## Re: Improper integral

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:

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