Distribution of $XY$ if $X \sim$ Beta$(1,K-1)$ and $Y \sim$ chi-squared with $2K$ degrees Suppose that $X$ has the beta distribution Beta$(1,K-1)$ and $Y$  follows a chi-squared with $2K$ degrees. In addition, we assume that $X$ and $Y$ are independent.
What is the distribution of the product $Z=XY$ .
Update
My attempt:
\begin{align}
f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{0}^{+\infty}  \frac{1}{B(1,K-1)2^K \Gamma(K)}  \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{0}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} [-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})]_0^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1,-z/2)
\end{align}
Is it correct? if yes, how we call this distribution? 

After some valuable remarks, I was able to find the solution:
We have $f_X(x)=\frac{1}{B(1,K-1)} (1-x)^{K-2}$ and $f_Y(y)=\frac{1}{2^K \Gamma(K)} y^{K-1} e^{-y/2}$.
Also, we have $0\le x\le 1$. Thus, if $x=\frac{z}{y}$, we get  $0 \le \frac{z}{y}  \le 1$ which implies that $z\le y \le \infty$.
Hence:
\begin{align}
f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{z}^{+\infty}  \frac{1}{B(1,K-1)2^K \Gamma(K)}  \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{z}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} \left[-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})\right]_z^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1) \\ &= \frac{1}{2} e^{-z/2}
\end{align}
where the last equality holds since $B(1,K-1)=\frac{\Gamma(1)\Gamma(K-1)}{\Gamma(K)}$.
So $Z$ follows an exponential distribution of parameter $\frac{1}{2}$; or equivalently, $Z \sim\chi_2^2$.
