I've been working on the classic problem of finding the path through which a body travels in least time between two points on the surface of the Earth, assuming that the body is allowed to fall through a spherically symmetric Earth with constant density and no dissipation present. The idea is to apply the methods of variational calculus to the basic integral
$\tau=\int_{1}^{2}dt$
where points 1 and 2 stand for the endpoints of the trajectory on the planet's surface and the differential time element is evaluated as $dt=\frac{d\ell}{v}$, where $d\ell$ is the line element and $v$ is the magnitude of the velocity. I'm neglecting any kind of rotational aspects outside the plane so the length element is parametrized in simple polar coordinates as $\sqrt{r^{2}+\dot{r}^{2}}d\theta$, where $r$ is the radius measured from the Earth's center and $\theta$ is the angle of rotation in the plane containing the end points of the trajectory. The dot notation signifies the derivative of $r$ with respect to $\theta$. From the conservation of energy $v$ follows simply as $v=\sqrt{\frac{g\left(R^{2}-r^{2}\right)}{R}}$, where $R$ is the radius of the Earth and $g$ is the gravitational acceleration on the surface.
With these considerations our integral has become
$\tau=\int_{1}^{2}\sqrt{\frac{R\left(r^{2}+\dot{r}^{2}\right)}{g\left(R^{2}-r^{2}\right)}}d\theta$
If we calculate something similar to the energy function $h$ for the above integrand denoted by $f$, we find the constant
$h=f-\dot{r}\frac{\partial f}{\partial \dot{r}}=r^{2}\sqrt{\frac{R}{g\left(R^{2}-r^{2}\right)\left(r^{2}+\dot{r}^{2}\right)}}$
We can solve for $\dot{r}^{2}$ to obtain what is apparently the equation of a hypocycloid, according to Wolfram's Mathworld.
$\dot{r}^{2}=\frac{R^{2}r^{2}}{r_{0}^{2}}\frac{r^{2}-r_{0}^{2}}{R^{2}-r^{2}}$
In the above, $r_{0}^{2}=\frac{h^{2}gR^{2}}{h^{2}g+R}$ and also equal, once again according to Mathworld, with $r_{0}=R-2a$, where $a$ is the radius of the smaller circle rotating about the circumference of the Earth. Furthermore, from the parametric representations of $x$ and $y$ on $\theta$ (Wikipedia's hypocycloid page or Mathworld), we have
$r^{2}=\left( R-a\right)^{2}+a^{2}-2a\left(R-a\right)\cos\left(\frac{R}{a}\theta\right)$
In this representation, I'm assuming that $\theta=0$ for the initial point, and consequently the second point is reached at an angle $\theta=\frac{a}{R}2\pi$. If we put all of these things together back in the formula for $\tau$, we obtain
$\tau=\frac{a}{r_{0}}\sqrt{\frac{R^{2}-r_{0}^{2}}{Rg}}\int_{0}^{2\pi}\frac{c+\cos\phi}{1-\cos\phi}d\phi$
where I've made the change of variables $\frac{R}{a}\theta=\phi$ and the constant $c=\frac{\left(R-a\right)^{2}+a^{2}}{2a\left( R-a\right)}$.
The final result should be either $\tau=\pi\sqrt{\frac{R^{2}-r_{0}^{2}}{Rg}}$, or equivalently $\tau=2\pi\sqrt{\frac{a\left( R-a\right)}{Rg}}$.
I can obtain either if I solve the problem in Cartesian coordinates, but I can't seem to get it right in this representation. I've attempted to solve the integral in the complex plane using the method of residues but I'm assuming I'm doing something wrong? Not really sure where else to look. Going that route I obtain the complex integral
$i\int_{C}\frac{z^{2}+2cz+1}{z\left(z-1\right)^{2}}dz$
which has a simple pole at $z=0$ and a second order pole at $z=1$. This would return a value of $2\pi i$ from the pole at $0$ and nothing for the pole at $z=1$, assuming the usual integration in the upper half plane.
Again, I can solve the problem just fine in Cartesian coordinates, but I find this method a bit more elegant and I would like to understand where the mistake or faulty judgement is. Thank you!
P.S.
For further reference, a detailed discussion of this problem exists in the book "Variational Methods In Optimization", by Donalds R. Smith, section 4.2. There, the author derives the above and other results, but ultimately resorts to the same Cartesian representation for the final solution. I don't have a problem with that, but at this point I'm just wondering what's wrong with the complex integral, since it would be such a nice solution in polar coordinates.
