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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 -> Because we use unilateral Laplace hence Fourier, see end of accepted answer !->).

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 -> Because we use unilateral Laplace hence Fourier, see end of accepted answer !).

(*)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 given at the end of the accepted answer

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).

(*)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 (with diracs only) one ! To see what this would be, please see the link given at the end of the accepted answer

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 -> Because we use unilateral Laplace hence Fourier, see end of accepted answer !->).

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) :

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 given at the end of the accepted answer

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 $-\omega_0$ (not accurate, see (*) just below), 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) :

(*)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 given at the end of the accepted answer