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I don't have that book, but the statement you cite (emphasis mine):

> ... **when s=jω** the integral on the RHS of eq. 2.48 does not converge for unstable systems

doesn't mean the transfer function for an unstable system doesn't exist.

The book makes a lot of confusion between Transfer Function (TF) H(s), which is the Laplace transform of the impulse response h(t), 
and Fourier Transform (FT) of h(t).


Let's take a simple unstable system, i.e. one that has a transfer function with a pole $p = 2$ in the RHP:

\[
 H(s) = \frac {1} {s-2}
\]

This corresponds to an impulse response:

\[
h(t) = e^{2t}u(t)
\]

The proof is simple:

\[
\begin{align}
    H(s) 
    = \int_{-\infty}^{+\infty}h(t)e^{-st}dt 
    = \int_{-\infty}^{+\infty}e^{2t}u(t)e^{-st}dt 
    = \\[2em]
    = \int_{0}^{+\infty}e^{-(s-2)t}dt
    = - \frac {1}{s-2} \biggl[e^{-(s-2)t}\biggr]_{0}^{+\infty}
    = \frac {1}{s-2} \biggl[e^{-(s-2)t}\biggr]_{+\infty}^{0} 
    = \\[2em]
    = \frac {1}{s-2} 
      \biggl[e^{-(s-2)\cdot 0} - e^{-(s-2)\cdot (+\infty)}\biggr]
    = \frac {1}{s-2} 
      \biggl[1 - e^{-(x + jy -2)\cdot (+\infty)}\biggr]
    = \\[2em]
    = \frac {1}{s-2} 
      \biggl[1 - 
          \underbrace{ e^{-(x-2)\cdot (+\infty)} }
             _{\substack{ 
               \text{converges} \\[4pt]
               \text{to 0 iff} \\[6pt]
               x - 2 > 0
              }}
          ~~\cdot
          \overbrace{ e^{-jy \cdot (+\infty)} }
             ^{\substack{ 
               \text{complex number} \\
               \text{with} \\
               \text{bounded} \\
               \text{magnitude} \\
              }}
      \biggr]
    = \frac {1}{s-2} 
      \biggl[1 - \underbrace{0}_{\text{iff x>2}} ~\biggr]
    = \\[2em]
    = \frac {1}{s-2} \quad \forall s=x+jy ~~\text{and}~~ x>2

\end{align}
\]

In other words, the integral that defines H(s) converges for every point in the complex plane to the right of the pole. That is its region of convergence.

This also means that the imaginary axis lies outside that convergence region, hence the Fourier transform of h(t) doesn't exist, i.e. ***the defining integral of the Fourier transform does not converge***.

So why does the Nyquist criterion for stability still works? Because that criterion doesn't depend on the Fourier transform. Simply substituting s with jω doesn't mean that you get a Fourier transform. 

In other words, **H(jω) is the Fourier transform of h(t) *only if the system is stable*** and H(s) region of convergence includes the imaginary axis. If this doesn't happen, H(jω) is still a valid complex function of a real variable.

In fact if you read [Wikipedia about Nyquist stability criterion](https://en.wikipedia.org/wiki/Nyquist_stability_criterion), you'll see that nowhere the Fourier transform is mentioned.

That's true even for other expressions which remain valid, like Y(s) and X(s) and the fact that H(s)=Y(s)/X(s) even for unstable systems.

The problem is that putting s=jω when you deal with an unstable system is misleading, since it could make you believe you are dealing with Fourier transforms, which is not the case.

The fact that "everything is OK" is a consequence that H(s) still exists and substituting jω inside the expressions before rather than after obtaining H(s) just doesn't change the end result when you apply Nyquist.

The real point here is that you can't interpret that ω as a real frequency existing in the system. You can't measure H(jω) for an unstable system.