Question
Let $f:[0,\infty) \rightarrow [0,\infty)$ be some continuously differentiable function and $a \in (0,1)$ then we define the function $H:[0,\infty) \rightarrow [0,\infty)$ by letting: $$H(x)=\int_0^x f(x-u) f(x-au) e^{-u} \, du.$$ Suppose we would like to compute $H$ numerically, then one way to do this would be to simply compute the integral in the right hand for each value of $x$. I am however looking for a more efficient method as is possible for the special case $a=0$.
Special case $a=0$
In this case we define $K(x) = \int_0^x f(x-u) e^{-u} \, du$ and we see that $H(x)=f(x) K(x)$. We now show that there is a simple method to compute K(x). We find by a simple change of variables $v=x-u$ that: $$ K(x)=\int_0^x f(v) e^{v-x} \, dv = \int_0^x f(v) e^{v}\, dv \cdot e^{-x}. $$ Applying the product rule we obtain $K'(x)=f(x) e^x e^{-x} - \int_0^x f(v)\, e^{v} \, dv\, e^{-x}$ applying the definition of $H$ and some rewriting we obtain: $$ K'(x)=f(x)-K(x). $$ This is a differential equation which can be solved much more quickly and $H(x)$ is easily obtained from $K(x)$.