Past exam of the elasticity and plasticity course of the University of Oxford / 2009 / 2 / b / iii

Multiplication of the preceding displacement formula by $-ik$ givesFor the supplied kernels, $\widehat{f_{1,x}}=-ik{e}^{-qy}$ and ${\hat{f}}_2=i(k/q)e^{-qy}$. The convolution theorem therefore givesSince $f_{1,x}(s,y)=-2ys/[\pi (s^2+y^2{)}^2]$ and $f_2(s,y)=s/[\pi (s^2+y^2)]$, combine the two fractions and divide by two: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.