The Laplace (and Fourier) Transform is a godsent

Transform of arbitrary signals

Suppose you have a time-varying signal $u(t)$, then there exists its Laplace or Fourier transform $U(s)$, where $s = \sigma + i\omega$ for the Laplace transform and $s = i\omega$ for the Fourier transform.

These transforms are defined as $$\mathcal{L}\{u(t)\}(s) = U(s) = \int_{0}^\infty u(t) e^{-st} \text{d}t,\; s \in \mathbb{C}$$ and $$\mathcal{F}\{u(t)\}(\omega) = U(\omega) = \int_{-\infty}^\infty u(t) e^{-i\omega t} \text{d}t,\; \omega \in \mathbb{R}.$$ And now you know why it's called the frequency domain: $\mathfrak{Re}(e^{-st})$ is a (dampened) oscillation! To be completely honest, this is way too big of a subject for me to explain. However, I don't have to because of legends like 3Blue1Brown, go watch these if you want to know more:

• watch this video on what Fourier transforms are
• and this one on Laplace transforms.
• Just watch his playlist on differential equations really, the man's a treasure.

But if you don't care about all that and just want to apply it, you need these key fact for the next derivation:

• The Fourier transform is very related to the Laplace transform and most things work for both so I'll just continue with the Laplace transform for now,
• These transforms basically say that any time signal is a sum of (dampened) sinusoids,
• Both transforms are linear.
• Derivatives and integrations are also linear operators.
• The order in which you apply linear operators doesn't matter.

Furthermore, some interesting mathematical fact: starting with the definition again $$\mathcal{L}\{u(t)\}(s) = U(s) = \int_{0}^\infty u(t) e^{-st} \text{d}t,\; s \in \mathbb{C}$$ and integrating by part (trust me bro) $$ \mathcal{L}\{u(t)\}(s) = \left[\frac{u(t) e^{-st}}{-s} \right]_{0}^\infty + \int_{0}^\infty\frac{1}{s} \frac{\text{d} u(t)}{\text{d}t} e^{-st} \text{d}t $$ $$ \qquad\qquad = -\frac{u(0) e^{0}}{-s} + \frac{1}{s} \mathcal{L}\left\{\frac{\text{d} u(t)}{\text{d}t}\right\}(s) $$ $$ \qquad\qquad = \frac{1}{s}u(0) + \frac{1}{s} \mathcal{L}\left\{\frac{\text{d} u(t)}{\text{d}t}\right\}(s) $$ $$ \implies U(s) = \frac{1}{s} \frac{\text{d} }{\text{d}t}U(s) + \frac{1}{s}u(0) $$ $$ \implies sU(s) = \frac{\text{d} }{\text{d}t}U(s) + u(0) $$ ... Amazing right! No but really this shows two major properties of the Laplace transform: $$ sU(s) = \frac{\text{d} }{\text{d}t}U(s) + u(0) \implies \frac{\text{d}}{\text{d}t} U(s) = s U(s) - u(0),$$ giving us the first amazing result about time derivatives. If we integrate both sides in time, we get the second amazing result about time integration $$ \int sU(s) \text{d}t = \int \frac{\text{d} }{\text{d}t}U(s) + u(0) \text{d}t \implies \int sU(s) \text{d}t = U(s) + u(0)t \implies \int U(s) \text{d}t = \frac1s U(s) + \frac1s u(0)t .$$ Then doing a pro engineering move: unless otherwise specified we'll neglect the terms $u(0)$ and $u(0)t$. We are then left with $$ \frac{\text{d}}{\text{d}t} U(s) = s U(s) \text{, and} $$ $$ \int U(s) \text{d}t = \frac{1}{s} U(s),$$ which are beautiful. As you'll see now!