find the value of the constant k that makes fxy(x,y) a valid joint PDF

Question

$$ fxy(x,y) = \begin{cases} ke^{-(3x+2y)}, & x \ge 0, y \ge 0 \\[2ex] 0, & \text{otherwise} \end{cases} $$

I've gotten to a point where i believe that the solution comes through the lagrange theorem bit still end up dividing by zero and i'm out of ideas

$$ \int_\infty^0\int_\infty^0ke^{−(3x+2y)}dxdy = 1 \\[2ex] k\int_\infty^0 \frac {e^{−(3x+2y)}}{-3x-2y} \Big|_0^\infty dy = 1 \\[2ex] k\int_\infty^0 \frac {e^{−(3\infty+2y)}}{-3\infty-2y} - \frac {e^{−(3(0)+2y)}}{-3(0)-2y} dy = 1 \\[2ex] k \frac {e^{−2y}}{2y} \Big|_0^\infty dy = 1 \\[2ex] $$

After this step I get a division by zero, and need a hint on how to proceed. I'm trying to get a numerical value for K

Answer 1

Hint:

• Use the property that pdf integrates to $1$ to solve for $k$.

Answer 2

Turns out I was using the wrong approach, and I had an error on step 2 of my solution.

The solution is a two step u - sustitution

$$ \int_\infty^0\int_\infty^0ke^{−(3x+2y)}dxdy = 1 \\[2ex] k\int_\infty^0\int_\infty^0\frac{e^{u}}{-3}dudy = 1 \\[2ex] \frac{k}{3}\int_\infty^0\int_\infty^0-e^{u}dudy = 1 \\[2ex] \frac{k}{3}\int_\infty^00-(-e^{-2y})dy = 1 \\[2ex] \frac{k}{3}\int_\infty^0e^{-2y}dy = 1 \\[2ex] \frac{k}{3}\int_\infty^0\frac{e^{u}}{-2}du = 1 \\[2ex] \frac{k}{6}\int_\infty^0e^{u}du = 1 \\[2ex] \frac{k}{6}[0-(-1)] = 1 \\[2ex] \frac{k}{6} = 1 \\[2ex] k = 6 $$