I stumbled into a paper about the Gieseking constant while reading about the Clausen function of order 2 and thus have fallen down a rabbit hole, digging in to hyperbolic geometry for the first time. As is mentioned in the paper, the area of an ideal triangle is always $\pi$. Below is my proof of this. Choosing two vertices $a$ and $b$ on the x-axis with b>a and then allowing the third vertex to be at infinity, we can then define the following variables
$$m=\frac{a+b}{2},\ d=\frac{b-a}{2}$$
and note that the geodesic which connects $a$ and $b$ is given by $f(x)=\sqrt{d^2-(x-m)^2}$. We may now solve the integrals below to find the hyperbolic area of the triangle.
$$\int_a^b\int_{f(x)}^{\infty}\frac{1}{y^2}\ dy\ dx=\int_a^b\frac{1}{\sqrt{d^2-(x-m)^2}}\ dx=\arcsin(\frac{b-m}{d})-\arcsin(\frac{a-m}{d})=\arcsin(1)-\arcsin(-1)=\pi$$
Furthermore, in the paper linked, they give a proof that for any triangle with one vertex at infinity and the other two on the unit circle, the hyperbolic area is $\arcsin(b)-\arcsin(a)$, thus showing that $\pi$ is the maximal area of such a triangle.
However, this is not the only way to construct an ideal triangle in the half-plane model. One could also create a triangle with zero defect by having all three vertices be on the x-axis.
Let us say two of our vertices are at the points $(-1,0)$ and $(1,0)$ with the third point at $(\gamma,0)$ with $-1<\gamma<1$. Then, like before, define the following variables
$$m=\frac{\gamma+1}{2},\ d=\frac{\gamma-1}{2}$$
and name similar functions $f_0(x)={\sqrt{1-x^2}},\ f_1(x)={\sqrt{m^2-(x-d)^2}}$ and $f_2(x)={\sqrt{d^2-(x-m)^2}}$. Then we may evaluate the following equation as we see fit:
$$V_T\ =\ \int_{-1}^1\int_0^{f_0(x)}\frac{1}{y^2}\ dy\ dx\ - \int_{-1}^{\gamma}\int_0^{f_1(x)}\frac{1}{y^2}\ dy\ dx\ -\ \int_{\gamma}^{1}\int_0^{f_2(x)}\frac{1}{y^2}\ dy\ dx$$
Normally, this would be highly problematic, as each summand requires you to take the reciprocal of zero at some point - none of the summands themself seem to converge. However, by vitue of the fact that this is an ideal triangle - i.e. a set of three vertices connected by geodesics with the angle measure of each vertex being zero - we know that it must ultimately equal $\pi$. Crisis averted.
Now consider a horocycle, a circle in the hyperbolic plane which, on the half-plane, is tangential to the x-axis. While dealing with it in rectangular co-ordinates would be troublesome, we can save ourselves a headache by converting to polar co-ordinates. Let our horocycle $C$ be the circle of radius $\frac{1}{2}$ centered at the point $(0,\frac{1}{2})$. Then with a fairly simple switch, we can evaluate its hyperbolic area thusly:
$$\int\int_C\ \frac{1}{y^2}\ dA\ =\ \int_0^{\pi}\int_0^{\sin(\theta)}\frac{1}{r\sin^2(\theta)}\ dr\ d\theta$$
And herein lie the trouble. In trying to solve the inner integral, we must evaluate $\ln(0)$. The integral diverges.
We may try to do a sort of inverse of the trick we did to divine the area of the ideal triangle with no vertex at infinity. If instead we make a horocycle that's tangential to infinity, we are left with a line of the form $y=a$ with $a>0$. If we try to approach the problem from this angle, we get
$$\int_{-\infty}^{\infty}\int_a^{\infty}\frac{1}{y^2}\ dy\ dx\ = \int_{-\infty}^{\infty}\ln(a)\ dx$$
Which is also divergent.
What's the hitch here? The ideal triangles are tangential to the x-axis at two or three locations, while the horocycle is tangential at only 1 by definition. Why should its are diverge while the ideal triangles converge? Does it have something to do with the fact that the points of tangency of the ideal triangles are formed by convex "seagull"-like points, while the horocycle's point of tangency is concave? Or is there an obvious trick I'm missing? Wikipedia gives a very vague "the area of a sector of a horocycle is finite", which feels very unsatisfying. Any insight would be much appreciated. Thank you!
The "area of a sector of a horocycle" refers to an area term like this: $$\int_{x=0}^{x=1} \int_{y=1}^{y=\infty} \frac{1}{y^2} dA $$ which is easily seen to be finite. But the horodisc is filled up by infinitely many non-overlapping sectors, with areas of the form $$\int_{x=n}^{x=n+1} \int_{y=1}^{y=\infty} \frac{1}{y^2} dA $$ and these areas are all clearly equal to each other. Thus, the "inside" of the horocycle, i.e. the horodisc $1 \le y < \infty$, must have infinite area, because it contains infinitely many nonoverlapping regions of equal positive area.
One point here is that the symmetry of the horodisc shares an important feature with the symmetry of a Euclidean half plane: the symmetry group can be identified with the additive group of real numbers. Taking the integer values of this group, we can use symmetry to find a bi-infinite sequence of disjoint, equal, positive areas inside the horodisc, just as we can use symmetry to find a bi-infinite sequence of disjoint, equal, positive areas inside a Euclidean half plane.