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.
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