The joint density function of $X$ and $Y$ is given by $f(x,y)=xe^{-x(y+1)}$ for $x>0,y>0$. find the conditional density of $Y$ given $X$.
I am close to get the answer but with a little problem
$\displaystyle f_{Y|X}(y|x)=\frac{f(x,y)}{f_x(x)} =\frac{xe^{-x(y+1)}}{\int_0^{\infty} xe^{-x(y+1)}dy}
=\frac{e^{-xy}}{\int_0^\infty e^{-xy}dy}$
but when it goes infinite it will approach 0 and the final conditional density wouldn't make any sense.....hope someone can guide me to the right approach....thanks
It seems like your integral and simplification is incorrect.
For now, I would leave $f(x,y)=xe^{-x(y+1)}=xe^{-xy}e^{-x} = \frac{x}{e^{xy}e^{x}}$
Then, $f_x(x) = \int_0^\infty xe^{-x(y+1)}dy = \int_0^\infty xe^{-xy}e^{-x}dy = \frac{-xe^{-xy}e^{-x}}{x}|_0^\infty = -e^{-xy}e^{-x}|_0^\infty = \frac{-1}{e^\infty} - \frac{-1}{e^x} = \frac{1}{e^x}$
And now your fraction is:
$f_{Y|X}(y|x) = \frac{\frac{x}{e^{xy}e^{x}}}{\frac{1}{e^x}} = \frac{xe^x}{e^{xy}e^{x}} = \frac{x}{e^{xy}}$
I'm not sure if that is correct, but based on the information given, it would appear to be.