The Laplace transform of the step function defined as \( f(t)=1 \) for \( 1 \le t \le 4 \) and \( f(t)=0 \) otherwise is calculated using the integral \( F(s)=\int_1^4 e^{-st} dt \). The discussion emphasizes that altering the function at a single point, such as excluding the endpoints, does not affect the value of the integral due to the properties of integrals over continuous intervals. This conclusion is supported by the mathematical principle that changes at countably many points do not influence the overall integral.
PREREQUISITESMathematicians, engineering students, and professionals working with differential equations and control systems will benefit from this discussion, particularly those interested in the applications of Laplace transforms.
Alexmahone said:Find the Laplace transform of $\displaystyle f(t)=1$ if $\displaystyle 1\le t\le 4$; $\displaystyle f(t)=0$ if $\displaystyle t<1$ or if $\displaystyle t>4$.
CaptainBlack said:Straight forward application of the definition:
\[ F(s)=\int_0^{\infty}f(t)e^{-st}\; dt=\int_1^4e^{-st}\;dt \]
CB
Alexmahone said:Thanks. How would the answer differ if one of the endpoints 1 or 4 (or both) were excluded?
Ackbach said:Do you mean if your function were defined as, for example, $f(t)=1$ if $1<t\le 4$; $f(t)=0$ if $t\le 1$ or if $t>4$? It would make no difference. The reason is that the changing of one point in a function does not alter the integral of that function. In fact, changing the function at countably many points does not change the value of the integral.
Alexmahone said:That's exactly what I meant. Thanks.