I need to evaluate the Laplace transform of the following integral:
$$ \phi(t)= \int_0^{t_0} K(t-x)f(x)dx $$
Note that the constant upper limit of the integral is different from the time variable, so that straightforward application of the Convolution Theorem is not feasible. I have an explicit functional form of the Laplace Transform of the Kernel $K()$, but not of the function $f()$ or its transform.
Any tips, pointers, hints or even answers would be deeply appreciated.
Thanks
Based on the excellent suggestion by Daniel Fischer above, I've attempted a solution below:
The Laplace Transform of
$$\phi(t)=\int_0^{t_0} K(t-x)f(x)dx$$
is
$$ L[\phi(t)]= \int_0^\infty e^{-st}\int_0^{t_0} K(t-x)f(x)dxdt =\int_0^{t_0}\Biggl[ \int_0^\infty e^{-st}K(t-x)dt\Biggr]f(x)dx $$
With the variable shift $y=t-x$, the inner integral can be written as
$$ \Lambda(s) = e^{-sx}\int_0^\infty e^{-sy}K(y)dy=e^{-sx}\bar K(s) $$
where the bar indicates a Laplace Transform. Substitution into the outer integral yields
$$ L[\phi(t)]=\bar K(s) \int_0^{t_0}e^{-sx}f(x)dx $$
Since $f(t)$ is zero outside of the interval $[0,t_0]$, the upper limit of the outer integral can be replaced by $t$, so that
$$ L[\phi(t)]=\bar K(s) \int_0^te^{-sx}f(x)dx=\bar K(s) \bar F(s) $$