I am trying to solve: $$f(x)=\delta(x)+\int_{-a}^\infty f(y)g(x-y)dy \quad a\in\mathbb{R}^{+}$$
Usually this integral equation is solved with Fourier when $a=\infty$, but here an Heaviside messes up the resolution.
The problem with writing $f(x)=\delta(x)+\int_{-\infty}^\infty u(y+a)f(y)g(x-y)dy$ is that we get $$\mathcal{F}[f]=1+\mathcal{F}[f]*\mathcal{F}[u(\cdot + a)]\cdot \mathcal{F}[g]$$ which is not suitable to isolate $\mathcal{F}[f]$.
I wonder if we can define a "truncated bilateral" Laplace transform and its inversion starting the integral from $-a$ instead of 0. What do you think?