Let :
- $U \sim \mathcal{U}[0,1]$
- $X \sim \gamma(n+1)$ that is to say :
$X= \sum_{k=0}^n T_k$ with $(T_k)_k$ iid and follow an exponential distribution with parameter $1$
- $Y=UX$
- $U$ and $X$ independent.
We want to prove that : $f_Y(y)= \frac{1}{n} \sum_{k=0}^{n-1} \frac{ y^k e^{-y} }{ k!}$ for $y >0$
My attempt :
$C= \ln(UX) = \ln U + \ln X$
$f_C(t)= \int_{ - \infty}^{\infty} e^{t-u} e^{u} f(e^u) \mathbb{1}_ {u <0} du = e^t \int_{t}^{+ \infty} f(e^u) du$
$ f_Y(t) = \frac{1}{t} f_C( \ln t) = \int_{ \ln t}^ { \infty} f(e^u) du = \int_{t}^{\infty}f(u) du $
\begin{align*} f_Y(t) & = \int_{t t} ^{ \infty} f_X(x) dx \\ &= \int_{ t} ^{ \infty} \frac{1}{n !} e^{-x}x^n dx \\ &= \frac{1}{n!} \frac{1}{t} ( t ) ^n + \int_{ t} ^{ \infty} \frac{1}{(n-1) !} e^{-x}x^{n-1} dx \\ \end{align*}
Setting
$$\begin{cases} y=ux\\ z=x \end{cases}\rightarrow\begin{cases} x=z\\ u=\frac{y}{z} \end{cases}$$
The Jacobian is $|J|=\frac{1}{z}$ and thus
$$f_{YZ}(y,z)=\frac{1}{n!}z^{n-1}e^{-z}\cdot\mathbb{1}_{[0;\infty)}(y)\mathbb{1}_{[y;\infty)}(z)$$
Thus
$$f_Y(y)=\frac{1}{n!}\int_y^{\infty}z^{n-1}e^{-z}dz\cdot\mathbb{1}_{[0;\infty)}(y)$$
You can leave the density in this way or observing that
$$f_Y(y)=\frac{\Gamma(n,y)}{n!}\cdot\mathbb{1}_{[0;\infty)}(y)$$
probably this expression can be expressed in what you are looking for...
Just as an example, with $n=3$ you get
$$f_Y(y)=\frac{1}{6}\int_y^{\infty}z^2e^{-z}dz=\frac{e^{-y}}{6}[y^2+2y+2]\cdot\mathbb{1}_{[0;\infty)}(y)$$
which is a nice density