Prove or disprove $\int_0^\infty {\frac{\sin ( 2 \omega x)}{{\sin x}}\frac{dx}{1 + x^2} = \frac\pi {e^2- 1}\frac{e^{2 \omega} - 1}{e^{2\omega - 1}}} $ 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. 
 A: Let $\omega = n \in \mathbb{N}$
$$
  \frac{\sin(2 \omega x)}{\sin(x)} = U_{2n-1}\left(\cos(x)\right)
$$
Then
$$
 \mathcal{I}(\omega) =  \int_0^\infty \frac{\sin(2 \omega x)}{\sin(x)} \cdot \frac{\mathrm{d} x}{1+x^2} = \int_0^\infty U_{2n-1}\left(\cos(x)\right) \frac{\mathrm{d} x}{1+x^2}
$$
By Chebyshev polynomial of the second kind for odd index can be written as a sum of polynomials of the first kind:
$$
   U_{2n-1}(z) = 2 \sum_{k=1}^{n} T_{2k-1}(z)
$$
Thus, using $T_n(\cos x) = \cos(n x)$ and parity
$$
   \mathcal{I}(n) = 2 \sum_{k=1}^n \int_0^\infty T_{2k-1}(\cos x) \frac{\mathrm{d} x}{1+x^2} =  \sum_{k=1}^n \int_{-\infty}^\infty \cos((2k-1) x) \frac{\mathrm{d} x}{1+x^2} =\\ \sum_{k=1}^n \pi \mathrm{e}^{-(2k-1)} = \frac{\pi}{\mathrm{e}^2-1} \frac{\mathrm{e}^{2n}-1}{\mathrm{e}^{2n-1}}
$$

Added One does not really need to use Chebyshev polynomials above. Instead we can use
$$
   \sum_{k=1}^n 2 \sin(x) \cos((2k-1)x) = \sum_{k=1}^n \left( \sin(2 k x) - \sin(2(k-1) x) \right) \stackrel{\text{telescope}}{=} \\ \sin(2nx) - \sin(0 x) = \sin(2n x)
$$
