Convolution integral with step function
Consider the following integral:

$$\int_{-\infty}^{\infty}\sigma(\tau)\sigma(t-\tau)f(\tau)d\tau\tag{1}$$

where \$f(t)\$ is an arbitrary function.

The integrand is zero for \$\tau&lt;0\$ (due to the factor \$\sigma(\tau)\$), and it is zero for \$\tau&gt;t\$ (due to the factor \$\sigma(t-\tau)\$).

Now it's important to note that replacing the integration limits by \$0\$ (lower limit) and \$t\$ (upper limit) is only justified for \$t&gt;0\$, because otherwise the factor \$\sigma(\tau)\sigma(t-\tau)\$ equals zero for all values of \$\tau\$, because \$\sigma(\tau)\$ and \$\sigma(t-\tau)\$ don't overlap in that case.

Consequently, the correct way to rewrite \$(1)\$ is

$$\sigma(t)\int_{0}^{t}f(\tau)d\tau\tag{2}$$

because without the factor \$\sigma(t)\$, the integral does not evaluate to zero for \$t&lt;0\$ as it should.

I don't see the relation of your second question to this one, so I suggest you formulate it as a separate question.