Finding phase from group delay
Note that if $F(x)$ is an antiderivative of $f(x)$, i.e., if $F'(x)=f(x)$, then from the fundamental theorem of calculus we have

$$\int_a^bf(x)dx=F(b)-F(a)\tag{1}$$

Since the phase $\phi(\omega)$ is an antiderivative of the negative group delay $-\tau(\omega)$ we get from $(1)$

$$-\int_a^b\tau(\Omega)d\Omega=\phi(b)-\phi(a)\tag{2}$$

With $b=\omega$ and $a=0$ we get from $(2)$

$$-\int_0^\omega\tau(\Omega)d\Omega=\phi(\omega)-\phi(0)\tag{3}$$

By adding $\phi(0)$ on both sides of $(3)$ we obtain the given formula:

$$\phi(\omega)=-\int_0^\omega\tau(\Omega)d\Omega+\phi(0)\tag{4}$$