Laplace transform property for convolutions of the type $$ \int_0^x f(x-y)g(y)\,dy $$ is very well known. There exists a Laplace transform property to calculate functions of the type below $$ \int_a^x f(x-y)g(y)\,dy, $$ when $a>0$?
2026-04-14 03:32:33.1776137553
Laplace transform property for modified convolution function
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You should look at the bilateral Laplace transform :
$$\mathcal{L}[f(t)1_{t > 0}](s) \ \mathcal{L}[g(t)1_{t > a}](s)= \left(\int_{-\infty}^\infty f(t)1_{t > 0} e^{-st} \, dt\right) \left( \int_{-\infty}^\infty g(t_2) 1_{t_2 > a} e^{-st_2} \, dt_2\right)$$ $$=\left(\int_0^\infty f(t) e^{-st}\,dt\right) \left( \int_a^\infty g(t_2) e^{-st_2} dt_2\right) = \int_a^\infty h(\tau) e^{-s\tau} \, d\tau$$ where $$h(\tau) = (f(t)1_{t > 0}) \ast (g(t)1_{t > a})(\tau) = \int_{-\infty}^\infty f(\tau-t)1_{\tau-t > 0}g(t)1_{t > a} \, dt = \int_a^\tau f(\tau-t) g(t) \, dt$$