Time Integral Property of The Laplace Transform

Electrical Circuit Analysis > The Laplace Transform > Properties of The Laplace Transform

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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

$$\mathcal{L}\left[\int_{0}^{t} t d t\right]=\mathcal{L}\left[\frac{t^{2}}{2}\right]=\frac{1}{s} \frac{1}{s^{2}}$$

$$\mathcal{L}\left[t^{2}\right]=\frac{2}{s^{3}}$$

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

$$f^{-1}\left(0^{-}\right)=\int_{-\infty}^{0^{-}} f(t) d t$$

Do you have any questions?

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