Regarding the integral, $$I=\int_0^\infty \frac{\sin(ax)}{b+x}~dx,~~~~a>0\land b>0$$
Let $t=ax$
$$I=\int_0^\infty \frac{\sin(t)}{ab+t}~dt$$
Let $\theta=ab+t$
$$\begin{align} I&=\int_{ab}^\infty \frac{\sin(\theta-ab)}{\theta}~d\theta\\ \\ &=\cos(ab)\int_{ab}^\infty \frac{\sin(\theta)}{\theta}~d\theta-\sin(ab)\int_{ab}^\infty \frac{\cos(\theta)}{\theta}~d\theta\\ \\ &=\cos(ab)\int_{0}^\infty \frac{\sin(\theta)}{\theta}~d\theta-\cos(ab)\int_{0}^{ab} \frac{\sin(\theta)}{\theta}~d\theta-\sin(ab)\int_{ab}^\infty \frac{\cos(\theta)}{\theta}~d\theta \end{align}$$
Define the sine integral and cosine integral as: $$\text{Si}(z)=\int_0^z \frac{\sin(\theta)}{\theta}d\theta,~~~~~\text{Ci}(z)=-\int_z^\infty \frac{\cos(\theta)}{\theta}d\theta$$
We get:
$$\boxed{\int_0^\infty \frac{\sin(ax)}{b+x}~dx=\frac\pi2\cos(ab)-\cos(ab)\text{Si}(ab)+\sin(ab)\text{Ci}(ab)~}$$
Question:
Is this the simplest form we can get for this integral?