The Laplace of 1 is:(adsbygoogle = window.adsbygoogle || []).push({});

$$\int_{0}^{\infty} 1 \exp(-st) dt = \left[ \frac{\exp(-st)}{-s} \right]_{0}^{\infty} = \frac{\exp(-s \infty) - \exp(-s 0)}{-s} = \frac{0 - 1}{-s} = \frac{1}{s}$$

It's result known, however, for this be true is assumed that s>0, because 0 = exp(-∞) = exp(-s∞). But we have a problem, s is a complex number (σ + iω), so you assume that s>0 thus you are saying that ω=0, but in laplace transform ω≠0. The most correct possible would be ##\exp(-s∞) = \exp(-(σ + iω)∞) = \exp(-σ∞ - iω∞) = \exp(-\text{sgn}(σ)∞ - i \text{sgn}(ω)∞) = \frac{\exp(-\text{sgn}(σ)∞)}{\exp(i \text{sgn}(ω)∞)}##, but it's become inpraticable... So what is correct form of approach this calculation of a simple way?

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# Conditions of transformations

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