3.7 Improper integrals  (Page 4/6)

Laplace transforms

In the last few chapters, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the Laplace transform and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and transforms them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.

The Laplace transform is defined in terms of an integral as

Note that the input to a Laplace transform is a function of time, $f(t),$ and the output is a function of frequency, $F(s).$ Although many real-world examples require the use of complex numbers (involving the imaginary number $i=\sqrt{\mathrm{-1}}),$ in this project we limit ourselves to functions of real numbers.

Let’s start with a simple example. Here we calculate the Laplace transform of $f(t)=t$ . We have

This is an improper integral, so we express it in terms of a limit, which gives

Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to t , so we treat the variable s as a constant. We have

Then we obtain

1. Calculate the Laplace transform of $f(t)=1.$
2. Calculate the Laplace transform of $f(t)=e^{\mathrm{-3}t}.$
3. Calculate the Laplace transform of $f(t)=t^2.$ (Note, you will have to integrate by parts twice.)
   Laplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let’s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.
   Let’s start with the definition of the Laplace transform. We have
   
   $L\left\{f(t)\right\}={\displaystyle {\int }_0^{\infty }{e}^{\text{−}st}f(t)dt}=\underset{z\to \infty }{\text{lim}}{\displaystyle {\int }_0^z{e}^{\text{−}st}f(t)dt}.$
4. Use integration by parts to evaluate $\underset{z\to \infty }{\text{lim}}{\displaystyle {\int }_0^z{e}^{\text{−}st}f(t)dt}.$ (Let $u=f(t)$ and $dv=e^{\text{−}st}dt.)$
   After integrating by parts and evaluating the limit, you should see that
   
   $L\left\{f(t)\right\}=\frac{f(0)}{s}+\frac{1}{s}\left[L\left\{f^{\prime }(t)\right\}\right].$
   
   Then,
   
   $L\left\{f^{\prime }(t)\right\}=sL\left\{f(t)\right\}-f(0).$
   
   Thus, differentiation in the time domain simplifies to multiplication by s in the frequency domain.
   The final thing we look at in this project is how the Laplace transforms of $f(t)$ and its antiderivative are related. Let $g(t)={\displaystyle {\int }_0^{t}f}(u)du.$ Then,
   
   $L\left\{g(t)\right\}={\displaystyle {\int }_0^{\infty }{e}^{\text{−}st}g(t)dt}=\underset{z\to \infty }{\text{lim}}{\displaystyle {\int }_0^z{e}^{\text{−}st}g(t)dt}.$
5. Use integration by parts to evaluate $\underset{z\to \infty }{\text{lim}}{\displaystyle {\int }_0^z{e}^{\text{−}st}g(t)dt}.$ (Let $u=g(t)$ and $dv=e^{\text{−}st}dt.$ Note, by the way, that we have defined $g(t),$ $du=f(t)dt.)$
   As you might expect, you should see that
   
   $L\left\{g(t)\right\}=\frac{1}{s}·L\left\{f(t)\right\}.$
   
   Integration in the time domain simplifies to division by s in the frequency domain.