Consider the density
$f_{X,Y}(x,y)=C\cdot e^{-(1+x^{2})(1+y^{2})},\quad -\infty <x,y<\infty$
$C$ is a normalizing constant.
What are the conditional densities $f_{X|Y=y}(x)$ and $f_{Y|X=x}(y)$?
This is what I did:
$f_{X}(x)=C\int_{-\infty}^{\infty}e^{-(1+x^{2})(1+y^{2})}\, dy$
$(1+x^{2})=b$
$I^{2}=\Big(\int_{-\infty}^{\infty}e^{-b(1+y^{2})}\,dy\Big)\Big(\int_{-\infty}^{\infty}e^{-b(1+w^{2})}\,dw\Big)\\ =\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}e^{-b(2+y^{2}+w^{2})}\, dy\Big)\, dw$
$y^{2}+w^{2}=r^{2},\quad dy\, dw=\, dA=r\, dr\, d\theta$
$I^{2}=\int_{0}^{2\pi}\Big(\int_{0}^{\infty}re^{-b(2+r^{2})}\, dr\Big)\, d\theta\\ =2\pi\int_{0}^{\infty}re^{-b(2+r^{2})}\, dr$
$u=b(2+r^{2})=2b+br^{2},\quad du=2br$
$2b\cdot I^{2}=2\pi\int_{0}^{\infty}2bre^{-b(2+r^{2})}\, dr\\ =2\pi\int_{2b}^{\infty}e^{-u}\, du\\ =2\pi\Big[-e^{-u}\Big]_{2b}^{\infty}\\ =2\pi\Big(0+e^{-2b}\Big)=2\pi e^{-2b}$
$I=\sqrt{\frac{\pi}{b}e^{-2b}}=\sqrt{\frac{\pi}{b}}e^{-b}$
$f_{X}(x)= C\sqrt{\frac{\pi}{(1+x^{2})}}e^{-(1+x^{2})}$
$f_{Y|X=x}(y)= \frac{C\cdot e^{-(1+x^{2})(1+y^{2})}}{C\sqrt{\frac{\pi}{(1+x^{2})}}e^{-(1+x^{2})}}\\ = \sqrt{\frac{(1+x^{2})}{\pi}}e^{-(1+x^{2})(1+y^{2})+(1+x^{2})}\\ =\sqrt{\frac{(1+x^{2})}{\pi}}e^{-(1+x^{2})y^{2}}$
$Y|X\in N(0,\frac{1}{2(1+x^{2})})$
$X|Y\in N(0,\frac{1}{2(1+y^{2})})$
Is this correct or have I missed something?