Can someone please derive and explain how the LHS is equal to the RHS using the fact that the function $f$ is symmetrical with respect to time variables $t_1$ and $t_2$. Here $t$ is some constant.
$\int_{0}^{t}dt_1 \int_{0}^{t} dt_2 f(t_2-t_1) = 2 \int_{0}^{t}dt_1 \int_{0}^{t_1} f(k) dk$
$f$ is not a function of two variables so I guess you mean $f$ is even $f(-x)=f(x)$. Change variables to $\{t_1,k\}=\{t_1,t_2-t_1\}$, Jacobian is 1.
You get $$\int_{0}^t dt_1\int_{-t}^t f(k)dk$$ since $f$ is even negative $k$ give exactly the same contribution as positive $k$ so that is the same as $$2\int_0^t dt_1\int_0^t f(k) dk$$
The function $f: t\mapsto f(t)$ as such is not "symmetrical with respect to $t_1$ and $t_2$", but is supposed to be an even function of its single variable $t$. This implies that the composed function $$g(t_1,t_2):=f(t_2-t_1)$$ of two variables is symmetric with respect to $t_1$ and $t_2$. It follows that the integral of $g$ over the square $Q:=[0,T]^2$ is twice the integral over the part below the main diagonal. We therefore have $$\int_Q f(t_2-t_1)\>{\rm d}(t_1,t_2)=\int_Q g(t_1,t_2)\>{\rm d}(t_1,t_2)=2\int_0^T\int_0^{t_1} f(t_2-t_1)\>dt_2\>dt_1\ .$$ Now $$\int_0^{t_1} f(t_2-t_1)\>dt_2=\int_0^{t_1} f(t_1-t_2)\>dt_2=\int_0^{t_1}f(t)\>dt\ .$$ One more thing: Please don't use the letter $k$ for a continuous variable.
