Integrating $f_{XY}(x,y) = \frac{2}{1-x}\boldsymbol{1}_{y\in(0, 1-x), x \in(0, 0.5)}$ over $x$

Question

I have the joint pdf $f_{XY}(x,y) =\frac{2}{1-x}\boldsymbol{1}_{y\in(0, 1-x), x \in(0, 0.5)}$. I want to obtain the marginal pdf for $Y$.

\begin{align} f_Y(y) = \int_0^{0.5} \frac{2}{1-x}\boldsymbol{1}_{y\in(0, 1-x), x \in(0, 0.5)}dx \\ \end{align}

How do I integrate this function with the indicator variable? Is it simply $$ f_Y(y) = \begin{cases} 1.386 \ \ \text{if } y \in (0, 1-x), x \in(0, 0.5) \\ 0 \ \ \text{otherwise} \end{cases} $$ This looks like a very odd marginal pdf to me. As a sanity check, I wanted to integrate this to ensure it comes out to be 1, but the upper limit of this integration would depend on $x$ within is $\in (0,0.5)$. How does this integration work? I feel like I did not obtain the correct $f_Y$ to begin with.

Answer 1

Note that $\mathbf 1_{x\in(0,1/2), y\in(0,1-x)} =\mathbf 1_{0< x< \min\{ 1/2,1-y\},0<y<1}$, then, $$f_{Y}(y)=\int_0^{ \min\{ 1/2,1-y\}}\frac{2}{1-x}\,\mathrm{d}x\quad,0\lt y\lt 1$$

Hence, $$f_{Y}(y)=\left\{ \begin{array}{ll} 2\ln2&,0<y<1/2\\ -2\ln y&,1/2<y<1 \end{array}\right.$$