Can the classical action for an electron in constant magnetic field be periodically infinity for different values of time? In Feynman's book, 'Quantum Mechanincs and Path integral', the following problem is given (page no. 64):
Now, we know that $Kernel\propto e^{iS_{cl}/h}$ if $S$ is quadratic. Here, $S_{cl}$ refers to the classical action.
So, the problem in the book implies that for an electon moving in a constant magnetic feild, $S_{cl}=\frac{m}{2}\left[\frac{(z_{b}-z_{a})^{2}}{T}+\left(\frac{w/2}{\tan(wT/2)}\right)\left[(x_{b}-x_{a})^{2}+(y_{b}-y_{a})^{2}\right]+w(y_{b}x_{a}-x_{b}y_{a})\right]$
Now, this makes no sense because we can notice that for some values of $T$, $\tan(wT/2)$ will become zero, making the action infinity. Since the classical path of an electron in constant magnetic field is circular or helical, the expression for action that I am getting has no scope of infinities in it.
So, is the expression for $S_{cl}$ as given in the book above correct? If not, is not the expression for kernel as given in the book wrong?
 A: The motion in the $x$, $y$ plane follows a circle. Let this circle have radius $R$. The equation of motion follows from the Lorentz force 
$$ \boldsymbol{F} = \frac{e}{c} \boldsymbol{v} \times \boldsymbol{B}. $$
The force is centripetal, so its magnitude is $mv^2/R$, and the equation of motion is
$$ \frac{m v^2}{R} = \frac{eBv}{c}. $$
Therefore the particle moves in the circle with velocity
$$ v = \frac{e B R}{m c}, $$
and the period for one cycle in the $x$, $y$ plane is
$$ \Delta t = \frac{2\pi R}{v} = \frac{2 \pi m c}{e B} = \frac{2\pi}{\omega}. $$
Integer multiples of this period are precisely the values of $T$ for which $\tan(\omega T/2) = 0$. Now consider the worrisome term in the classical action:
$$ \frac{\omega/2}{\tan(\omega T/2)} \left( (x_b - x_a)^2 + (y_b - y_a)^2 \right). $$
When $T = n\Delta t = 2\pi n/\omega$, this term does not necessarily diverge, because the particle has returned to its starting point where $x_a = x_b$ and $y_a = y_b$, so this term actually approaches the indeterminate form $0/0$.
To simplify this indeterminate form, let us choose coordinates such that $t_a = 0$ and $t_b = T$, and the motion takes the form
$$ x(t) = \cos(\omega t), \quad y(t) = \sin(\omega t). $$
Then 
\begin{align}
(x_b - x_a)^2 + (y_b - y_a)^2 &= (\cos(\omega T)-1)^2 + \sin(\omega T)^2 \\ 
&= 2 - 2 \cos(\omega T).
\end{align}
Therefore the term in the action is
$$ \frac{\omega}{2} \frac{2 - 2 \cos(\omega T)}{\tan(\omega T/2)} = \omega\sin(\omega T), $$
which is finite for all $T$. The problem you identified at $T = 2\pi n/\omega$ is just a removable discontinuity, not an actual divergence.
