Question: Given $f(x,y)=e^{-\lambda(x+y)}$ for $x,y\geq0$ and $\lambda>0$, what constant must $f(x,y)$ be multiplied by to get a joint density function?
Work: I obtained $$ \int_{0}^1\int_0^\infty e^{-\lambda x-\lambda y}\,dx\,dy, $$ but I am unsure about the upper bounds. For the inner integral I obtained $$ \frac{1}{\lambda}\int_0^\infty e^{-u}\,du=\frac{1}{\lambda} $$ by $u$-substitution, and for the outer integral I obtained $$ \int_0^1 \frac{1}{\lambda}\,dy=\frac{1}{\lambda} $$ but I'm not sure this is quite right. Should the upper bound for my outer integral be $\infty$? But then the outer integral would diverge. Any ideas as to how to effectively calculate here? What exactly am I missing?
Observe that we can factor the joint density as $$ f(x,y)=e^{-\lambda x}e^{-\lambda y}; \quad (x,y\geq 0)\tag{1}. $$ Let $$ I=\int_0^\infty\int_0^\infty f(x,y)\,dx\,dy $$ and note that $$ I=\int_0^\infty\int_0^\infty e^{-\lambda x}e^{-\lambda y}\, dx \,dy =\int_0^\infty e^{-\lambda y}\left[\int_0^\infty e^{-\lambda x}\, dx \right]\,dy =\left[\int_0^\infty e^{-\lambda x}\, dx \right]\left[\int_0^\infty e^{-\lambda y}\, dy \right] $$ whence $$ I=\left[\int_0^\infty e^{-\lambda x}\, dx \right]^2=\frac{1}{\lambda^2} $$ Thus $\lambda ^2I=1$ so $f$ must be multiplied by $\lambda ^2$ to obtain a pdf.