What is the distribution of $(1+X^2)e^{-X^2/2}$ when $X$ is Cauchy? 
If $X$ is a Cauchy random variable with $f(x)=\frac{1}{\pi}\frac{1}{1+x^{2}}$, what is the distribution of $Y= (1+X^2)e^{-X^2/2}$ ?

What I tried:
I was thinking I may be able to use Jacobian, but I am unable to invert the function $Y=f(X)$ to find solutions. Any Idea how to go about this problem? 
 A: Simplifying the problem: This is quite a complicated problem, which necessitates a detailed examination of the transformation used to get the latter random variable.  To simplify the problem slightly, we first note that $F \equiv X^2 \sim \text{F-Dist}(1,1)$ and the random variable $Y$ depends on $X$ only through $F$.  Thus, your problem reduces to:
$$Y = T(F) \quad \quad \quad \quad \quad F \sim \text{F-Dist}(1,1).$$
where the transformation $T: [0, \infty] \rightarrow \mathbb{R}$ has the following form:
$$T(F) = (1+F) \cdot \exp \Big( -\frac{F}{2} \Big).$$
With a bit of calculus it can be shown that $T$ is a strictly quasi-concave function with maximising value $\hat{F}=1$, and its derivatives can be written as:
$$\begin{equation} \begin{aligned}
T'(F) &= - \frac{1}{2} (F-\hat{F}) \cdot \exp \Big( -\frac{F}{2} \Big), \\[10pt]
T''(F) &= - \frac{1}{4} (3-F) \cdot \exp \Big( -\frac{F}{2} \Big). \\[6pt]
\end{aligned} \end{equation}$$
The important output values for the function are:
$$T(0) = 1 \quad \quad \quad \quad \quad 
T(1) = 2 \cdot e^{-1/2} \quad \quad \quad \quad \quad 
T(\infty) = 0.$$
Thus, we seen that the function $T(F)$ starts at the value $T(0)=1$ and then increases up to the point where $F=1$, where it takes the value $T(1) = 2 \cdot e^{-1/2}$, and then it subsequently decreases down to the limiting value $T(\infty) = 0$.  (Have a look at a plot of this function and you will see this form clearly.)

Solving the problem: In order to solve your problem, consider the range of possible values $0 \leqslant y \leqslant 2 \cdot e^{-1/2}$ and define the functions:
$$L(y) \equiv \begin{cases}
0 & & \text{for } 0 \leqslant y \leqslant 1, \\[6pt]
\min \{ 0 \leqslant r \leqslant 1 | T(r) \geqslant y \} & & \text{for } 1 < y \leqslant 2 \cdot e^{-1/2}, \\[6pt]
\end{cases}$$
and:
$$U(y) \equiv \begin{cases}
0 & & \text{for } y = 0, \\[6pt]
\max \{ r \geqslant 1 | T(r) \geqslant y \} & & \text{for } 0 < y \leqslant 2 \cdot e^{-1/2}. \\[6pt]
\end{cases} \quad \quad$$
These functions give lower and upper bounds which satisfy:
$$T(F) \geqslant y \quad \quad \quad \iff \quad \quad \quad L(y) \leqslant F \leqslant U(y).$$
So the cumulative distribution function for $F$ is:
$$\begin{equation} \begin{aligned}
F_Y(y) 
&\equiv \mathbb{P}(Y \leqslant y) \\[12pt]
&= 1-\mathbb{P}(Y \geqslant y) \\[12pt]
&= 1-\mathbb{P} ( L(y) \leqslant F \leqslant U(y) ) \\[12pt]
&= 1- \int \limits_{L(y)}^{U(y)} \text{F}(r|1,1) \ dr \\[6pt]
&= 1- \int \limits_{L(y)}^{U(y)} \frac{1}{\pi} \cdot \frac{1}{\sqrt{r}} \cdot \frac{1}{1+r} \ dr \\[6pt]
\end{aligned} \end{equation}$$
Differentiating and applying Leibniz integral rules gives the corresponding density of $Y$, which is:
$$\begin{equation} \begin{aligned}
f_Y(y) 
&\equiv \frac{dF_Y}{dy}(y) \\[12pt]
&= - \frac{d}{dy} \int \limits_{L(y)}^{U(y)} \frac{1}{\pi} \cdot \frac{1}{\sqrt{r}} \cdot \frac{1}{1+r} \ dr \\[6pt]
&= \frac{1}{\pi} \Bigg[ \frac{1}{\sqrt{L(y)}} \cdot \frac{1}{1+L(y)} \cdot L'(y) - \frac{1}{\sqrt{U(y)}} \cdot \frac{1}{1+U(y)} \cdot U'(y) \Bigg]. \\[6pt]
\end{aligned} \end{equation}$$
For $0 \leqslant y \leqslant 1$ we have $L'(y) = 0$ and $U'(y) = 1/T'(U(y))$ so that:
$$\begin{equation} \begin{aligned}
f_Y(y) 
&= - \frac{1}{\pi} \frac{1}{\sqrt{U(y)}} \cdot \frac{1}{1+U(y)} \cdot \frac{1}{T'(U(y))} \\[6pt]
&= \frac{1}{\pi} \frac{1}{\sqrt{U(y)}} \cdot \frac{1}{1+U(y)} \cdot \frac{2}{(U(y)-1) \cdot \exp(-U(y)/2)}. \\[6pt]
\end{aligned} \end{equation}$$
For $1 < y \leqslant 2 \cdot e^{-1/2}$ we have $L'(y) = 1/T'(L(y))$ and $U'(y) = 1/T'(U(y))$ so that:
$$\begin{equation} \begin{aligned}
f_Y(y) 
&= \frac{1}{\pi} \Bigg[ \frac{1}{\sqrt{L(y)}} \cdot \frac{1}{1+L(y)} \cdot \frac{1}{T'(L(y))} - \frac{1}{\sqrt{U(y)}} \cdot \frac{1}{1+U(y)} \cdot \frac{1}{T'(U(y))}  \Bigg] \\[6pt]
&= \frac{1}{\pi} \Bigg[ \frac{1}{\sqrt{L(y)}} \cdot \frac{1}{1+L(y)} \cdot \frac{2}{(1-L(y)) \cdot \exp(-L(y)/2)} - \frac{1}{\sqrt{U(y)}} \cdot \frac{1}{1+U(y)} \cdot \frac{2}{(U(y)-1) \cdot \exp(-U(y)/2)} \Bigg]. \\[6pt]
\end{aligned} \end{equation}$$
This gives you the form for the density in terms of the underlying implicit functions $L$ and $U$.
