xi_2) &= 2i\pi\int_{0}^{+\infty} \frac{\xi_1}{\xi} J_1(\xi r) \delta(r-a)rdr \\
\bar{F}(\xi_1,\xi_2) &= 2i\pi\frac{\xi_1}{\xi} aJ_1(\xi a)
\end{align*}
My problem is that this result is different from the paper I read it where the Fourier transform found is \begin{equation*} \bar{F}(\xi_1,\xi_2) = -i\pi\frac{\xi_1}{\xi} aJ_1(\xi a) \end{equation*}
Am I wrong somewhere or did the authors of the paper made a mistake?