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{\Large\bf The Laplace Transform of a Function}

\vspace{.3in}

The Laplace Transform of a function y(t) defined for t>0
is defined by

$$
Y(s)=L[y(t)](s)=\int_{0}^{\infty}e^{-st}y(t)dt
$$

if the integral exists. The functions y(t) and Y(s) are
partner functions. 

\vspace{.3in}

{\bf Examples}

\vspace{.2in}

Let y(t)=exp(t). We have

$$
Y(s)=\int_{0}^{\infty}e^{-st}e^{t}dt=
\int_{0}^{\infty}e^{-(s-1)t}dt=\frac{1}{s-1}
$$

The integral converges if s>1. The functions exp(t) and
1/(s-1) are partner functions.

Let y(t)=cos(3t). We have

$$
Y(s)=\int_{0}^{\infty}e^{-st}\cos(3t)dt=\frac{s}{s^2+9}
$$

The integral converges for s>0. The integral can be
computed by doing integration by parts twice. 

\vspace{.3in}

{\bf Existence of the Laplace Transform}

\vspace{.2in}

If y(t) is piecewise continuous for t>=0 and of exponential
order, then the Laplace Transform exists for some
values of s. A function y(t) is of exponential order c
if there is exist constants M and T such that

$$
|y(t)|\leq Me^{ct} \hspace{.2in} t\geq T
$$

All polynomials, simple exponentials (exp(at), where
a is a constant), sine and cosine functions, and products
of these functions are of exponential order. An
example of function not of exponential order is 
exp(t^2). This function grows too rapidly. The integral

$$
\int_{0}^{\infty}e^{-st}e^{t^2}dt 
$$

does not converge for any value of s. 

\vspace{.3in}

{\bf Table of Laplace Transforms} 

\vspace{.2in}

The following table lists the Laplace Transforms for a
selection of functions

\vspace{.3in}

{\bf Rules for Computing Laplace Transforms of Functions}

\vspace{.2in}

There are several formulas and properties of the Laplace 
transform which can greatly simplify calculation of the
Laplace transform of functions. We summarize them
below. The properties can verified using the integral
formula for the Laplace Transform and can be found in
any textbook.

\vspace{.2in}

{\bf Linearity}

\vspace{.2in}

Like differentiation and integration the Laplace
transformation is a {\it linear} operation. What does
this mean? In words, it means that the Laplace
transform of a constant times a function is the
constant times the Laplace transform of the function.
In addition the Laplace transform of a sum of
functions is the sum of the Laplace transforms.

Let us restate the above in mathspeak. Let Y_1(s) and 
_2(s) denote the Laplace transforms of y_1(t) and y_2(t),
respectively,  and let c_1 be a constant. Recall that
L[f(t)](s) denotes the Laplace transform of f(t).
We have

$$
L[c_1y_1(t)](s)=c_1L[y_1(t)](s)=c_1Y_1(s)
$$

$$
L[y_1(t)+y_2(t)](s)=L[y_1(t)](s)+L[y_2(t)](s)=Y_1(s)+Y_2(s)
$$

Here are several examples:

$$
L[3e^{t}](s)=3L[e^{t}](s)=3\cdot\frac{1}{s-1}=\frac{3}{s-1}
$$

Here we have used the results in the table for the
Laplace transform of the exponential. Here are a
couple of more examples:

$$
L[e^t+\cos 2t](s)=L[e^t](s)+L[\cos 2t](s)=\frac{1}{s-1}+\frac{s}{s^2+4}
$$

$$
L[3e^t+4\cos 2t](s)=L[3e^t](s)+L[2\cos 2t](s)=3L[e^t](s)+4L[\cos
2t](s)= \frac{3}{s-1}+\frac{4s}{s^2+4}
$$

\vspace{.3in}

{\bf Translation Property}

\vspace{.2in}

The translation formula states that Y(s) is the Laplace
transform of y(t), then 

$$
L[e^{at}y(t)](s)=Y(s-a),
$$

where a is a constant. Here is an example. The Laplace
transform of y(t)=t is Y(s)=1/s^2. Hence:

$$
L[e^{2t}t](s)=Y(s-2)=\frac{1}{(s-2)^2}
$$

\vspace{.2in}

{\bf Laplace Transform of the Derivative}

\vspace{.2in}

Suppose that the Laplace transform of y(t) is Y(s). Then
the Laplace Transform of y'(t) is

$$
L[y'(t)](s)=sY(s)-y(0)
$$

For the second derivative we have

$$
L[y''(t)](s)=s^2Y(s)-sy(0)-y'(0)
$$

For the n'th derivative we have

$$
L[y^{(n)}(t)](s)=s^nY(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\ldots -
y^{(n-1)}(0)
$$

\vspace{.2in}

{\bf Derivatives of the Laplace Transform}

\vspace{.2in}

Let Y(s) be the Laplace Transform of y(t). Then

$$
L[t^ny(t)](s)=(-1)^nY^{(n)}(s)
$$

Here is an example. Suppose we wish to compute
the Laplace transform of tsin(t). The Laplace
transform of sin(t) is 1/(s^2+1). Hence, we have 

$$
L[t\sin t]=-\frac{d}{ds}[\frac{1}{s^2+1}]=\frac{2s}{(s^2+1)^2}
$$

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