An identity derived from the Laplace transform

Question

It seems that $$\int_0^t \int_0^l f(\tau) ~d \tau ~d l = \int_0^t z f(t-z) dz $$ since the Laplace transform of both sides is $F(s)/s^2$, where $F(s)$ is the Laplace transform of $f(t)$: the left-hand side is integration corresponds to multiplication by $1/s$ in the Laplace domain while the right-hand side because convolution in the time domain is multiplication in the Laplace domain.

Is it possible to prove this identity directly without appealing to the Laplace transform?

Answer 1

Hint: interchange order of integrations on the left, then a change of variable.