How to prove that
$$\int^\infty_0 \frac{\mathrm{ci}(px)}{q^2-x^2}\,dx = \frac{\pi}{2q}\mathrm{si}(pq)$$
The integral was taken from Table of integrals , series and products by Daniel Zwillinger.
It says that the equality holds for $p>0,q>0$. But, I am not sure if that is correct. If I set $p=q=1$ wolframalpha says it is divergent.
The functions are defined as
$$\mathrm{ci}(x) = -\int^\infty_x \frac{\cos(t)}{t}\,dt$$
$$\mathrm{si}(x) = -\int^\infty_x \frac{\sin(t)}{t}\,dt$$
I can prove that
$$\int^\infty_0 \frac{\mathrm{ci}(px)}{q+x}\,dx = -\frac{1}{2}\left\{ \mathrm{si}^2(pq)+\mathrm{ci}^2(pq)\right\}$$
Find the proof here.
I am not sure about
$$\int^\infty_0 \frac{\mathrm{ci}(px)}{q-x}\,dx $$
For computing the principal values of such integrals, you may exploit the identities: $$ \mathcal{L}(\text{ci}(x)) = -\frac{\log(1+s^2)}{s^2}\,\qquad \mathcal{L}(\text{si}(x))=\frac{1}{s}\arctan\left(\frac{1}{s}\right)\tag{1} $$ given by the Laplace transform and differentiation under the integral sign. For instance, since $$ \mathcal{L}^{-1}\left(\frac{1}{q+x}\right)=e^{-qs}\tag{2} $$ we have: $$ J(p,q)=\int_{0}^{+\infty}\frac{\text{ci}(px)}{q+x}\,dx = -\int_{0}^{+\infty}\frac{\log\left(1+\frac{s^2}{p^2}\right)}{2s e^{qs}}\,ds\tag{3}$$ where $\lim_{p\to +\infty}J(p,q)=0$ and $$ \frac{d}{dp}\,J(p,q)=\frac{1}{p}\int_{0}^{+\infty}\frac{s e^{-qs}}{s^2+p^2}\,ds=\frac{1}{p}\left(\frac{\pi}{2}\sin(pq)-\sin(pq)\text{si}(pq)-\cos(pq)\text{ci}(pq)\right).\tag{4} $$ Integrating with respect to $p$ we get: $$\int_{0}^{+\infty}\frac{\text{ci}(px)}{q+x}\,dx = -\frac{\text{si}^2(pq)+\text{ci}^2(pq)}{2}\tag{5}$$ as wanted. Let we apply $(1)$ to the first integral in OP's question, too: $$\text{PV}\int_{0}^{+\infty}\frac{\text{ci}(px)}{q^2-x^2}\,dx = \int_{0}^{+\infty}\frac{\log\left(1+\frac{s^2}{p^2}\right)\sinh(qs)}{2qs}\,ds \tag{6}$$ but the RHS is blatantly divergent. If we consider, instead: $$ J(p,q)=\int_{0}^{+\infty}\frac{\text{ci}(px)}{q^2+x^2}\,dx = -\frac{1}{2}\int_{0}^{+\infty}\log\left(1+\frac{s^2}{p^2}\right)\text{sinc}(qs)\,ds\tag{7}$$ we have $\lim_{p\to +\infty}J(p,q)=0$ as before and $$ \frac{d}{dp}J(p,q)=\frac{1}{p}\int_{0}^{+\infty}\frac{s^2\text{sinc}(qs)}{s^2+p^2}\,ds =\frac{\pi}{2pq}e^{-pq}\tag{8}$$ for any $p,q>0$, giving that the LHS of $(7)$ is an exponential integral, $\frac{\pi}{2q}\,\text{Ei}(-pq)$.