 Laplace Transforms (Improper Integrals)

Before we examine the definition of the Laplace Transform, let's quickly recall some basic knowledge of Improper Integrals.

Definition of Improper Integrals:

An improper integral is a limit of integrals over finite intervals that is used to define an unbounded interval :

$$\int_{a}^{\infty} f(t) dt = \lim_{A\to\infty} \int_{a}^{A} f(t) dt \qquad(definition \; 1)$$

IF the integral exists AND the limit from A to infinity exists, THEN the improper integral converges to that limiting value. Otherwise it diverges or fails to exist. Let's look at some examples.

1) Evaluate the following improper integral:

$$\int_{0}^{\infty} e^{ct} dt$$

By definition (1) we know that this integral is equal to:

$$=\lim_{A\to\infty} \int_{0}^{A} e^{ct} dt$$

$$= \lim_{A\to\infty} \frac{1}{c}e^{ct} \Big|_0^A$$

$$= \lim_{A\to\infty} \frac{1}{c} (e^{cA} - 1)$$

We note that the evaluation of the above limit depends on the value of c. If c<0, then

$$= \lim_{A\to\infty} \frac{1}{c} (e^{cA} - 1) =\frac{-1}{c} \;,\; (converges \; to \; \frac{-1}{c} \;for \; c<0)$$

...otherwise

$$= \lim_{A\to\infty} \frac{1}{c} (e^{cA} - 1) = \infty = undefined \;,\; (diverges \;for \; c\geq0)$$

2) Evaluate the following improper integral:

$$\int_{1}^{\infty} \frac{1}{t} dt$$

Again, by definition (1) we know that this integral is equal to:

$$=\lim_{A\to\infty} \int_{1}^{A} \frac{1}{t} dt$$

$$= \lim_{A\to\infty} ln(t) \Big|_1^A$$

$$= \lim_{A\to\infty} ln(A)-ln(1)$$

$$= \lim_{A\to\infty} ln(A)$$

$$= \infty = undefined \;,\; diverges$$

Now that we have briefly gone over the concepts of the improper integral, let's begin to discuss the Laplace Transform itself

Continue on to Laplace Transform (part2/intro)