At the problem of energy transfer from the two-dimensional crystal molecule to arise such here integrals: $$\int\limits_{-\infty}^{\infty} \dfrac{\sqrt{Q^2+k^2} (E-\sqrt{Q^2+k^2} ) } {(k-a)(k+a)}e^{ikg} dk ,где g,Q \geqslant 0$$
Breaking integral into the sum of two and counting the one in the sense of the principal value of residues $$-\int\limits_{-\infty}^{\infty} \dfrac{ ({Q^2+k^2} ) } {(k-a)(k+a)}e^{ikg} dk=\dfrac{\pi(a^2+Q^2)}{a}\sin(ga)$$ [url]http://prntscr.com/djachg[/url]
But in another integral
$$E\cdot\int\limits_{-\infty}^{\infty} \dfrac{\sqrt{Q^2+k^2}} {(k-a)(k+a)}e^{ikg} dk$$
There are two branch points k = -iQ and k = iQ (if not mistaken). If I'll contour here so Contour of integration
The integral over such a closed circuit is nulyu.I then remains to estimate the integral over Q branch points at $r\to0,$ And the integral of singularities at $r\to0 $ And the integral over the semi-major axis radius R($R\to\infty $).
Have I chosen bypass circuit. And it seemed to me that some of them are zero (if, in general, equal to) and find the answer?