Let $f$, $g$, and $h$ denote probability distributions such that $$f(x) = \int_0^x g(x')h(x-x')~dx'$$ where $x$ only takes non-negative values. Now, suppose we want to perform a change of variable from $x$ to $y=q(x)$ in the above equation. My question is: how does the convolution equation change when we perform the change of variables. In other words, after changing the variable to $y$, suppose the distributions corresponding to $f(x)$, $g(x)$, and $h(x)$ are $F(y)$, $G(y)$, and $H(y)$, then can we write the equation satisfied by $F$, $G$, and $H$?
Take for example $y=\log x$. Here is my attempt: We can see that $F(y) = e^yf(e^y)$ where $F(y)$ denotes the distribution of $y$ obtained by performing a change of variable $x$ to $y$ in $f$. Similarly, we define $y' = \log x'$, and write $G(y') = e^{y'}g(e^{y'})$. Now we come to the the part where I am facing some difficulty: dealing with $h(x-x')$.
Approach:
Define $z = x-x'$, and $w = \log z$. Now, $h(z) = H(w)e^{-w}$, and substituting $w = \ln(e^y - e^{y'})$ allows us to write $$h(x-x') = \frac{H(\ln(e^y - e^{y'}))}{e^y-e^{y'}}$$
This would mean, our original convolution can now be written as: $$e^{-y}F(y) = \int_{-\infty}^{e^y} e^{-y'}G(y')\cdot\frac{H(\ln(e^y - e^{y'}))}{e^y-e^{y'}}~ e^{y'} dy'$$ which gives $$F(y) = \int_{-\infty}^{e^y} e^{y}~G(y')\cdot \frac{H(\ln(e^y - e^{y'}))}{e^y-e^{y'}}~ dy'.$$
Here, I have literally written each term in the first equation in terms of the corresponding distribution of $y$. Does this seem correct? If not, can someone point me in the right direction?