On can show that if a real function $t \mapsto f(t)$ is Laplace-transformable, as well is $\displaystyle\int_0^tf(\tau)\,d\tau$ . The latter is also continuous for $t > 0$ and by the Newton-Leibniz formula, has the derivative equal $f(t)$ . Hence we may apply the formula for Laplace transform of derivative, obtaining $$F(s) \;=\; \mathcal{L}\{f(t)\} \;=\; s\,\mathcal{L} \left\{\int_0^t\!f(\tau)\,d\tau\right\}-\int_0^0\!f(t)\,dt \;=\; s\,\mathcal{L} \left\{\int_0^t\!f(\tau)\,d\tau\right\},$$ i.e.

(1)

Application. We start from the easily derivable rule $$\frac{1}{s} \;\curvearrowright\; 1,$$ where the curved arrow points from the Laplace-transformed function to the original function. The formula (1) thus yields successively $$\frac{1}{s^2} \;\curvearrowright\; \int_0^t\!1\,d\tau \;=\; t,$$ $$\frac{1}{s^3} \;\curvearrowright\; \int_0^t\!\tau\,d\tau \;=\; \frac{t^2}{2!},$$ $$\frac{1}{s^4} \;\curvearrowright\; \int_0^t\!\frac{\tau^2}{2!}\,d\tau \;=\; \frac{t^3}{3!},$$ etc. Generally, one has

(2)