Is the integral $$I = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \frac{e^{i a x}}{\sqrt{x^2+y^2}}$$ convergent for real $a$?
I have an idea to calculate it but I am not sure if it is correct:
First I consider
$$\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \frac{e^{i a x}}{\sqrt{(c_1-x)^2+(c_2-y)^2}}=\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \ f_1(c_1-x,c_2-y)f_2(x,y)$$ with $f_1(x,y)=1/\sqrt{x^2+y^2}$ and $f_2(x,y)=e^{i a x}$. Their Fourier transforms are $\mathcal F(f_1)=F_1(k_1,k_2)=1/\sqrt{k_1^2+k_2^2}$ and $\mathcal F(f_2)=F_2(k_1,k_2)=2\pi\delta(a+k_1)\delta(k_2)$.
From the convolution theorem, I think I can conclude that $$(2\pi)\mathcal F(I)=F_1(k_1,k_2)F_2(k_1,k_2)=\frac{2\pi\delta(a+k_1)\delta(k_2)}{\sqrt{k_1^2+k_2^2}}$$ and therefore after taking the inverse Fourier transform $$I=2\pi\frac{e^{i a c_1}}{|a|}.$$ Setting $c_1\rightarrow0$ would lead me to $$I=\frac{2\pi}{|a|}.$$ Is this reasoning correct? Is there a way to figure out if the integral is convergent from some other argument?