Marginal probability density of $f_{X,Y}(x,y)=\frac{1}{x}\mathbf 1_{\{0I'm working through "Introduction to probability theory and statistical inference" by Larson, and I'm stuck on one of the practice problems. Hoping someone could help point me in the right direction.

Assume $(X,Y)$ is a two-dimensional continuous random variable with density: 
  $$f_{X,Y}(x,y)=\frac{1}{x}\;\;\;\;\; 0<y<x,\:\:\: 0<x<1$$
  Find the marginal densities of $X$ and $Y$.

So my calculus is very rusty, but this is what I thought the answer should be:
\begin{align}f_X(x) &= \int_0^x{\frac{1}{x}}dy = 1 \\[0.2cm] f_Y(y) &= \int_0^1{\frac{1}{x}}dx = \ln(1) - \ln(0)\end{align}
which is obviously wrong. The solutions manual says that that $$f_Y(y) = -\ln(y) \;\;\; \text{ for } \;\;\; 0<y<1$$
Can't seem to reverse engineer this answer in a way that makes sense to me. Any help would be great

You ignored the restriction on the domain of $x$. For any given $0<y<1$, $x$ ranges from $y$ to $1$, so $$f_Y(y):=\int_{X}f_{XY}(x,y)dx=\int_{y}^1\frac{1}{x}dx=[\ln x]^1_{y}=\ln1-\ln y=-\ln y$$ for all $0<y<1$. Similarly $$f_X(x):=\int_{Y}f_{XY}(x,y)dy=\int_{0}^x\frac{1}{x}dy=\frac1x[y]^x_{0}=1$$ for any $0<x<1$, meaning that $X\sim U(0,1)$, i.e. it has the uniform distribution on $(0,1)$.
