Normalizing a Joint PDF

Detailed explanation of solving $\int_{0}^{1}\int_{0}^{\infty }Cxye^{-y^{2}}dydx=1$ :

$u=-y^{2}$

$du=-2ydy$

$ydy=-\frac{1}{2}du$ .

Substituting we get:

$\int_{0}^{1}\int_{0}^{-\infty }-\frac{1}{2}Cxe^{u}dudx=1$

Notice that integration bounds have changed also. That's because $\lim _{y\to\infty } u=\lim _{y\to\infty } -y^{2}=-\infty$

Solving inner integral:

$\begin{aligned} & \int_{0}^{-\infty }\underbrace{-\frac{1}{2}Cx}_{const.}e^{u}du \\ & = \left [ -\frac{1}{2}Cxe^{u} \right ]^{-\infty }_{0}=-\frac{1}{2}Cxe^{-\infty }-(-\frac{1}{2}Cxe^{0}) \\ & = 0-(-\frac{1}{2}Cx) \\ & =\frac{1}{2}Cx \end{aligned}$

Now, we have:

$\begin{aligned} & \int_{0}^{1}\frac{1}{2}Cxdx=1 \\ & \Leftrightarrow \frac{1}{2}C\int_{0}^{1}xdx=1 \\ & \Leftrightarrow \frac{1}{2}C\left [ \frac{x^{2}}{2} \right ]^{1}_{0}=1 \\ & \Leftrightarrow C\times \frac{1}{2}=2 \\ & \Leftrightarrow C=4 \end{aligned}$

This is computed directly by integration:

$\int_0^1 \int_0^{\infty} Cxye^{-y^2} dy dx = \int_0^1 \frac{Cx}{2} dx = \frac{C}{4} \implies C=4.$