My Laplace Transform vs. My Desmos laplace Transform

Question

If I were to calculate the Laplace Transform of 1 I would do the following: \begin{equation}\mathcal{L}\{1\}=\int_{0}^{\infty}e^{-st}dt \end{equation} From here I would go on evaluate the improper integral:\begin{align}&=-\frac{1}s[e^{-st}]\Bigg\vert_0^\infty \\&=\frac{1}s[e^{-s\cdot0}-e^{-\infty \cdot s}]\end{align} So the end result would be that I would get $\mathcal{L}\{1\}=\frac{1}s$. I plotted this on desmos. I used a really big upper limit for the approximation and saw it matches the positive x axis part of only of $\frac{1}x$. Why is that? Does this have heavy implications differential equations and/or calculus if so please let me know?

Answer 1

Your evaluation of this improper integral is valid only for $s>0$. For $s \le 0$ the integral diverges to $\infty$. By replacing $\infty$ by some large number $R$, the integral becomes $$ \frac{1}{s} \left(1 - e^{-Rs}\right)$$ When $s< 0$ the $e^{-Rs}$ is a very large number.