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For the two semi-infinite cracks, choose insteadwith cuts on the two crack segments. Approaching from above givesThese opposite signs on the two cracks are essential. Multiplying the Hilbert problem for an antiplane crack by $\chi$ givesThe Sokhotski–Plemelj theorem therefore yields the antiplane ligament Cauchy solutionFor $-a<x=x_1<a$, $\chi (x)=\sqrt{a^2-x^2}$ and the kernel has no pole on either crack. HenceAgain this is the additional stress, and ${\sigma }_{13}=0$ on the intact ligament. The integrals have the Cauchy principal value interpretation wherever required, and their convergence at infinity is assumed. For a point on the right crack, the upper value of $H$ has imaginary part $+\sqrt{t^2-a^2}p(t)$; division by $-i\sqrt{t^2-a^2}$ gives real part $-p(t)$. On the left crack both signs reverse and the same cancellation follows.

The homogeneous ambiguity here is physically different from the finite-crack displacement period. A real termhas zero real traction on the crack faces but adds ligament shear with resultantIts potential is $(C/\mu )\arcsin (z/a)$, producing logarithmic displacement growth at infinity. Thus it represents a separately imposed load transfer between the two sides, even though its stress tends to zero pointwise at infinity. The induced correction with no added remote resultant selects $C=0$, giving the boxed formula. Merely requiring pointwise decay of stress would not fix this constant; the far-field load normalization is needed.