the transform of an integral
If differentiating a function corresponds to multiplying its transform by s, you might guess that integrating — the opposite operation — corresponds to dividing by s. That guess is exactly right, and it gives the partner rule to the derivative rule: a clean way to transform an accumulated quantity.
The rule is L{ integral from 0 to t of f(tau) d tau } = F(s) / s. In words: if g(t) is the running integral of f (the area under f from 0 up to the current time t), then the transform of g is simply the transform of f divided by s. Where differentiation pulled a factor of s up front, integration pushes a factor of s down into the denominator — the mirror image. There is no leftover boundary term here, because the running integral starts at 0 and so has value 0 at t = 0; the initial condition is automatically zero.
This rule is handy in two directions. Forward, it transforms equations or systems that involve an integral of the unknown — for instance an integro-differential equation from a circuit with a capacitor, where charge is the integral of current. Backward, it gives an inversion trick: if you can invert F(s) to get f(t), then F(s)/s inverts to the integral of f, so an extra 1/s factor in the s-domain means 'integrate the answer.' That turns some awkward inversions into a familiar integral.
Invert 1/(s(s^2 + 1)). Reading it as [1/(s^2 + 1)] / s, and knowing 1/(s^2 + 1) inverts to sin(t), the extra 1/s means integrate: integral from 0 to t of sin(tau) d tau = 1 - cos(t). So the inverse is 1 - cos(t).
An extra 1/s in F(s) means 'integrate the answer from 0 to t.'
The rule is for the integral that starts at 0. An integral with a different lower limit picks up a constant, which shifts the result — keep the lower limit at 0 (or absorb the constant) for the clean F(s)/s form.