4.1.4. Derivatives and Integrals

Assume the Laplace transform of \(x(t)\) is \(X(s)\). Then the Laplace transform of the derivative \(x'(t)\) equals:

\[\int_{0}^{\infty} x'(t) e^{-st} dt = \left[ x(t)e^{-st} \right]_{0}^{\infty} + s \int_{-\infty}^{\infty} x(t) e^{-st} dt = -x(0) + s X(s)\]

Assuming \(x(0)=0\) we get the transform pair:

\[x'(t) \ltarrow s X(s)\]

It is easy to prove (see It’s so Easy) that in general:

\[x^{(n)}(t) \ltarrow s^n X(s)\]

where \(x^{(n)}(t)\) is the \(n\)-th order derivative of \(x(t)\). In this case it must be true that \(x(0)=x'(0)=\cdots=x^{(n-1)}(0)=0\).

We define the signal \(y(t)\) as:

\[y(t) = \int_0^{t} x(u) du\]

then we have:

\[Y(s) = \int_0^{\infty} y(t) e^{-st} dt = \int_0^\infty \left(\int_0^t x(u) du \right) e^{-st} dt = \frac{1}{s} X(s)\]

In general we have:

\[\int_0^t x(u)du \ltarrow \frac{1}{s} X(s)\]