Convolving two unit step functions yields
$$u(t) \ast u(t) = \int_{-\infty}^{\infty} u(\tau)u(t-\tau) \ \text{d}\tau = tu(t)$$
This result is identical to $\int_{-\infty}^t u(\tau) \ \text{d}\tau= tu(t)$ which suggests that convolving a function with a unit step function is the same as integrating the function.
I have not been able to find any proofs or mentions of this, however.
Can someone confirm/disprove that convolving with a unit step function is the same as integrating?
If $f:\mathbb{R} \to \mathbb{R}$ and the integral $\int_{-\infty}^t f(\tau)d\tau$ exists for any $t \in \mathbb{R}$, then you obviously have
$(u*f)(t) = \int_{-\infty}^{\infty}f(\tau)u(t-\tau)d\tau = \int_{-\infty}^tf(\tau)d\tau. $
Is this what you mean?