= Solution
{c}
For general $\alpha>0$, put $S(\lambda)=\lambda^{-4}+2\lambda^2$. The reduced <strain energy density> and its derivative are
$$
h(\lambda)=\frac\mu2\left[S(\lambda)^\alpha-3^\alpha\right],\qquad
h'(\lambda)=2\mu\alpha S(\lambda)^{\alpha-1}(\lambda-\lambda^{-5}).
$$
The same factorization as in part (i) yields
$$
\frac{h'(\lambda)}{\lambda^3-1}
=2\mu\alpha S(\lambda)^{\alpha-1}
\left(\lambda^{-2}+\lambda^{-5}\right).
$$
This form removes the apparent singularity at $\lambda=1$: its limit is $4\mu\alpha3^{\alpha-1}$, finite for every fixed $\alpha>0$. Thus only the upper endpoint can obstruct convergence of the <improper integral>.
As $\lambda\to\infty$, $S(\lambda)\sim2\lambda^2$, giving
$$
\frac{h'(\lambda)}{\lambda^3-1}
\sim\mu\alpha2^\alpha\lambda^{2\alpha-4}.
$$
For $\mu>0$, the <comparison test> with a power <integral> implies convergence exactly when $2\alpha-4<-1$. Therefore
$$
\boxed{0<\alpha<\frac32}
$$
is 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.
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