I believe I have now also succeeded at evaluating this integral by use of the residue theorem: ## \\ ## The integral can be rewritten as ## I=-\frac{1}{2} \int Re[ \frac{e^{i2z}[1-e^{-i2(z-x_o)}]}{(z-x_o)(z+x_o)}] \, dz ##. ## \\ ## (Using ## sin(x-x_o)sin(x+x_o)=-\frac{1}{2}[cos(2x)-cos(2x_o)] ## and then writing ## cos(2x) ## as the real part of ## e^{i2x} ##). ## \\ ## A Taylor expansion shows the only pole is at ## z=-x_o ## (from the ## (z+x_o) ## term in the denominator. ## \\ ## (Note that ## e^{-i2(z-x_o)}=1-2i(z-x_o)+higher \, order \, terms \, in \, (z-x_o) ##. Thereby, there is no pole at ## z=x_o ##). ## \\ ## The contour will go along the x-axis with an infinitesimal semi-circle loop over the pole at ## z=-x_o ##, and will be closed in the upper half complex plane. (I needed to consult my complex variables book for this next part:). The infinitesimal semi-circle loop over the pole in the clockwise direction results in ## -B_o \pi i ## where ## B_o ## is the residue of the function at ## z=-x_o ##. ## \\ ## Upon evaluating the residue ## B_o ##, this results in ## -\pi \frac{sin(2x_o)}{2x_o} ## for the portion containing the infinitesimal semi-circle loop over the pole (which is not part of the integration of the function along the x-axis, where the entire function is considered to be well-behaved.). Since the contour doesn't enclose any poles, the complete integral around it must be zero, so the functional part along the x-axis must be equal to ## \pi \frac{sin(2x_o)}{2x_o} ##, which is the result that we anticipated.
