= Solution
Multiplication of the preceding <displacement> formula by $-ik$ gives
$$
\boxed{2\mu\widehat{v_x}=\widehat P\left[-ikye^{-qy}-2i(1-\nu)\frac{k}{q}e^{-qy}\right].}
$$
For the supplied kernels, $\widehat{f_{1,x}}=-ike^{-qy}$ and $\widehat f_2=i(k/q)e^{-qy}$. The <convolution theorem> therefore gives
$$
2\mu v_x=P*\left[yf_{1,x}-2(1-\nu)f_2\right].
$$
Since $f_{1,x}(s,y)=-2ys/[\pi(s^2+y^2)^2]$ and $f_2(s,y)=s/[\pi(s^2+y^2)]$, combine the two fractions and divide by two:
$$
\boxed{\mu v_x(x,y)=-\frac1\pi\int_{-\infty}^\infty\frac{(1-\nu)s^2+(2-\nu)y^2}{(s^2+y^2)^2}\,P(x-s)s\,ds.}
$$
The sign is tied to the positive forward Fourier phase used here. This is a <displacement> derivative, so an additive <displacement> reference drops out. At $y>0$ the kernel is regular near $s=0$; the usual decay/integrability of the applied load controls the tails. A boundary-limit singular integral, if wanted at $y=0$, would require its own principal-value interpretation.
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