I'd like to know how to prove or disprove that
$\int\limits_0^\infty {\frac{{\sin \left( {2 \omega x} \right)}}{{\sin x}}\frac{{dx}}{{1 + {x^2}}} = \frac{\pi }{{{e^2} - 1}}\frac{{{e^{2 \omega}} - 1}}{{{e^{2\omega - 1}}}}} $
I always try to solve this problems with differential equations but this one yields
\int\limits_0^\infty {\frac{{\sin \left( {2wt} \right)}}{{\sin t}}dt} = I\left( w \right) - \frac{{I''\left( w \right)}}{4}
...and the integral of the LHS is not defined.