I want to find the marginal mass function of $\ x $ given
$$\ f_{x,y} (x,y) = \begin{cases} \frac{1}{x} & 0< y<x < 1 \\ \\0 & else \end{cases} $$
if I draw the area of the function
I was thinking I could just integrate $\ y $ to find the pmf of $\ x $. so $\ \int_0^x \frac{1}{x} \ \ dy = 1 $ but it seems wrong to me.
The value of $\ f_{x,y} $ is very high when both $\ x,y $ close to $\ 0 $ (approaching $\ \infty $ ) so when I cancel y over $\ x$, the $\ f_x $ function should have very high values when $\ x $ also closer to infinity and approach to $\ 1 $ as $\ x $ approach $\ 1 $
Note that you know the joint density function $$f_{x,y}(x,y)=\frac{1}{x}\mathbf{1}_{\{0<y<x<1\}}=\left\{\begin{aligned}\frac{1}{x}, \quad 0<y<x<1\\ 0, \quad \text{otherwise} \end{aligned} \right.$$ Now, we need to calcule the marginal function of $x$,by definition we have $$\color{blue}{f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy}$$ since that $S_{X,Y}=\{(x,y): 0<y<x<1\}$, where $S_{X,Y}$ denote the support of $(X,Y)$. So, we have $$f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy=\int_{0}^{x}\frac{1}{x}dy=\left.\frac{1}{x}y\right|_{0}^{x}=\frac{x}{x}=1$$ So, we can conclude that $$\boxed{f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy=1\cdot \mathbf{1}_{\{0<x<1\}}}$$