 Solution

For general $\alpha >0$, put $S(\lambda )={\lambda }^{-4}+2{\lambda }^2$. The reduced strain energy density and its derivative areThe same factorization as in part (i) yieldsThis form removes the apparent singularity at $\lambda =1$: its limit is $4\mu \alpha 3^{\alpha -1}$, finite for every fixed $\alpha >0$. Thus only the upper endpoint can obstruct convergence of the improper integral.

As $\lambda \to \infty$, $S(\lambda )\sim 2{\lambda }^2$, givingFor $\mu >0$, the comparison test with a power integral implies convergence exactly when $2\alpha -4<-1$. Thereforeis precisely the range with finite critical pressure, the cavitation criterion for power-law strain energy. At $\alpha =3/2$ the divergence is logarithmic; for larger $\alpha$ it is a power divergence. These materials require an unbounded tensile loading magnitude in this ideal model. The qualification $\mu >0$ excludes the degenerate zero-energy material.