Finding conditionally expected $y$ given a specific $x$ from a joint distribution function!

Question

I want to determine expected $y$, given $x=2$ given joint pdf shown below

$$\frac{1}{2y} * e^{-\frac{y^2 + \frac{x}{2}}{y}}$$

for $x,y \gt 0$ and $0$ otherwise

I believe this means I want to find $f_{Y \mid X}(y \mid x)= \frac{f_{X,Y}(x,y)}{f_X(x)}$, where I calculate $f_X(x)$ by integrating the joint pdf from $0$ to $x$ with respect to $y$, but this I can't seem to compute(refer to prior edits).

Is there an alternative way that simplifies this? Thank you for your time!${}$${ }$

Answer 1

$f_Y(y)=\int_{0}^{\infty} \frac{1}{2y}e^{-(y+\frac{x}{2y})}dx = e^-y$
$f_Y(2)=e^{-2}$
$f_{X|Y}(x|2)= \frac{1}{4}e^{-(2+\frac{x}{4})}$

$\int_{0}^{\infty}\frac{f_{X|Y}(x|2)}{f_Y(2)} = \int_{0}^{\infty} \frac{1}{4}e^{-\frac{x}{4}} dx = 1$