I'm currently studying the textbook An Introduction to Laplace Transforms and Fourier Series, second edition, by Phil Dyke. Chapter 2.2 Derivative Property of the Laplace Transform says the following:
The function $$Si(t) = \int_0^t \dfrac{\sin(u)}{u} \ du$$ defines the Sine Integral function which occurs in the study of optics. The formula for its Laplace Transform can now be easily derived as follows.
Let $F(t) = \sin(t)$ in the result $$\mathcal{L} \left( \dfrac{F(t)}{t} \right) = \int_s^\infty f(u) \ du$$ to give $$\begin{align*} \mathcal{L} \left( \dfrac{\sin(t)}{t} \right) &= \int_s^\infty \dfrac{du}{1 + u^2} \\ &= \left[ \tan^{-1}(u) \right]_s^\infty \\ &= \dfrac{\pi}{2} - \tan^{-1}(s) \\ &= \tan^{-1} \left( \dfrac{1}{s} \right). \end{align*}$$
So, in the last part, we have that $\dfrac{\pi}{2} - \tan^{-1}(s) = \tan^{-1} \left( \dfrac{1}{s} \right) \Rightarrow \tan^{-1} \left( \dfrac{1}{s} \right) + \tan^{-1}(s) = \dfrac{\pi}{2}$. Researching this identity, I found it said that
$$\tan^{-1}(x) + \tan^{-1} \left( \dfrac{1}{x} \right) = \begin{cases} \dfrac{\pi}{2}, & x > 0 \\ -\dfrac{\pi}{2}, & x < 0 \end{cases}$$
And we know that, in this context of the Laplace transform, $s$ is a complex number. But I don't think the general Laplace transform requires that $s > 0$, right? But, in the above case, since we have that $\tan^{-1} \left( \dfrac{1}{s} \right) + \tan^{-1}(s) = \dfrac{\pi}{2}$, does this not imply that $s > 0$?