Finding phase from group delay I know that group delay is defined as :

$$\begin{align} \tau(\omega) = grd [H(e^{j\omega})] &amp;=-\frac{d[\arg[H(e^{j\omega})]]}{d\omega}\end{align}$$

But if what is available is group delay itself, is it possible to recover the phase?

According to Discrete-Time Signal Processing 3rd edition by Alan Oppenheim equation 5.86 the expression would look like this:

$$\begin{align} \arg[H(e^{j\omega})] = -\int\limits_{0}^{\omega} \end{align} grd [ H(e^{j\phi})]\, d \phi + \arg[H(e^{j0})] $$

and my doubt is:

Where did the second term $$\arg[H(e^{j0}) ]$$ came from?
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:

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