 Solution

Use equal small excisions on the two sides of each real pole. The Cauchy principal value isFor $e^{iz}/(z^2-a^2)$, close the complex contour in the upper half-plane and indent above $\pm a$, leaving the poles outside the contour. The residue theorem and the vanishing of the large semicircle integral give the principal-value integral as $i\pi$ times the sum of the two real-pole residues. Each clockwise indentation contributes $-i\pi$ times its residue. Sincethe sum is $i\sin a/a$. Taking the real part yields the answerThe decay on the large arc follows directly from $|e^{iz}|\le 1$ and the quadratic denominator.