Making suitable assumptions wherever necessary, what is the Laplace Transform $\mathcal{L}(S(t))$ where $S(t)=\int_{0}^{t}\int_{0}^{t}f(t-s_1,t-s_2)g(s_1)h(s_2)ds_1ds_2$.
I tried using the Double Laplace theorem but I am not sure if that is applicable here since $S(t)$ is a function of one argument.
I was thinking of the following:
Choosing a specific form of $f(t_1,t_2)=f(\frac{t_1+t_2}{2})$ , $f(t)=0 \text{ }\forall t<0$ and the Heaviside function $H(t)$
$\hat{S}(p) =\mathcal{L}(S(t))=\int_{0}^{\infty}S(t)e^{-pt}dt\\ \Rightarrow \hat{S}(p) =\int_0^{\infty}e^{-pt}dt\int_0^{t}\int_0^{t}f(t-s_1,t-s_2)g(s_1)h(s_2)ds_1ds_2\\ \Rightarrow \hat{S}(p) = \int_0^{\infty}e^{-pt}dt\int_0^{\infty}\int_0^{\infty}f(t-\frac{s_1+s_2}{2})H(t-\frac{s_1+s_2}{2})g(s_1)h(s_2)ds_1ds_2\\ \Rightarrow \hat{S}(p) =\int_0^{\infty}g(s_1)ds_1\int_0^{\infty}h(s_2)ds_2\int_{\frac{s_1+s_2}{2}}^{\infty}f(t-\frac{s_1+s_2}{2})e^{-pt}dt\\ \text{with a change of variable } t-\frac{s_1+s_2}{2}=\tau\\ \Rightarrow \hat{S}(p) =\int_0^{\infty}g(s_1)e^{-s_1/2}ds_1\int_0^{\infty}h(s_2)e^{-s_2/2}ds_2\int_{0}^{\infty}f(\tau)e^{-p\tau}d\tau\\ \Rightarrow \hat{S}(p) = \hat{f}(p)\hat{g}(p/2)\hat h(p/2) \qed\\ \text{Please let me know if there is a better way to solve this without taking the specific form of }f(t_1,t_2)$.