Laplace transform : integral vs poles & zeros
The definition of the Laplace transform you're using is called the bilateral Laplace transform, which is less common than the unilateral Laplace transform. The difference between the two is that the first has a lower integration limit of $-\infty$ whereas the second has a lower limit of $0$. This difference becomes irrelevant if the signal under consideration is zero for $t&lt;0$. This is the case in the example of the book. Note, however, that the bilateral Laplace transform of the impulse response $h(t)$ that you defined in your question doesn't exist for any value of $s$. It does exist if you set $h(t)$ to zero for $t&lt;0$ (i.e., multiply it with a unit step $u(t)$).
The figures in the book refer to a causal notch filter. Note that the region of convergence (ROC) of the Laplace transform of that impulse response is to the right of the poles. Consequently, the Laplace transform evaluated at any fixed value of $s$ to the right of the poles will be finite, simply because we're inside the ROC, i.e., the integral converges. If we choose $s$ exactly at the zero of the filter then we evaluate the filter's response at the notch frequency, which must be zero, simply by the definition of &quot;notch frequency&quot;. The filter's response to a signal at that frequency must be zero. Finally, if we choose the value of $s$ exactly at or to the left of the filter's poles, we are outside the ROC, in which case the integral doesn't converge.