Let $X$ be a variable that follows a Cauchy distribution. I.e., the density function is given by $$f_X(t) = \frac{1}{\pi(1+t^2)}$$
Then, there is asked to find the probability distribution function $F_Z: \mathbb{R} \to [0,1]$
where $$F_Z(x) = P(Z \leq x)$$ and $Z = X^{-1}$
I did the following: for $x \neq 0$
$$F_Z(x) = P(Z \leq x) = P(1/X \leq x) = P(X \geq 1/x) = 1 - P(X \leq 1/x)$$ $$= 1 - \int_{- \infty}^{1/x} \frac{1}{\pi(1+t^2)} dt = 1/2 - \frac{1}{\pi}\arctan(1/x)$$
and analoguously $F_Z(0) = 1/2$
but here is my problem:
This can't be the correct answer, because this function is not monotonic, the function is not $1$ at $+ \infty$ and the function is not $0$ at $-\infty$.
Where did I go wrong?
Hint: $1/X \leq -1$ is not equivalent to $X \geq -1$. Instead, split up into $x > 0$ and $x < 0$ cases.