If Laplace transform is expressed as :

$$\int_{-\infty}^{+\infty} h(t)e^{-st}dt $$

with :

$$s = \sigma + j\omega$$

and $h(t)$ an impulse response expressed as :

$$h(t) = Ae^{-\sigma_0t}\cos(\omega_0t+\phi) = e^{-\sigma_0t}\cos(\omega_0t)$$ ($A=1$ and $\phi = 0$ for simplification, $h(t)=0$ if $t<0$)

  

Then, each vertical line (parallel to the imaginary axis) in the $s$ plane corresponds to the Fourier transform of $f(t) = h(t)e^{-\sigma t}$ for a fixed $\sigma$.

For $\sigma = -\sigma_0$, the decaying exponential of $h(t)$ is canceled and we get the Fourier transform of $h(t) = \cos(\omega_0t)$, that is : diracs at $\omega_0$ and $-\omega_0$, hence two poles : $-\sigma_0 + j\omega_0$ and $-\sigma_0 - j\omega_0$ as in the following picture (illustration only, poles not located correctly) :

[![Poles][1]][1]

Indeed, we can understand that :

*Please, note that the following is not accurate : since $h(t) = 0$ if $t<0$, we should use the unilateral Laplace transform, not bilateral ! So here we would get the unilateral Fourier transform of a sinusoid, not the bilateral one (with diracs only) ! To see what this would be, please see the [link][2] given at the end of the accepted answer*

$$\int_{-\infty}^{+\infty} h(t)e^{-j\omega t}dt $$
$$= \int_{-\infty}^{+\infty} \cos(\omega_0t)e^{-j\omega t}dt$$
$$= \int_{-\infty}^{+\infty} \frac{e^{j\omega_0t}-e^{-j\omega_0t}}{2}e^{-j\omega t}dt$$
$$= \frac{1}{2}\int_{-\infty}^{+\infty} e^{j(\omega_0-\omega)t}-e^{-j(\omega_0+\omega)t}dt$$

If $\omega = \omega_0$ or $-\omega_0$, then the integral would blow up due to the $$\int_{-\infty}^{+\infty} e^0dt $$ member, hence the poles in the s plane.

So as shown in ch.32, p.24 of [The Scientist and Engineer's Guide to DSP][3] (see figures below), with Laplace transform we multiply $h(t)$ with $e^{-st}$ = $e^{-\sigma}e^{-j\omega}$, that is we multiply $h(t)$ with sinusoids that are either :

 - (a) Exponentially decaying ($\sigma$ > 0)
 - (b) Stable ($\sigma = 0$)
 - (c) Exponentially growing slower than our impulse response decay ($ -\sigma_0 < \sigma < 0$)
 - (d) Exponentially growing, compensating our impulse response decay ($\sigma = -\sigma_0$) : OK, as studied above.
 - (e) Exponentially growing quicker ($\sigma < - \sigma_0$ and $\sigma < 0$)

(letters correspond to pairs of points in the s plane shown in figures below, always at a fixed $\omega$ or $-\omega$ value)

[![Different values of s...][4]][4]
[![Lead to different values of the integral][5]][5]

I understand case d : since we cancel the exponential part, we get only the *(unilateral !!)* Fourier transform of a sinusoid. That is : infinite at $\omega_0$ and $-\omega_0$ hence the poles (though I don't know why we have a continuous function of omega with infinite values at $\omega_0$ and $-\omega_0$ instead of diracs as in the original Fourier transform of a sinusoid). 

Case a, c and e are intuitive. In case a, we multiply $h(t)$ with a decaying exponential. The integral will be some finite complex value (for all values of $\sigma > 0$. In case c, we multiply by an exponential growing slower than the decaying exponential of $h(t)$, hence some finite complex value for the integral (for all values of $-\sigma_0 < \sigma < 0$). In case e, we multiply the $h(t)$ by an exponential that grows quicker than exponential of $h(t)$ decays : hence integral does not converge (for all values of $\sigma < -\sigma_0$).

**But for case b, I can't get the intuition of why the integral would be zero as shown with the area under the curve (red in the above figures) ? In other words, I understand the vertical line in the s plane at $\sigma = -\sigma_0$, it is the Fourier transform of $h(t)e^{-\sigma_0 t}$ so it is Fourier transform of $h(t)$ once its exponential component is removed, hence 2 poles due to sinusoid. We get poles whenever $e^{-st}$ is identical (compensates) to the impulse response. But what would cause Fourier transform of $h(t)e^{-\sigma t}$ to be 0 at some $\omega$ ? For which $h(t)$ and how it would impact the area under the curve (integral) ?** 


  [1]: https://i.sstatic.net/iprcE.gif (maximintergrated.com)
  [2]: http://www.thefouriertransform.com/pairs/rightSidedSinusoids.php
  [3]: http://www.dspguide.com/CH32.PDF
  [4]: https://i.sstatic.net/yL74t.png
  [5]: https://i.sstatic.net/kL5ke.png