Laplace Transform: zeros and corresponding impulse response $h(t)$ Poles and the impulse response
If our impulse response is in the form :
$$h(t) = e^{-\sigma_0 t}\cos(\omega_0 t) \, u(t)$$
(where $u(t)$ is the unit step function)
And its Laplace transform is :
$$H(s) = \frac{N(s)}{D(s)} = \int_{0}^{+\infty} h(t)e^{-st}dt$$
$$s = \sigma + j\omega$$
Poles are values of $s$ so that $$D(s) = 0 \rightarrow H(s) = +\infty $$
But to understand this, I prefer to look at the integral : it will go to infinity (poles) when $s$ reflects components of $h(t)$. In a way, $e^{-st}$ &quot;probes&quot; $h(t)$. Indeed :

A single real pole ($s = -\sigma_0$) means $h(t) = e^{-\sigma_0t}u(t)$ because : $$\int_{0}^{+\infty} e^{-\sigma_0t}e^{-(-\sigma_0)t}dt = \int_{0}^{+\infty} 1dt = +\infty $$.

Complex conjugate poles ($s = -\sigma_0 \pm j\omega_0$) mean $h(t)$ is an exponentially decaying sinusoid (say $h(t) = e^{-\sigma_0t}\cos(\omega_0t)$) because : $$\int_{0}^{+\infty} e^{-\sigma_0t}\cos(\omega_0t)e^{-(-\sigma_0)t}e^{-j\omega t}dt = \int_{0}^{+\infty}\cos(\omega_0t)e^{-j\omega t}dt $$ which is infinite at $\omega = \pm\omega_0$ (Fourier transform of $h(t)$ without its exponential component, which is a sinusoid).

Complex conjugate poles with $\sigma = 0$ ($s = \pm j\omega_0$) mean $h(t)$ has no decaying component (say $h(t) = \cos(\omega_0t) u(t)$) because : $$\int_{0}^{+\infty} \cos(\omega_0t)e^{-j\omega t}dt$$ which is infinite at $\omega = \pm\omega_0$ (Fourier transform of $h(t)$ which is a sinusoid).


Zeros : a dirac in the impulse response ?
Now, let's look at $H(s)$ for a Notch filter, as shown in ch.32,p.17 of &quot;The Scientist and Engineer's Guide to DSP&quot; and see if similar reasoning on the integrals can be done.

Let's use the following filter (figure above for illustration only, I use different poles and zeros here) :
$$H(s) = \frac{s^2+1}{(s-(-1+i))(s-(-1-i))}$$
This filter has 2 poles and 2 zeros :

Zeros : $z_1,z_2 =\pm i$
Poles : $p_1,p_2 =-1 \pm i$

Let's find $h(t)$ and see why the integral would indeed go to 0 or $+\infty$ for these values of zeros and poles, respectively.
If it makes sense, this tool gives the following inverse Laplace transform for $H(s)$ :
$$h(t) = \delta(t) - 2e^{-t}\cos(t) u(t) + e^{-t}\sin(t) u(t)$$

Poles : for $s=p_1$ or $p_2$ in the Laplace transform, the exponentials of h(t) get canceled and remain the Fourier transform of some sinusoid which is indeed infinite at $\omega = \pm 1$ (I'm not discussing the $\delta(t)$ but I suppose it won't change this result).

Zeros : for $s=z_1$ or $z_2$ in the Laplace transform, the result is 0 if real part and imaginary of the Laplace transform are 0. Real part is :


$$\int_{0}^{+\infty} (\delta(t) - 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt$$
$$=\int_{0}^{+\infty} \delta(t)\cos(t)dt +  \int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt$$
with
$$\int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt = -1$$
Imaginary part is :
$$\int_{0}^{+\infty} \delta(t)\sin(t)dt +  \int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\sin(t)dt$$
with
$$\int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\sin(t)dt = 0$$
Questions

If the inverse Laplace transform is correct, how to handle $\int_{0}^{+\infty} \delta(t)\cos(t)dt$ and $\int_{0}^{+\infty} \delta(t)\sin(t)dt$ to show that $H(s)$ is indeed 0 at $z_1$ and $z_2$ ?
If all of this is correct, what does it (physically) mean for an impulse response to have a dirac in its expression ? I thought impulse response of most physical systems was only a combination of decaying exponentials and sinusoids ?
For you first question you can use the following
$$
\int_{-\infty}^{\infty} \delta (t-a)\,f(t)\,dt = f(a),
$$
with $f(t)$ any function. In your case those integrals would thus yield the values one and zero respectively.
For your second question I will only considering linear time invariant systems. In that case the impulse response of such a system can only contains a Dirac delta function if the transfer function of that system has a numerator of the same order as the denominator. Namely, any transfer function of the form
$$
G(s) = \frac{b_n\,s^n + b_{n-1}\,s^{n-1} + \cdots + b_1\,s + b_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0},
$$
with $b_n \neq 0$ can also be written as
$$
G(s) = b_n + \frac{b'_{n-1}\,s^{n-1} + \cdots + b'_1\,s + b'_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0},
$$
with $b'_k = b_k - b_n\,a_k$. The inverse Laplace transform of the constant $b_n$ would contribute a Dirac delta term. For the remaining part of the transfer function one could use partial fraction expansion to show that it can't contribute a Dirac delta term.
If a physical system would have a numerator of the same order as the denominator then it would require that the output of the system is directly affected by the input. An example of such physical system might be some electrical motor where you input a voltage and measure the angular position with some voltage leakage from the input signal to the output. However, most physical systems have numerator of a lower order as the denominator. It is more likely that you might encounter equal order numerators and denominators in digital filters (though, those would be z-domain and not s-domain, but roughly the same argument holds) such as notch filters. Those filters are however often used in series with physical systems, so their combined transfer function would also have a lower order numerator.