It is given that $$f(x,y) = \frac{1}{y}e^{-(y+\frac{x}{y})} \text{ for } x,y > 0$$ Find the marginal distribution of $X$.
I can find the marginal distribution of $Y$ which is $f_{Y}(y)=e^{-y}$. And also, the conditional distribution of $X|Y$ which is $f_{X|Y}(x|y)=\frac{1}{y} e^{-\frac{x}{y}}$. But how to find the marginal distribution of $X$?
Integrating out the $y$ variable, for $x>0$ we have $$ f_X(x)=\int_0^{\infty}e^{-(y+\frac{x}{y})}\;\frac{dy}{y}=\int_0^{\infty}e^{-\sqrt{x}(\frac{y}{\sqrt{x}}+\frac{\sqrt{x}}{y})}\;\frac{dy}{y}$$ and setting $t=\frac{y}{\sqrt{x}}$, this becomes $$ f_X(x)=\int_0^{\infty}e^{-\sqrt{x}(t+\frac{1}{t})}\;\frac{dt}{t}=2K_0(2\sqrt{x})$$ for $x>0$, where $$ K_s(x)=\frac{1}{2}\int_0^{\infty}t^se^{-\frac{x}{2}(t+\frac{1}{t})}\;\frac{dt}{t}$$ is a $K$-bessel function.