We know,

\int^{\infty}_{0}\frac{\sin{yx}}{x}dx=\frac{\pi}{2  } (y>0)

That is uniformly converges relatively to y for y\geqslant y_0>0, so:


\int^{\infty}_{0}\frac{dx}{x}\int_a^b\sin{yx}\;{dy  }=\int_0^{\infty} \frac{\cos{ax}-\cos{bx}}{x^2}dx=\frac{\pi}{2}(b-a)