The integrand is even and decays at both infinity and zero, since $a,b>0$. On $x>0$ set $u=\sqrt{a}x-\sqrt{b}/x$, a strictly increasing bijection onto the whole real line. ThenUsing evenness for the negative half-axis givesThe second summand is an integrable odd function, so its integral vanishes. The remaining Gaussian integral givesFor the second integral use $x=1/t$ separately on each half-axis. The factor $x^{-2}dx$ becomes $-dt$, and reversing the transformed limits yields $J(a,b)=I(b,a)$. ThusThese identities are instances of the Gaussian integral with an inverse-square term. Splitting at zero before either change of variables avoids treating a discontinuous transformation as a single bijection.

Write $\mathbf{\Delta }=\mathbf{d}-\mathbf{b}$. At an intersection, $\mathbf{\Delta }=\lambda \mathbf{a}-\mu \mathbf{c}$, so its scalar triple product with $\mathbf{a},\mathbf{c}$ vanishes. The expression in the question is exactlyUnder a shift of origin by $\mathbf{h}$, the direction vectors $\mathbf{a},\mathbf{c}$ stay fixed, while $\mathbf{b},\mathbf{d}$ both change by $-\mathbf{h}$. Thus $\mathbf{\Delta }$ and the condition are invariant. For nonparallel lines this is also sufficient for intersection; for parallel lines it only asserts coplanarity and need not imply intersection.

Put $\delta \lambda ={\lambda }_2-{\lambda }_1$ and $\delta \mu ={\mu }_2-{\mu }_1$. Then ${\mathbf{x}}_2-{\mathbf{x}}_1=\delta \lambda \mathbf{a}$ and ${\mathbf{x}}_4-{\mathbf{x}}_1=({\mathbf{x}}_3-{\mathbf{x}}_1)+\delta \mu \mathbf{c}$. Bilinearity of the cross product givesThe tetrahedron volume is one sixth of the absolute value of this triple product. Sliding either vertex pair along its line preserves $|\delta \lambda |$ or $|\delta \mu |$, because its separation is fixed and the direction vectors are fixed. Thus the volume is unchanged, including the zero-volume coplanar case. This is tetrahedron volume under sliding opposite edges.

The dot product is $\mathbf{a}\cdot \mathbf{b}=2-1+2=3$, and both vectors have length $\sqrt{6}$. The angle between vectors therefore satisfies $\cos \theta =3/6=1/2$, giving

Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the circle. The diameter bisects the angle at $A$, so the two right triangles obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp \sin \theta ,-\cos \theta )$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower circular segment of central angle $2\theta$. The triangle $ABC$ has area $\sin \theta (1+\cos \theta )$, while the segment has area $\theta -\sin \theta \cos \theta$, by subtracting the central triangle from its circular sector. Adding giveswith $\theta$ measured in radians. This is the unit-radius case of the area enclosed by two chords and a circular arc.