I am trying to calculate this integral:
$$f(y)=\int_{\mathbb{R}}2e^{-x-y}\textbf{1}_{(0,\infty)}(x)\textbf{1}_{(x,\infty)}(y)dx.$$
This is obviously equal to
$$2e^{-y}\int_0^\infty e^{-x}\textbf{1}_{(x,\infty)}(y)dx.$$
But… How do I integrate this? The characteristic function has the variable $x$ in the interval, so I can't get the characteristic function out of the Integral. The function $f$ is only depending on $y$… This doesn't really make sense… If I try to imagine this function $g(x,y):=2e^{-x-y}\textbf{1}_{(0,\infty)}(x)\textbf{1}_{(x,\infty)}(y) $ and especially the domain of this function, it should be a function whose domain is in the first quadrant of the $x,y$-level but only above the graph $x=y$. The positive $y$-axis is contained but the set $\{(x,y,z)\in\mathbb{R}^3\big|x=y,z=0\}$ is not contained. So if we now integrate along the $x$-axis as above, there should only be one function value because the $x$-axis is not contained in the domain either. So this integral should just be $0$…?
Thank you for your help! Kind regards, Max.
Best Answer
Hint: Notice that $\mathbf{1}_{(x,\infty)}(y) = \begin{cases} 1, & y \in (x,\infty) \\ 0, & y \in (-\infty,x] \end{cases}$. So, if we want to change this to be a function of $x$ instead of $y$, we see that \begin{align*} \mathbf{1}_{(x,\infty)}(y) &= \begin{cases} 1, & y \in (x,\infty) \\ 0, & y \in (-\infty,x] \end{cases} \\ &= \begin{cases} 1, & x \in (-\infty,y) \\ 0, & x \in [y,\infty) \end{cases} \\ &= \mathbf{1}_{(-\infty,y)}(x). \end{align*}
Therefore, the product becomes $\mathbf{1}_{(0,\infty)}(x) \mathbf{1}_{(x,\infty)}(y) = \mathbf{1}_{(0,\infty)}(x) \mathbf{1}_{(-\infty,y)}(x) = \mathbf{1}_{(0,\infty) \cap (-\infty,y)} (x)$.