two dimensional change of variable

Question

\begin{equation} \frac{1}{(2\pi)^2}\int{\frac{\sin^2(q.(x_1-x_2))}{q^2}d^2q} \end{equation} The notation $d^n q_k$ to indicate integral over the $n$ coordinates of a point in $\mathbb{R}^n$.

How to perform the integration of the integral above?

Answer 1

In polar coordinates, $q^2\equiv \vec q \cdot \vec q$, $$ \int d^2 q= \int_0^\infty \!\! q ~dq ~\int _0^{2\pi} d\theta $$ and use $(\vec x_1-\vec x_2)$ to fix the arbitrary origin of $\theta$, so $\vec q \cdot (\vec x_1-\vec x_2)=q |(\vec x_1-\vec x_2)| \cos \theta$, hence, as virtually always, \begin{equation} \frac{1}{(2\pi)^2}\int{\frac{\sin^2(\vec q\cdot (\vec x_1-\vec x_2))}{q^2}d^2q} \\ = \frac{1}{(2\pi)^2}\int _0^{2\pi} d\theta ~\int_0^\infty \!\! \frac{dq}{q} ~ \sin^2( q ~|(\vec x_1-\vec x_2)| ~\cos\theta ) . \end{equation}

Now the scale $ |(\vec x_1-\vec x_2)| \cos\theta$ may be absorbed into the definition of q and the integral in $\theta$ be performed, so the above reduces to $$ \frac{1}{2\pi } \int_0^\infty \!\! \frac{dq}{q} ~ \sin^2 q = \infty . $$