Differentiation
The Laplace transform of the derivative of a signal will be used widely. Consider
$$\mathfrak{L}\left\{\frac{d}{dt},x,\left(t\right)\right\}={\int}_{{0}^{}}^{\infty}{x}^{\text{'}}\left(t\right){e}^{st}dt$$
this can be integrated by parts:
$$\begin{array}{cc}u={e}^{st}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & {v}^{\text{'}}={x}^{\text{'}}\left(t\right)\hfill \\ {u}^{\text{'}}=s{e}^{st}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\hfill & v=x\left(t\right)\hfill \end{array}$$
which gives
$$\begin{array}{ccc}\hfill \mathfrak{L}\left\{\frac{d}{dt},x,\left(t\right)\right\}& =& {\left(u,v\right}_{{0}^{}}^{\infty}{\int}_{{0}^{}}^{\infty}{u}^{\text{'}}vdt\hfill \\ \hfill & =& {\left({e}^{st},x,\left(t\right)\right}_{{0}^{}}^{\infty}+{\int}_{{0}^{}}^{\infty}sx\left(t\right){e}^{st}dt\hfill \\ \hfill & =& x\left({0}^{}\right)+sX\left(s\right)\hfill \\ \hfill \end{array}$$
therefore we have,
$$\frac{d}{dt}x\left(t\right)\leftrightarrow sX\left(s\right)x\left({0}^{}\right)$$
Higher order derivatives
The previous derivation can be extended to higher order derivatives. Consider
$$y\left(t\right)=\frac{dx\left(t\right)}{dt}\leftrightarrow sX\left(s\right)x\left({0}^{}\right)$$
it follows that
$$\frac{dy\left(t\right)}{dt}\leftrightarrow sY\left(s\right)y\left({0}^{}\right)$$
which leads to
$$\frac{{d}^{2}}{d{t}^{2}}x\left(t\right)\leftrightarrow {s}^{2}X\left(s\right)sx\left({0}^{}\right)\frac{dx\left({0}^{}\right)}{dt}$$
This process can be iterated to get the Laplace transform of arbitrary higher order derivatives, giving
$$\begin{array}{ccc}\hfill \frac{{d}^{n}x\left(t\right)}{d{t}^{n}}& \leftrightarrow & {s}^{n}X\left(s\right){s}^{n1}x\left({0}^{}\right)\sum _{k=2}^{n}{s}^{nk}\frac{{d}^{k1}x\left({0}^{}\right)}{d{t}^{k1}}\hfill \end{array}$$
where it should be understood that
$$\frac{{d}^{m}x\left({0}^{}\right)}{d{t}^{m}}\equiv {\left(\frac{{d}^{m}x\left(t\right)}{d{t}^{m}}\right}_{t={0}^{}},m=1,...,n1$$
Integration
Let
$$g\left(t\right)={\int}_{{0}^{}}^{t}x\left(\tau \right)d\tau $$
it follows that
$$\frac{dg\left(t\right)}{dt}=x\left(t\right)$$
and
$g\left({0}^{}\right)=0$ . Moreover, we have
$$\begin{array}{ccc}\hfill X\left(s\right)& =& \mathfrak{L}\left\{\frac{dg\left(t\right)}{dt}\right\}\hfill \\ \hfill & =& sG\left(s\right)g\left({0}^{}\right)\hfill \\ \hfill & =& sG\left(s\right)\hfill \end{array}$$
therefore
$$G\left(s\right)=\frac{X\left(s\right)}{s}$$
but since
$$G\left(s\right)=\mathfrak{L}\left\{{\int}_{{0}^{}}^{t},x,\left(\tau \right),d,\tau \right\}$$
we have
$${\int}_{{0}^{}}^{t}x\left(\tau \right)d\tau \leftrightarrow \frac{X\left(s\right)}{s}$$
Now suppose
$x\left(t\right)$ has a nonzero integral over negative values of
$t$ . We have
$${\int}_{\infty}^{t}x\left(\tau \right)d\tau ={\int}_{\infty}^{{0}^{}}x\left(\tau \right)d\tau +{\int}_{{0}^{}}^{t}x\left(\tau \right)d\tau $$
The quantity
${\int}_{\infty}^{{0}^{}}x\left(\tau \right)d\tau $ is a constant for positive values of
$t$ , and can be expressed as
$$u\left(t\right){\int}_{\infty}^{{0}^{}}x\left(\tau \right)d\tau $$
it follows that
$${\int}_{\infty}^{t}x\left(\tau \right)d\tau \leftrightarrow \frac{{\int}_{\infty}^{{0}^{}}x\left(\tau \right)d\tau}{s}+\frac{X\left(s\right)}{s}$$
where we have used the fact that
$u\left(t\right)\leftrightarrow \frac{1}{s}.$
The initial value theorem
The initial value theorem makes it possible to determine
$x\left(t\right)$ at
$t={0}^{+}$ from
$X\left(s\right)$ . From the derivative property of the Laplace transform, we can write
$$\mathfrak{L}\left\{\frac{dx\left(t\right)}{dt}\right\}=sX\left(s\right)x\left({0}^{}\right)$$
Taking the limit
$s\to \infty $
$$\begin{array}{cc}\hfill \underset{s\to \infty}{lim}{\int}_{{0}^{}}^{\infty}\frac{dx\left(t\right)}{dt}{e}^{st}dt& =\underset{s\to \infty}{lim}\left[s,X,\left(s\right),,x,\left({0}^{}\right)\right]\hfill \\ \hfill {\int}_{{0}^{}}^{\infty}\underset{s\to \infty}{lim}\frac{dx\left(t\right)}{dt}{e}^{st}dt& =\underset{s\to \infty}{lim}\left[s,X,\left(s\right),,x,\left({0}^{}\right)\right]\hfill \end{array}$$
