Finding marginal probability density functions from joint probability density functions

Question

Question

Let $X$ and $Y$ have a joint continuous density function of $$f(x, y) = c\frac y x, \quad 0 < x < 3, \quad 0 \leq y \leq \frac 1 3 x,$$ where $c$ is a constant to be determined.

$(a)\quad$ Determine the value of $c$.

$(b)\quad$ Are $X$ and $Y$ independent? Why or why not?

$(c)\quad$ Find the marginal density function of Y.

My working

$$\begin{aligned} \int^3_0\int^{\frac 1 3 x}_0 c \frac y x\ \mathrm{d}y\ \mathrm{d}x & = \int^3_0 \frac 1 {18} cx\ \mathrm{d}x \\[5 mm] & = \frac 1 4 c \\[5 mm] & = 1 \end{aligned}$$

$$\implies c = 4$$

X and Y are not independent as the range of $y$ depends on $x$.

$$f(y) = \int^3_0 4 \frac y x\ \mathrm{d}x$$

However, this is where I am stuck. The integral above cannot be evaluated at $x = 0$, so I know I must have gone wrong, though I know not where.

Any intuitive explanations will be greatly appreciated :)

Edit

As it turns out, I made a simple careless mistake, as pointed out by some really keen eyes!

Answer 1

You have

$$f_Y(y)=\int_{0}^3 f(x,y) dx = \int_0^3 c\frac{y}{x}\chi_{[3y,3)}(x) dx = c\int_{3y}^3 \frac{y}{x} dx = c y \log\left(\frac{1}{y}\right).$$

Answer 2

the correct integral bouds are the following

$$f_Y(y)=4y\int_{3y}^{3}\frac{1}{x}dx$$

here is a drawing of the joint support