We have a density function: $p_{X,Y}(x,y)=ce^{-y}, 0\leq x$$p_{XY}(x, y)= ce^{-y},\quad 0\leq x \lt y\lt (x+1)$$
I need to find $c$.
My try:
$$\int_0^{\infty}\int_x^{x+1}ce^{-y}dydx = \int_0^{\infty}\int_0^{y}ce^{-y}dxdy = \ldots =1$$
Is the idea correct, are the bounds of the integral correct (I did it both ways, to see if I understand it).
Let's say that I would be interested in $P(X<1)$ I would do:
$$P(X<1) = \int_0^1\int_x^{x+1}ce^{-y}dy$$
Are the above two processes of solving correct?
 A: Please note that $0 \lt x \lt y \lt x + 1 \lt \infty$
So the first integral is correct. For the second, when $y \in (0,1), 0 \lt x \lt y$ and for $y \in (1, \infty), y-1 \lt x  \lt y$
So the integral should be,
$ \displaystyle \int_0^1 \int_0^y  c \cdot e^{-y} ~dx ~dy + \int_1^\infty \int_{y-1}^y  c \cdot e^{-y} ~dx ~dy ~= 1$
A: Why reverse the order of integration in solving for $c$?
\begin{aligned}\int_0^{\infty}\int_x^{x+1}ce^{-y}dydx =\int_0^\infty \left(-ce^{-x-1}+ce^{-x}\right) dx &=\lim_{t \to \infty} \left(ce^{-x-1}-ce^{-x}\right) \Big]_0^t\\&=\left(ce^{-1} -c\right)\cdot\lim_{t \to \infty}e^{-t} - ce^{-1}+c \\&=c(1-e^{-1})\end{aligned}
\begin{aligned}\int_0^1 \int_0^y  c  e^{-y} dx dy + \int_1^\infty \int_{y-1}^y  c e^{-y} dx dy &=\int_0^1 ce^{-y}y dy+ \int_1^\infty c e^{-y}dy
\\&=\left[-ce^{-y}y\right]_0^1+\left[-ce^{-y}\right]_0^1+\lim_{t \to \infty} \left[-ce^{-y} \right]_1^t
\\&=-ce^{-1}-ce^{-1}+c+\lim_{t \to \infty} -ce^{-t} +ce^{-1}\\&= c(1-e^{-1})\end{aligned}
Seems like extra work to do more work. Pardon my candor, I just point this out as something to consider.
