Derivatives and Integrals
=========================

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

.. math::
   \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:

.. math::
   x'(t) \ltarrow s X(s) 

It is easy to prove (see :ref:`easytoprove`) that in general:

.. math::
   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:

.. math::
   y(t) = \int_0^{t} x(u) du 

then we have:

.. math::
   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:

.. math::
   \int_0^t x(u)du \ltarrow \frac{1}{s} X(s)