Inverse Fourier Transform of the Hdamard's finite part - 2 dimensions I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.
In two dimensions, we defined the finite part $\text{PF}\left(1/k^2\right)$, where $k\equiv|\vec{k}|=\sqrt{k_x^2+k_y^2}$, as the distribution such that for each test function $f$
\begin{equation}
\left\langle \text{PF}\left(\frac{1}{k^2}\right),f\right\rangle =\text{PF}\iint\!d^2\vec{k}\frac{f(\vec{k})}{k^2} =\iint_{k<1}\!\!\!d^2\vec{k}\frac{f(\vec{k})-f(0)}{k^2}+\iint_{k>1}\!\!\!d^2\vec{k}\frac{f(\vec{k})}{k^2}
\end{equation}
Quoting my professor:

Sloppily then we can compute the inverse Fourier transform as
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{(2\pi)^2}\text{PF}\iint\!d^2\vec{k}\frac{e^{i\vec{k}\cdot\vec{x}}}{k^2} = \frac{1}{(2\pi)^2}\text{PF}\int_0^\infty\!dk \frac{k}{k^2}\int_0^{2\pi}d\vartheta\,e^{ikr\cos{\vartheta}}
\end{equation}
  Where we used spherical coordinates in two dimensions, and we called $r\equiv |\vec{x}|=\sqrt{x^2+y^2}$. Using Bessel $J_n$ functions, one eventually get
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{(2\pi)^2}\text{PF}\int_0^\infty\!dk\, (2\pi)\frac{J_0(kr)}{k} \tag{1}
\end{equation}
  Using now the definition of finite part, and changing variable $u\equiv rk$, it can be showed that
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{2\pi}\left\{-\log(r)+\int_0^1\!du\frac{J_0(u)-J_0(0)}{u}+\int_1^\infty\!du\frac{J_0(u)}{u}\right\} \tag{2}
\end{equation}

First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.
Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
\begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{2\pi}\int_0^{1/r}\!du\,\frac{J_0(u)-J_0(0)}{u} +\frac{1}{2\pi}\int_{1/r}^\infty\!du\,\frac{J_0(u)}{u}
\end{equation}
And i really can't understand how to go on, and get the $\log(r)$ term.
 A: I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^{i k \cdot w}$ by definition as if $e^{i k \cdot w}$ were a test function. In the resulting integral, make the change of variables $k_x = \rho \cos \phi, k_y = \rho \sin \phi$:
$$\left< \frac 1 {|k|^2}, e^{i k \cdot w} \right> =
\iint_{|k| < 1} \frac {e^{i k \cdot w} - 1} {|k|^2} dk_x dk_y +
 \iint_{|k| > 1} \frac {e^{i k \cdot w}} {|k|^2} dk_x dk_y = \\
\int_0^1 \int_0^{2\pi}
  \frac {\exp(i \rho (w_x \cos \phi + w_y \sin \phi)) - 1}
  \rho d\phi d\rho + \\
 \int_1^\infty \int_0^{2\pi}
  \frac {\exp(i \rho (w_x \cos \phi + w_y \sin \phi))} \rho d\phi d\rho.$$
We know that
$$w_x \cos \phi + w_y \sin \phi =
r \left( \frac {w_x} r \cos \phi + \frac {w_y} r \sin \phi \right) =
r \cos(\phi + \phi_0)$$
for some $\phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $\phi_0$, and
$$\int_0^{2\pi} e^{i \rho r \cos \phi} d\phi =
2 \pi J_0(\rho r) \Rightarrow \\
\frac 1 {2\pi} \left< \frac 1 {|k|^2}, e^{i k \cdot w} \right> =
\int_0^1 \frac {J_0(\rho r) - 1} \rho d\rho +
 \int_1^\infty \frac {J_0(\rho r)} \rho d\rho = \\
\int_0^r \frac {J_0(u) - 1} u du +
 \int_r^\infty \frac {J_0(u)} u du = \\
\left( \int_0^1 \frac {J_0(u) - 1} u du +
 \int_1^r \frac {J_0(u) - 1} u du \right) + \\
 \left( \int_1^\infty \frac {J_0(u)} u du +
 \int_r^1 \frac {J_0(u)} u du \right) = \\
-\ln r + \int_0^1 \frac {J_0(u) - 1} u du +
 \int_1^\infty \frac {J_0(u)} u du = \\
-\ln r - \gamma + \ln 2.$$
The last step is for extra points.
