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Work first with $|k|<1$, so the real integrands and their parameter derivatives are smooth on every finite amplitude interval. For the elliptic integral of the second kind,On the other hand, subtracting the integrands of the second-kind and first-kind elliptic integrals givesConsequentlyThis is the modulus derivative of the incomplete elliptic integral of the second kind. At $k=0$ its right-hand side has a removable limit zero, agreeing with $E_k(\theta ,0)=0$. Both $E$ and $F$ here use amplitude $\theta$ and modulus $k$; differentiating with respect to the alternative parameter $k^2$ would give a different prefactor.

Use the integration by substitution $x=\sin \theta$, with $-\pi /6\le \theta \le \pi /6$. On this interval $\cos \theta >0$, so it cancels the positive denominator square root. The transformed integral isIts integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the elliptic integral of the second kind isThe modulus is $2/3$, not the squared parameter $4/9$.

Differentiate the worldline with respect to $\theta$:Since $|\tan \theta |\le 1$ and $0<\kappa <1$, this is a future-directed timelike curve. The proper time element satisfies $c^{2}d{\tau }^2=c^{2}d{t}^2-d{x}^2$, henceAlice's total proper time is thereforeThis definite integral is an exact answer; it is an elliptic integral of the second kind. In notation $E(\alpha \mid m)={\int }_0^{\alpha }\sqrt{1-m{\sin }^{2}u}du$, it is $2\sqrt{2}TE(\pi /4\mid 1+{\kappa }^2)$. Although this parameter exceeds one, the integrand is real on the stated interval because its minimum squared value is $(1-{\kappa }^2)/2>0$.

The endpoint events have $(t,x)=(0,0)$ and $(2T,0)$. Sam stays at $x=0$ in his inertial frame, so his elapsed proper time is $2T$. For $\kappa >0$, the integrand is strictly less than $\cos \theta$ except at $\theta =0$, and ${\tau }_A<2T$.