Double Integral Over The Whole Plane

Note : I define here the fourier transform of $f(x)$ as $\mathscr{F}\{f\}(\xi) = \int \ f(x) e^{-i x \xi}$ dx

Rearranging slightly :

$I \equiv \frac{1}{2 \pi \hbar}\iint (q-\frac{1}{2}x) k e^{\frac{i}{\hbar}(p-k)x}dx dk = \frac{1}{2 \pi \hbar} \int (q-\frac{1}{2}x) e^{\frac{i}{\hbar}px} \int \ k e^{-\frac{i}{\hbar}kx} dk dx$

Now, we interpret the second integral as (up to a constant) the Fourier transform of the function $k$ . From standard property of FT, we know that : $\forall u \in C^{\infty}(\mathbb{R}), \mathscr{F}\{xu\}(\xi) = i \partial_{\xi} \mathscr{F}\{u\} (\xi)$ .

Thus, in our case, the second integral yields :

$\int \ k e^{-\frac{i}{\hbar}kx} dk = \hbar^2 \mathscr{F}\{k\}(x) = i \hbar^2 \partial_{x} \mathscr{F}\{1\}(x) = 2\pi i \hbar^2 \partial_x \delta_0(x)$

Ergo, returning back to our integral above, we have :

$I = i \hbar \int (q-\frac{x}{2})e^{\frac{i}{\hbar}px} \partial_x \delta_0(x) dx= -i \hbar [ \partial_x \{(q-\frac{x}{2})e^{\frac{i}{\hbar}px} \} ] \Bigr|_{x=0} = pq + \frac{i \hbar}{2}$

PS : if more details are asked, I can expand the argument here. By writing down the solution, I noticed that this is a quite tricky problem! Well done if you managed to get anywhere with this!