I found the following question kind of counterintuitive (it is a question from a probability theory book). I feel it is hard to claim $\mathbb{E}[\frac{1}{X}]$ has an infinite expectation without knowing the distribution of $X$ (the only thing we know is that $X$ has a continuous density function):
Let $\xi$ be a random variable with a continuous density $p_{\xi}$ such that $p_{\xi}(0) \geq 0$. Find the density of $\eta$, where \begin{equation*} \eta(\omega) = \begin{cases} 1/\xi(\omega) \ \text{ if } \xi(\omega) \neq 0 \\ 0 \ \ \ \ \ \ \ \ \ \text{ if } \xi(\omega) = 0 \\ \end{cases} \label{eq: state} \end{equation*}
Prove that $\eta$ does not have a finite expectation.
Also, any suggestions on how to find the distribution of $\eta$?
Use the Jacobian method. Let $f_X(x)$ be the density of $X=\epsilon$ and $f_Y (y)$ be the density of $Y = \eta$ (I find writing greeks tiresome...)
Then $Y = X^{-1}$ which can be inverted to obtain $X = Y^{-1}$
This transforation is one-to-one and continuous almost everywhere (It's only discontinuous/undefined at $X=0$ which has probability $0$). Hence
$J = \frac{d}{dy} y^{-1} = -y^{-2}$
And so the pdf of $Y$ is given by
$$f_Y (y) = |-y^{-2}| f_X (y^{-1}) = y^{-2}f_X (y^{-1})$$
To show $Y$ has non-finite expectation, since $P(X \neq Y)=0$ I think you should write it as
$$E[Y] = \int x^{-1} f_X (x)dx$$
and make an argument that $E[Y]$ must be infinite since $x^{-1}f_X(x)$ diverges at $0$ (but is defined since $Y = 0$ at $X=0$ by construction). Since we are told $f_X (x) > 0$ then you can show the limit doesn't exist fairly easily, since $f(x)$ converges to a finite positive constant, while $x^{-1}$ diverges and is undefined at $0$.