# Question on determining the variance of joint probability distribution

$(X,Y)$ has joint distribution $f_{X,Y}(x,y)=\frac{1}{2x}e^{-x}$, for $x>0, -x\le y\le x$, so determine the $Var[X], Var[Y]$ and correlation coefficient

But I don't know how to get one of marginal functions. $h(y)=\int_{0}^{\infty}\frac{1}{2x}e^{-x}dx$, so it's hard to continue to find $\sigma$

-
Notice that the joint density factors $f_{X,Y}(x,y)= f_{Y|X}(y) f_X(x) = (\frac{1}{2 x} \mathbf{1}_{-x\le y \le x}) \cdot \exp(-x) \mathbf{1}_{x \ge 0}$, so that $(X,Y) = (E, E U)$, where $U$ is uniform $\mathcal{U}(-1,1)$, and $E$ is exponential with unit mean. – Sasha Oct 7 '11 at 20:26

Michael Hardy answered the question you seemingly asked: "I don't know how to get one of marginal functions" but as you discovered here, the marginal density of $Y$ seems to be a special function. On the other hand, computing the moments of $Y$ (which is what really you want to do) does not require that you first find the marginal density of $Y$. We have \begin{align*} E[Y^n] &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} y^n f(x,y)\ \mathrm dy\ \mathrm dx\\ &= \int_{x = 0}^{\infty}\int_{y = -x}^{y = x} y^n \ \mathrm dy \frac{1}{2x} \exp(-x)\mathrm dx\\ &= \int_{x = 0}^{\infty} \left . \frac{y^{n+1}}{n+1}\right\vert_{-x}^x \frac{1}{2x} \exp(-x)\mathrm dx\\ &= \begin{cases} \int_{x = 0}^{\infty} \frac{1}{n+1} x^n \exp(-x)\mathrm dx = \frac{\Gamma(n+1)}{n+1}, & n ~\text{even},\\ \quad & \\ \quad & \\ 0, & n ~\text{odd}, \end{cases}\\ \\ \end{align*} from which you can get the variance of $Y$ and the covariance of $X$ and $Y$.
$h(y)$ is not $\int_0^\infty \frac{1}{2x} e^{-x}\;dx$. It is $$\int_{|y|}^\infty \frac{1}{2x} e^{-x}\;dx.$$ That is because of the inequality $-x \le y \le x$, which is the same as $|y| \le x$. In other words, the density is positive only for those values of $x$ that are $\ge |y|$.
The fact that no "$y$" appears in the expression you had, $\int_0^\infty \frac{1}{2x} e^{-x}\;dx$, should arouse suspicions. It would mean that the marginal density does not depend on $y$, i.e. is constant. Probability density functions on the whole line are never constant, since the integral of a constant over the whole line is either $0$ or $\infty$ or $-\infty$, never $1$.