Proof some 2 D Fourier transforms Here are several Fourier transforms I used, I would like to prove those identity.
I took some times to figure out how they are derived, I tried the residue theorem and other methods, but I failed, maybe I was on a wrong way, please help me get the solutions.
the basic one
$$\begin{align}
\iint_{-\infty}^{+\infty}\! \frac{1}{\xi_1^2 + \xi_2^2}e^{\mathrm{i}\vec{\xi} \cdot \vec{x}}\, \mathrm{d}\xi_1\mathrm{d}\xi_2 = - 2\pi \log|\vec{x}|  \tag{1}
\end{align}$$ 
and
$$\begin{align}
\frac{1}{\pi} \iint\! \frac{\xi_1 \xi_2}{(\xi_1^2 + \xi_2^2)^2}e^{\mathrm{i}\vec{xi}\cdot \vec{x}}\,\mathrm{d}\xi_1\mathrm{d}\xi_2 &= - \frac{x_1 x_2}{|\vec{x}|^2}   \tag{2} \\
\frac{1}{\pi} \iint\! \frac{\xi_1^2}{(\xi_1^2 + \xi_2^2)^2} e^{\mathrm{i}\vec{\xi} \cdot \vec{x}}\, \mathrm{d}\xi_1\mathrm{d}\xi_2 = =-\log|\vec{x}| - \frac{x_1^2}{|\vec{x}|^2} \tag{3} \\
\frac{1}{\pi} \iint \frac{e^{- \mathrm{i}\vec{\xi} \cdot \vec{x}}}{|\vec{\xi}|^2 + |\beta|^2}\, \mathrm{d}\xi_1\mathrm{d}\xi_2 = K_0(|\vec{\beta}| |\vec{x}|) \tag{4}\\
\frac{1}{2 \pi} \iint \frac{e^{- \mathrm{i}\vec{\xi} \cdot \vec{x}}}{(|\vec{\xi}|^2 + |\beta|^2)^2}\, \mathrm{d}\xi_1\mathrm{d}\xi_2 = \frac{|\vec{x}|}{|\beta|}K_1(|\vec{\beta}| |\vec{x}|) \tag{5}
\end{align}$$
those results come from the Appendix A (A1- A6) of this PDF
Thanks a lot for any help and suggestions.
 A: Some comments on the first one:
It's as far as i can see not an good idea to switch to polar coordinates. Regularisation of the expression after performing one of the integrations seems horrible
An indirect way is much faster here:
First: It's known that 
$G(\vec{x})=-2\pi\log(|\vec{x}|)\quad $ (1)
is the Greensfunction of the $2$-D Poisson equation.
$$
(\partial^2_x+\partial^2_y)G(\vec{x})=-4\pi^2\delta(x)\delta(y) 
$$(2)
Looking at Poisson eq. in Fourier space yields
$$
-\int d\vec{\xi}(\xi_x^2+\xi_y^2)\tilde{G}(\vec{\xi})e^{-i \vec{x}\vec{\xi}}=-\int d\vec{\xi}e^{-i \vec{x}\vec{\xi}}\\\rightarrow\tilde{G}(\vec{\xi})=\frac{1}{\xi_x^2+\xi_y^2}
$$
it follows that $G(\vec{x})=\int d\vec{\xi}e^{-i \vec{x}\vec{\xi}}\frac{1}{\xi_x^2+\xi_y^2}=-2\pi\log(|\vec{x}|)$
Done!
Edit:
Sketchy Proof  of (1) :
Integrate (2) inside the Ball with Radius $R$, we get
$$
-4 \pi^2=\int_{B_R}\nabla (\nabla G(\vec{x}))\underbrace{=}_{
Gauss Law}\int_{\partial B_R}\nabla (G(\vec{x}))\underbrace{\vec{n}}_{=e_r}=\int_{\partial B_R}\partial_R G(R)=  2\pi R \partial_R G(R)\\\rightarrow G(R)=-(2\pi) \log(R)+\text{const}
$$
Second Edit:
I want to add the the integrals 2 and 3 are easily derived from the first one by noting that we can rewrite the first integral using IBP (wrt to $\xi_1$)as
$$
I_1=-\frac{1}{-i x_1}\int_{R^2}\frac{2\xi_1}{(\xi_1^2+\xi_2^2)^2}e^{-i \vec{x}\vec{\xi}} \quad (3)
$$
differanting wrt to  to $x_2$ yields
$$
(-x_1\partial_{x_2} I_1)/2=\int_{R^2}\frac{\xi_1\xi_2}{(\xi_1^2+\xi_2^2)^2}e^{-i \vec{x}\vec{\xi}}
$$
which yields the desired result...
For the third integral use again (3) but differentiate with respect to $x_1$
The integrals 4 and 5 can be obtained by the same techniques: 
Using the Helmholtz equation $(\partial^2_x+\partial^2_y+\beta^2)G(\vec{x})=-4\pi^2\delta(x)\delta(y)$ to get integral 4. Then differentiate with respect to $\beta$ to obtain integral 5
