Distribution of $Y = X \bmod 2\pi$ with $X$ being a Cauchy distribution

Question

Let $X$ be a Cauchy distribution with parameter $\theta$, that is to say, its density function is: $$ f(x;\theta) = \frac{\theta}{\pi(x^2 + \theta^2)} $$ I'm asked to get the distribution of $Y$ where $Y = X \bmod 2\pi$. I've tackled this problem but I feel like I'm stuck. My attempt so far has been the following: Let's compute the distribution function for $Y$: $$ F(y) = P(-\pi \leq Y \leq y) = P(\bigcup_{n=-\infty}^{\infty} \{ -\pi + 2\pi n \leq X \leq y + 2\pi n \} ) $$ Integrating $X$'s density function I reach the following series: $$ F(y) = \sum_{n = -\infty}^{\infty} \int_{-\pi + 2\pi n}^{y + 2\pi n} \frac{\theta}{\pi (x^2 + \theta^2)} = \sum_{n = -\infty}^{\infty} \frac{1}{\pi}\left[ \arctan{\frac{2\pi n + y}{\theta}} - \arctan{\frac{2\pi(n-1)}{\theta}} \right] $$ I can now differentiate to get the density function of $Y$: $$ f(y) = \sum_{n=-\infty}^{\infty} \frac{1}{\pi}\frac{1}{1+\left(\frac{2\pi n + y}{\theta} \right)^2} $$ However, my professor has told me that he expects a closed form solution. In order to do that he has told me that I should use the Poisson kernel, but I'm not exactly sure how should I proceed with that information. I've tried using the logarithmic expression for the inverse tangent to no avail. I've noticed that the density function of $X$ IS the Poisson kernel for the upper half-plane, but I'm not entirely how that could help. I've tried expressing either of the two series I've provided in a form that allows to apply the Poisson kernel summation formula (the one for the circle) but I haven't had success with that approach. If you could help me in any way I would really appreciate it. Finally, I know it is a trope but I have to say: English is not my mother tongue, so I'm sorry if there are any typos. Thanks for your attention.
Edit: I wish I could accept both answers because they are helpful in different ways, but the system is what it is. Thank you both for your thorought responses!

Answer 1

The probability is simply $$\sum_{n=-\infty}^{\infty}f(x+2\pi n;\theta) = \frac\theta\pi\sum_{n=-\infty}^{\infty}\frac{1}{(x+2\pi n)^2 + \theta^2}$$ We can use the Poisson summation formula, but first we need to compute the Fourier Transform $$\frac1{(2\pi t)^2+\theta^2}\longleftrightarrow\frac{e^{-\theta|f|}}{2\theta}$$ $$\frac1{(2\pi t+x)^2+\theta^2}\longleftrightarrow\frac{e^{-\theta|f|-ifx}}{2\theta}$$ Therefore, $$\sum_{n=-\infty}^{\infty}\frac{1}{(x+2\pi n)^2 + \theta^2}=\sum_{n=-\infty}^{\infty}f(n)=\sum_{n=-\infty}^{\infty}\hat f(n)\\=\frac{1}{2\theta}\sum_{n=-\infty}^{\infty}e^{-\theta|n|-inx}=\frac{\sinh\theta}{2\theta(\cosh\theta-\cos x)}$$ Finally, $$\sum_{n=-\infty}^{\infty}f(x+2\pi n;\theta) =\frac{\sinh\theta}{2\pi(\cosh\theta-\cos x)},0\le x\le2\pi$$

Answer 2

The Poisson kernel (See Rudin , Real and complex analysis, page 223) is for $0<r<1$ $$P_r(x)=\sum_{n\in Z}r^{|n|}e^{inx}=\frac{1-r^2}{1-2r\cos x +r^2}$$ and therefore $$r^{-|n|}=\frac{1}{2\pi}\int_0^{2\pi} e^{-inx}P_r(x)dx.\ (*)$$ The Fourier transform of the Cauchy variable $X$ estimated on the integer $n$ is $$e^{-\theta|n|}=\frac{1}{\pi}\int_{R}\frac{\theta}{\theta^2+x^2}e^{inx}dx=\frac{1}{\pi}\sum_{n\in Z}\int_{0}^{2\pi}\frac{\theta}{\theta^2+(x-2\pi n)^2}e^{in x}\ \ (**)$$ Comparing (*) and (**) the density of $X$ modulo $2\pi$ is $\frac{1}{2\pi}P_{e^{-\theta}}(x)$ since these two functions have the same Fourier coefficients.