Suppose given the input-output relation $$y(t) = \int_{-\infty}^{t} x(\tau + 3)d\tau$$
I know $$y(t) =x(t)\ast u(t)= \int_{-\infty}^{\infty} x(\tau)u(t-\tau)d\tau = \int_{-\infty}^{t} x(\tau)d\tau,$$ where $u(t)$ is a unit step function. So here $h(t) = u(t)$
My question is how to deal with $+3$?
My idea is $h(t) = u(t) \ast \delta(t+3)$ (use echo property), but looks not correct.
With the simple substitution $\tau'=\tau+3$, $$y(t) = \int_{-\infty}^{t} x(\tau + 3)\,\mathrm{d}\tau=\int_{-\infty}^{t+3} x(\tau')\,\mathrm{d}\tau'.$$ Now our integrand is simply $x(\tau')$, so $y(t)=x(t+3)*u(t+3)$, where $u$ is the unit step function.