# Why does the unit step function not have infinite poles?

Let's say I have a transfer function which is a unit step function.

$$H(s) = \int_{0}^{∞}e^{-st}dt$$

But when we write,

$$H(s) = 1/s$$ it is only true when $$Re(s) > 0$$

So after we derive the Laplace Transform for the unit step, we forget all about how we derived it and how $$H(s) = \infty$$ for all $$Re(s) \leq 0$$

What makes the pole $$s=0$$ so special. It does not lie in the ROC like every other point in the left half plane, so why is it the only pole for our system?

Shouldn't the system have infinite poles then on the left half of the s-plane? I guess the more generalized question would be isn't every point outside the ROC of a transfer function its pole?

EDIT:

If you can accept that the above proposition is TRUE then consider the following contradiction.

Consider a simple RC filter whose transfer function is

$$H(s) = \frac{1}{s+1}$$

This system is said to have 1 pole at $$s=-1$$

Applying the inverse Laplace transform, we now the corresponding time domain function

$$h(t) = \exp(-t)$$

But we know that Laplace Transforms for exponential such as $$\exp(at)$$ is only defined when $$s>a$$

In our case the Laplace transform could only be defined for $$s>-1$$. Therefore, to say that there is only one pole at $$S=-1$$ would be incorrect as everything to the left of $$s=-1$$, including the line at $$Re(s)=1$$ would also be undefined and out of the ROC.

So the question again becomes why do we, when solving such systems only consider them to have ONLY one pole at $$s=-a$$?

Is that not an incomplete answer?