the transform of a periodic function
Many real forcing signals repeat: a square wave that flips on and off, a sawtooth, the rectified output of a power supply, any periodic pulse train. Computing the Laplace transform of such a signal by integrating from 0 to infinity looks daunting — but because the signal repeats, you only ever need to integrate over ONE period, then a tidy formula handles the infinite repetition for you.
Suppose f(t) repeats with period T, meaning f(t + T) = f(t) for all t. The integral from 0 to infinity breaks into a sum over successive periods, and each later period is just the first one shifted, which by the second shifting idea contributes an extra factor e^(-sT). Summing the resulting geometric series gives the periodic-function formula: L{f(t)} = (1/(1 - e^(-sT))) times the integral from 0 to T of e^(-st) f(t) dt. In words: transform just the first period, then divide by (1 - e^(-sT)) to account for all the copies. The factor 1/(1 - e^(-sT)) is exactly the sum 1 + e^(-sT) + e^(-2sT) + ... that stacks up every repetition.
This turns an intimidating infinite integral into a short, finite one over a single cycle, plus one division. It is how you feed clocked signals, switching power inputs, and repeating mechanical loads into the Laplace machinery. Combined with the step function (to describe the shape within one period), it covers essentially every periodic input an engineer meets.
For a square wave that is 1 on 0 ≤ t < 1 and 0 on 1 ≤ t < 2 with period T = 2, the integral over one period is integral from 0 to 1 of e^(-st) dt = (1 - e^(-s))/s, so L{f} = (1 - e^(-s))/(s(1 - e^(-2s))).
Transform one period, then divide by (1 - e^(-2s)) to stack the infinite repetition.
The formula needs true periodicity for ALL t ≥ 0, with the same shape each cycle. If a signal repeats only after some startup transient, split it: transform the transient separately and apply the periodic formula only to the steadily repeating tail.