Cfrgtkky

Is there a simplified result for Laplace Transform of $f(x) theta(x) star f(x) theta(-x)$, where $f(x)$ is an even function, $theta(x)$ is Heaviside function and $star$ is convolution?

For $f(x) theta(x) star f(x) theta(x)$, LT simplifies this to $LT[f(x)] * LT[f(x)]$.

$f(x) theta(x) star f(x) theta(-x)$ is a function that is non-zero for $x <0$ and $x >=0$. I am interested in the one-sided LT of this expression.

I tried plugging in the expansion of convolution to get the following.

$$LT [f(x) theta(x) star f(x) theta(-x)]$$

$$= intlimits_{p=0}^{infty} Big[intlimits_{x=-infty}^{infty} f(x) theta(x) f(p-x) theta(x-p) dx Big] e^{-ps} dp$$

$$= intlimits_{x=0}^{infty} f(x)Big[ intlimits_{p=0}^{infty} f(p-x) theta(x-p) e^{-ps} dp Big] dx$$

$$=intlimits_{x=0}^{infty} f(x) Big[ intlimits_{y=-x}^{0} f(y) e^{-ys} dy Big] e^{-sx} dx$$

And because f(x) is even, this can be converted to

$$=intlimits_{x=0}^{infty} f(x) Big[ intlimits_{y=0}^{x} f(y) e^{ys} dy Big] e^{-sx} dx$$

Is there a further simplification where the LT of this convolution can be expressed in terms of LT of the individual terms?