Marginal Probability Density Functions Q) Let X and Y have the joint p.d.f. 
\begin{equation}
f(x,y)=\begin{cases}
e^{-y}, \text{if } 0<x<y<\infty \\
0, \text{otherwise }
\end{cases}
\end{equation}
Obtain the marginal p.d.f's for X and Y.
A)I know that $F_{x}(s)=\lim_{t \to \infty} F_{x,y}(s,t)$ So have found f(x) to be 
\begin{equation}
f(x)=\begin{cases}
\lim_{y \to \infty} e^{-y} = 0, \text{if } 0<x<y<\infty \\
\lim_{y \to \infty} 0 = 0, \text{otherwise }
\end{cases}
\end{equation}
and f(y) to be 
\begin{equation}
f(y)=\begin{cases}
\lim_{x \to \infty} e^{-y} = 0, \text{if } 0<x<y<\infty \\
\lim_{x \to \infty} 0 = 0, \text{otherwise }
\end{cases}
\end{equation}
(since if x is going to infinity then y is going to infinity.)
However, the next question is asking about calculating the co-variance matrix of X+Y and X-Y which makes me think that my first answer was wrong. I don't know how to fix it so any hints would be really helpful, thanks!
 A: The following figure depicts the joint density which is zero outside the red region.

So the marginals are
$$f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\ dy=\int_x^{\infty}e^{-y}\ dy=e^{-x},$$
if $x\ge 0$
and
$$f_Y(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\ dx=\int_0^{y}e^{-y}\ dx=ye^{-y},$$
if $y\ge 0$.
EDIT
$$E[XY]=\int_0^{\infty}\int_0^{\infty}xyf_{X,Y}(x,y)\ dxdy=$$
$$=\int_0^{\infty}x\int_x^{\infty}ye^{-y}\ dy\ dx.$$
For the internal integral, the antiderivative is $-e^{-y}(1+y)$. This antiderivative is gained by integrating by parts and using $v'=e^{-y}$ and $u=y$.
And $\left[-e^{-y}(1+y)\right]_x^{\infty}=e^{-x}(1+x).$ Then
$$E[XY]=\int_0^{\infty}xe^{-x}(1+x)\ dx=\int_0^{\infty}xe^{-x}\ dx+\int_0^{\infty}x^2e^{-x}\ dx$$
and you can use the antiderivative formula given above. But without that, you know that, the first integral is $1$ bacause we compute the mean of the exponential distribution of $\lambda=1$, also, the other integral is $2$ because we compute the second momentum of the same. So,
$$E[XY]=3.$$
