Proving valid joint pdf is valid with unknowns Question
Suppose Suppose that $Y$ and $Z$ are continuous random variables with joint probability density function $f_{y,z}(y,z) = sy^{2}e^{-(\alpha + \beta z)y}$ where $y >0 \;,\; z > 0$ and $\alpha >0 \;,\beta > 0$. Find the value of $s$ that would allow for $f_{y,z}(y,z)$ to be a valid pdf.
My approach
For $f_{y,z}(y,z)$ to be a valid distribution, the following condition must be met.
$$
\iint_{A}f_{y,z}(y,z) \; \mathrm{d}z\mathrm{dy}=1
$$
In context, for the given bounds, I have come up with this result with the given bounds.
\begin{align}
    \int_{0}^{\infty}\int_{0}^{\infty} sy^{2}e^{-(\alpha + \beta z)y} \; \mathrm{d}z \; \mathrm{d}y
\end{align}
Integrating for $\mathrm{d}z$, I have obtained the following:
\begin{align}
    \int_{0}^{\infty} sy^{2}e^{-(\alpha + \beta z)y} \; \mathrm{d}z &= sy^{2}\int_{0}^{\infty} e^{-(\alpha + \beta z)y} \; \mathrm{d}z \\
     &= sy^{2}\int_{0}^{\infty} e^{-y\alpha + \beta zy} \mathrm{d}z \\
    &= sy^{2} \cdot -\frac{1}{\beta y} \cdot e^{-y\alpha + \beta zy} \\
\end{align}
Combining and evaluating for the bounds
\begin{align}
    -\frac{sye^{-y(\alpha + \beta z)}}{\beta} \Biggr|_{0}^{\infty} &= -\frac{sye^{-y(\alpha + \beta (\infty))}}{\beta} - \left(-\frac{{sye^{-y(\alpha + \beta (0))}}}{\beta}\right) \\
    &= \frac{sy}{e^{y\alpha + \beta (\infty)}\beta} + \frac{sy}{e^{y(\alpha)}\beta}
\end{align}
As, $\lim_{z\rightarrow\infty}$, the first fraction becomes $0$. Thus, leaving only the second result $\frac{sy}{e^{y(\alpha)}\beta}$.
Integrating for $\mathrm{d}y$ over the bounds using integration by parts.
\begin{align}
    \int_{0}^{\infty} \frac{sye^{-y\alpha}}{\beta} \; \mathrm{d}y &= \frac{s}{\beta} \int_{0}^{\infty} ye^{-y\alpha} \; \mathrm{d}y \\
    &= \frac{s}{\beta}\cdot\left(\frac{e^{-y(\alpha)}(\alpha y + 1)}{a^2} \Biggr|_{0}^{\infty}\right)
\end{align}
Whereby, this yields (I am very much unsure about this component)
\begin{align}
  \frac{s}{\beta}\cdot\frac{e^{-y(\alpha)}(\alpha y + 1)}{a^2} \Biggr|_{0}^{\infty}  &= \frac{s}{\beta}\cdot\left(\frac{e^{-\infty\alpha}(\alpha \infty)}{a^2} - \frac{e^{-\alpha(0)}(\alpha(0) + 1)}{a^2}\right)
\end{align}
To satisfy the condition for a valid pdf, this must be equal to 1.
\begin{align}
    \frac{s}{\beta}\cdot - \frac{1}{a^2} &= 1 \\
    -\frac{s}{\alpha^2\beta} &= 1 \\ 
    \alpha^2\beta &= -s \\
    \therefore s &= -\alpha^2\beta
\end{align}
I am very unsure if my approach is correct and my integration is not the greatest aspect. Any input would mean the world to me.
Thanks in advance!
 A: Everything looks correct except the sign is wrong.  If $\alpha > 0$ and $\beta > 0$, then your choice $s = -\alpha^2 \beta$ would be negative, which in turn would make the joint density negative.
The marginal density of $Y$ that you obtained is correct:  $$f_Y(y) = \frac{s}{\beta} y e^{-\alpha y} \, y > 0.$$  Its integral can be computed via integration by parts, or by recognizing it as a gamma function, or by recognizing it as a gamma density:
$$\begin{align}
1 &= \frac{s}{\beta} \int_{y=0}^\infty ye^{-\alpha y} \, dy \\
&= \frac{s}{\beta} \left( \left[ -\frac{1}{\alpha} y e^{-\alpha y} \right]_{y=0}^\infty + \int_{y=0}^\infty \frac{1}{\alpha} e^{-\alpha y} \, dy \right) \\
&= \frac{s}{\beta} \left( -0 + 0 + \left[-\frac{1}{\alpha^2} e^{-\alpha y} \right]_{y=0}^\infty \right) \\
&= \frac{s}{\beta} \left( -0 + \frac{1}{\alpha^2} \right) \\
&= \frac{s}{\alpha^2 \beta}.
\end{align}$$
Equivalently, the substitution $$u = \alpha y, \quad du = \alpha \, dy$$ gives
$$\int_{y=0}^\infty ye^{-\alpha y} \, dy = \int_{u=0}^\infty \frac{u}{\alpha} e^{-u} \cdot \frac{1}{\alpha} \, du = \frac{1}{\alpha^2} \int_{u=0}^\infty ue^{-u} \, du = \frac{1}{\alpha^2} \Gamma(2) = \frac{1}{\alpha^2}$$ where we have used the gamma function $$\Gamma(z) = \int_{u=0}^\infty u^{z-1} e^{-u} \, du.$$
Finally, we can also note that $f_Y(y)$ is proportional to a gamma density with shape parameter $a = 2$ and rate parameter $b = \alpha$; i.e., $$f_Y(y) \propto \frac{b^a y^{a-1} e^{-by}}{\Gamma(a)} = \alpha^2 y e^{-\alpha y}.$$  Therefore we require $s/\beta = \alpha^2$, or $s = \alpha^2 \beta$.
