# Time Integral Property of The Laplace Transform

If $F(s)$ is the Laplace transform of $f(t)$, the Laplace transform of its integral is
$$\mathcal{L}\left[\int_{0}^{t} f(t) d t\right]=\int_{0^{-}}^{\infty}\left[\int_{0}^{t} f(x) d x\right] e^{-s t} d t$$
To integrate this by parts, we let
$$u=\int_{0}^{t} f(x) d x, \quad d u=f(t) d t$$
and
$$d v=e^{-s t} d t, \quad v=-\frac{1}{s} e^{-s t}$$
Then
\begin{aligned}\mathcal{L}\left[\int_{0}^{t} f(t) d t\right]=& {\left.\left[\int_{0}^{t} f(x) d x\right]\left(-\frac{1}{s} e^{-s t}\right)\right|_{0^{-}} ^{\infty} } \\&-\int_{0^{-}}^{\infty}\left(-\frac{1}{s}\right) e^{-s t} f(t) d t\end{aligned}
For the first term on the right-hand side of the equation, evaluating the term at $t=\infty$ yields zero due to $e^{-s \infty}$ and evaluating it at $t=0$ gives $\frac{1}{s} \int_{0}^{0} f(x) d x=0$. Thus, the first term is zero, and
$$\mathcal{L}\left[\int_{0}^{t} f(t) d t\right]=\frac{1}{s} \int_{0^{-}}^{\infty} f(t) e^{-s t} d t=\frac{1}{s} F(s)$$
or simply,
$$\mathcal{L}\left[\int_{0}^{t} f(t) d t\right]=\frac{1}{s} F(s) \tag{1}$$
As an example, if we let $f(t)=u(t)$, from Example 1, $F(s)=$ 1/s. Using Eq. (1),
$$\mathcal{L}\left[\int_{0}^{t} f(t) d t\right]=\mathcal{L}[t]=\frac{1}{s}\left(\frac{1}{s}\right)$$
Thus, the Laplace transform of the ramp function is
$$\mathcal{L}[t]=\frac{1}{s^{2}}$$
Applying Eq. (1), this gives
$$\mathcal{L}\left[\int_{0}^{t} t d t\right]=\mathcal{L}\left[\frac{t^{2}}{2}\right]=\frac{1}{s} \frac{1}{s^{2}}$$
or
$$\mathcal{L}\left[t^{2}\right]=\frac{2}{s^{3}}$$
Repeated applications of Eq. (1) lead to
$$\mathcal{L}\left[t^{n}\right]=\frac{n !}{s^{n+1}}$$
Similarly, using integration by parts, we can show that
$$\mathcal{L}\left[\int_{-\infty}^{t} f(t) d t\right]=\frac{1}{s} F(s)+\frac{1}{s} f^{-1}\left(0^{-}\right)$$
where
$$f^{-1}\left(0^{-}\right)=\int_{-\infty}^{0^{-}} f(t) d t$$