Consider the joint pdf of $(X, Y)$ given by
$$f(x, y) = \begin{cases} e^{-y} & \text{ if } 0 < x < y < \infty \\ 0, & \text{ otherwise} \end{cases} $$
(a) Find the marginal PDF of $Y$
(b) Find the conditional PDF of $X$ given $Y = y$
(c) Find $\mathbb{E}[X \mid Y = y]$
My attempt:
(a)
$$f_{Y}(y) = \int_{0}^{y} e^{-y} \mathop{dx} $$
$$= xe^{-y} \big|_{0}^{y} \implies \boxed{f_{Y}(y) = \begin{cases} ye^{-y} & \text{ for } y > 0 \\ 0 & \text{ otherwise } & \end{cases}}$$
(b)
$$f_{X\mid Y = y}(x, y) = \frac{f(x, y)}{f_{Y}(y)} = \frac{e^{-y}}{ye^{-y}} = \frac{1}{y} \implies \boxed{f_{X \mid Y = y}(x,y) = \begin{cases} 1/y & \text{ if } 0 < x < y < \infty \\ 0 & \text{ otherwise} \end{cases}}$$
(c) Using what I have from (b),
$$\mathbb{E}[X \mid Y = y] = \int_{0}^{\infty} x \cdot \frac{1}{y} \mathop{dx} $$
$$= \left(\frac{x^2}{2y}\right)\Big|^{\infty}_{0} = \infty$$
Where did I go wrong?