Finding $E[X]$ from the joint density function of $X$ and $Y$. The joint density function of $X$ and $Y$ is given by
$$f(x,y)=\frac{1}{y}e^{-(y+\frac{x}{y})},\quad x>0,y>0.$$
Find $E[X]$, $E[Y]$ and $Cov\left(X,Y\right)$.
Calculating $E[Y]$ was easy for me.
\begin{align}
f_{Y}(y)&=\int_{0}^{\infty} \frac{1}{y}e^{-(y+\frac{x}{y})}dx\\
&=e^{-y}, \quad y>0
\end{align}
Therefore $Y$ is an exponential random variable with parameter $1$ so
$$E[Y]=1$$

Now, I am stuck in calculating the density function of X. I tried to calculate it from the usual integral but I couldn't. I looked at my calculus book to revise the integrals chapter but I haven't stumble accross something like it. Any help would be much appreciated.

 A: Thanks to @sudeep5221 for the suggestion. We do not need to explicitly calculate the density function of $X$.
\begin{align}
E[X]&=\int_{0}^{\infty}\int_{0}^{\infty}xf(x,y)\ dx\ dy\\
&=\int_{0}^{\infty}e^{-y}\int_{0}^{\infty}\frac{x}{y}e^{-x/y}\ dx\ dy
\end{align}
It is noticeable that $\int_{0}^{\infty}\frac{x}{y}e^{-x/y}\ dx$ is the mean of an exponential random variable with parameter $1/y$, and thus equals to $y$.
$$
E[X]=\int_{0}^{\infty}\int_{0}^{\infty}ye^{-y}dy=1
$$
To end, calculate $Cov(X,Y)$.
$$Cov(X,Y)= E[XY]-E[X]E[Y]$$
Also
\begin{align}
E[XY]&=\int_{0}^{\infty}\int_{0}^{\infty}xyf(x,y)\ dx\ dy\\
&=\int_{0}^{\infty}ye^{-y}\left(\int_{0}^{\infty}\frac{x}{y}e^{-x/y}dx\right)\ dy\\
&= \int_{0}^{\infty}y^{2} e^{-y}dy
\end{align}
We can now integrate by parts where $U=y^2$ and $dV= e^{-y}dy$.
\begin{align}
E[XY]&=\int_{0}^{\infty}y^2e^{-y} dy= \left(-y^2 e^{-y}|_{0}^{\infty}\right)+\int_{0}^{\infty}2ye^{-y}dy\\
&=2E[Y]\\
&=2
\end{align}
As a result,
\begin{align}Cov(X,Y)&=2-(1)(1)\\
Cov(X,Y)&=1
\end{align}