There are two cases, the first is when
$x\left(t\right)$ is continuous at
$t=0$ . In this case it is clear that
$\frac{dx\left(t\right)}{dt}{e}^{st}\to 0$ as
$s\to \infty $ , so
[link] can be written as
$$0=\underset{s\to \infty}{lim}\left[s,X,\left(s\right),,x,\left({0}^{}\right)\right]$$
Since
$x\left(t\right)$ is continuous at
$t=0$ ,
$x\left({0}^{}\right)=x\left({0}^{+}\right)$ , the Initial Value Theorem follows,
$$x\left({0}^{+}\right)=\underset{s\to \infty}{lim}sX\left(s\right)$$
The second case is when
$x\left(t\right)$ is discontinuous at
$t=0$ . In this case, we use the fact that
$${\left(\frac{dx\left(t\right)}{dt}\right}_{t=0}=\left[x,\left({0}^{+}\right),,x,\left({0}^{}\right)\right]\delta \left(t\right)$$
For example, if we integrate the righthand side of
[link] with
$x\left({0}^{}\right)=0$ and
$x\left({0}^{+}\right)=1$ , we get the unit step function,
$u\left(t\right)$ . Proceeding as before, we have
$$\underset{s\to \infty}{lim}{\int}_{{0}^{}}^{\infty}\frac{dx\left(t\right)}{dt}{e}^{st}dt=\underset{s\to \infty}{lim}\left[s,X,\left(s\right),,x,\left({0}^{}\right)\right]$$
The lefthand side of
[link] can be written as
$$\underset{s\to \infty}{lim}{\int}_{{0}^{+}}^{{0}^{}}\left[x,\left({0}^{+}\right),,x,\left({0}^{}\right)\right]\delta \left(t\right){e}^{st}dt+\underset{s\to \infty}{lim}{\int}_{{0}^{+}}^{\infty}\frac{dx\left(t\right)}{dt}{e}^{st}dt$$
From the sifting property of the unit impulse, the first term in
[link] is
$$\left[x,\left({0}^{+}\right),,x,\left({0}^{}\right)\right]$$
while the second term is zero since in the limit, the real part of
$s$ goes to infinity. Substituting these results into the lefthand side of
[link] again leads to the initial value theorem, in
[link] .
The final value theorem
The Final Value Theorem allows us to determine
$$\underset{t\to \infty}{lim}x\left(t\right)$$
from
$X\left(s\right)$ . Taking the limit as
$s$ approaches zero in the derivative property gives
$$\underset{s\to 0}{lim}{\int}_{{0}^{}}^{\infty}\frac{dx\left(t\right)}{dt}{e}^{st}dt=\underset{s\to 0}{lim}\left[s,X,\left(s\right),,x,\left({0}^{}\right)\right]$$
The lefthandside of
[link] can be written as
$${\int}_{{0}^{}}^{\infty}\underset{s\to 0}{lim}\frac{dx\left(t\right)}{dt}{e}^{st}dt={\int}_{{0}^{}}^{\infty}\frac{dx\left(t\right)}{dt}dt=x\left(\infty \right)x\left({0}^{}\right)$$
Substituting this result back into
[link] leads to the Final Value Theorem
$$x\left(\infty \right)=\underset{s\to 0}{lim}sX\left(s\right)$$
which is only valid as long as the limit
$x\left(\infty \right)$ exists.
Laplace Transform properties.
Property 
$y\left(t\right)$ 
$Y\left(s\right)$ 
Linearity 
$\alpha {x}_{1}\left(t\right)+\beta {x}_{2}\left(t\right)$ 
$\alpha {X}_{1}\left(s\right)+\beta {X}_{2}\left(s\right)$ 
Time Delay 
$x(t\tau )$ 
$X\left(s\right){e}^{s\tau}$ 
sShift 
$x\left(t\right){e}^{at}$ 
$X(s+a))$ 
Multiplication by
$t$ 
$tx\left(t\right)$ 
$\frac{dX\left(s\right)}{ds}$ 
Multiplication by
$t}^{n$ 
${t}^{n}x\left(t\right)$ 
$(1)}^{n}\frac{{d}^{n}X\left(s\right)}{d{s}^{n}$ 
Convolution 
$x\left(t\right)*h\left(t\right)$ 
$X\left(s\right)H\left(s\right)$ 
Differentiation 
$\frac{dx\left(t\right)}{dt}$ 
$sX\left(s\right)x\left({0}^{}\right)$ 

$\frac{{d}^{2}x\left(t\right)}{d{t}^{2}}$ 
$s}^{2}X\left(s\right)sx\left({0}^{}\right)\frac{dx\left({0}^{}\right)}{dt$ 

$\frac{{d}^{n}x\left(t\right)}{d{t}^{n}}$ 
$s}^{n}X\left(s\right){s}^{n1}x\left({0}^{}\right)\sum _{k=2}^{n}{s}^{nk}\frac{{d}^{k1}x\left({0}^{}\right)}{d{t}^{k1}$ 
Integration 
${\int}_{\infty}^{t}x\left(\tau \right)d\tau$ 
$\frac{{\int}_{\infty}^{{0}^{}}x\left(\tau \right)d\tau}{s}+\frac{X\left(s\right)}{s}$ 
Initial Value Theorem 
$x\left({0}^{+}\right)=\underset{s\to \infty}{lim}sX\left(s\right)$ 
Final Value Theorem 
$x\left(\infty \right)=\underset{s\to 0}{lim}sX\left(s\right)$ 