It is known that the distribution function of a product of two random variables $Z = XY$ with $X \sim f_X$ and $Y \sim f_Y$ is
$$ f_Z(z) = \int_{-\infty}^\infty f_X(x)f_Y(\frac{z}{x})\frac{1}{|x|}dx $$
This assumes that both $X$ and $Y$ can take values in $\mathbb{R}$.
Say Y is a standard uniform distribution (constant density equal to $1$ over $[0; 1]$), then the integral can be shown to become
$$ f_Z(z) = \int_{z}^\infty f_X(x)\frac{1}{|x|}dx $$
Essentially, because the support for $f_Y$ is compact, the PDF is non-zero only for values of $\frac{z}{x} \in [0; 1] \Leftrightarrow 0 \leq \frac{z}{x} \leq 1 \Leftrightarrow 0 \leq \frac{1}{x} \leq \frac{1}{z}$. This can be interpreted as $z \leq x \leq +\infty$, if I am correct.
My question is, would it be possible to come up with an arbitrary distribution for $Y$ with compact support and at the same time get rid of the $\frac{1}{| x |}$ term? I was thinking in the line of
$$ f_Y(y) = \begin{cases} \frac{1}{b}|y|, & \text{if} -b \leq y \leq b \\ 0, & \text{otherwise} \end{cases} $$
And set $b = 1$ so that the PDF of $Z$ becomes
$$ f_Z(z) = \int_{-z}^{z} f_X(x)dx $$
But could this be called a PDF?