The joint probability density function of $(X,Y)$ is given by \begin{align*} f_{X,Y}(x,y) = \begin{cases} c(y^{2} - x^{2})e^{-y} & \text{if}\,\,\,-y\leq x \leq y\,\,\text{and}\,\,0 < y < +\infty\\\\ 0 & \text{otherwise} \end{cases} \end{align*}
(a) Determine the value of $c$
(b) Determine the marginal probability density functions $f_{X}$ and $f_{Y}$
(c) Calculate $\textbf{E}(X)$ and $\textbf{E}(Y)$
MY SOLUTION
(a) According to the definition of joint probability density function, we have \begin{align*} c\int_{0}^{\infty}\int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})\mathrm{d}x\mathrm{d}y & = c\int_{0}^{\infty}\left(2y^{3}e^{-y} - \frac{2y^{3}}{3}e^{-y}\right)\mathrm{d}y\\ & = \frac{c}{3}\int_{0}^{\infty}4y^{3}e^{-y}\mathrm{d}y = 8c = 1 \end{align*}
(b) Once again, it results immediately from the definition that \begin{align*} f_{X}(x) & = \int_{0}^{\infty}f_{X,Y}(x,y)\mathrm{d}y = \frac{1}{8}\int_{0}^{\infty}(y^{2}e^{-y} - x^{2}e^{-y})\mathrm{d}y = \frac{2-x^{2}}{8}\\\\ f_{Y}(y) & = \int_{-y}^{y}f_{X,Y}(x,y)\mathrm{d}x = \frac{1}{8}\int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})\mathrm{d}x = \frac{y^{3}e^{-y}}{6} \end{align*}
Where the support of $Y$ is given by $S_{Y} = \textbf{R}_{\geq0}$. Nonetheless, I am still having troubles in calculating the support of $X$. Can someome help me?
EDIT
According to Did's observation, the actual probability density function is given by \begin{align*} f_{X}(x) = \int_{|x|}^{\infty}(y^{2}e^{-y} - x^{2}e^{-y})\mathrm{d}y = \frac{e^{-|x|}(|x| + 1)}{4} \end{align*}
where the support of $X$ is given by $S_{X} = \textbf{R}$.
(c) Finally, we have \begin{align*} \textbf{E}(X) & = \int_{-\infty}^{+\infty}xf_{X}(x)\mathrm{d}x = \frac{1}{4}\int_{-\infty}^{+\infty}xe^{-|x|}(|x|+1)\mathrm{d}x = 0\\\\ \textbf{E}(Y) & = \int_{0}^{\infty}yf_{Y}(y)\mathrm{d}Y = \frac{1}{6}\int_{0}^{\infty}y^{4}e^{-y}\mathrm{d}y = 4 \end{align*}
If there is any additional mistake, do not hesitate to comment.