Difficulty in finding marginal distribution Let $X=(X_{1},X_{2})$ have joint pdf.$$f(x_{1},x_{2})=\begin{cases}\frac{e^{-\frac{x_{2}^2}{2}}}{x_{2}\sqrt{2\pi}},\ &\text{if}\ 0<|x_{1}|\le x_{2}<\infty.\\0,\  &\text{otherwise} \end{cases}$$ Then the variance of $X_1$, is for finding out the variance I must need marginal dist. of $X_1$ $f(x_{1})$? but I am stuck in finding this integral $$\int_{x_{1}}^{\infty}\frac{e^{-\frac{x_{2}^2}{2}}}{x_{2}\sqrt{2\pi}}$$ even if need to use total variance formula I need $f_{x_{1}}$ but  I think that will be much more clumsy rather than finding variance by moments.

We don't need to know the marginal distribution of $X_1$ to compute the variance of $X_1$. That is a good thing, since the required integral cannot be evaluated using elementary functions. 
By symmetry the mean of $X_1$ is $0$. So the variance is $E(X_1^2)$, which is
$$\int_{x_2=0}^\infty \left(\int_{x_1=-x_2}^{x_2} x_1^2\frac{e^{-x_2^2/2}}{x_2\sqrt{2\pi}}\,dx_1\right) \,dx_2.$$
The inner integral is, after some cancellation,
$$\frac{2}{3\sqrt{2\pi}}x_2^2 e^{-x_2^2/2}.$$ We need to integrate this from $0$ to $\infty$. The integral can be found using the fact that the standard normal has variance $1$.
