Find the Laplace transform

[Alexmahone]

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]

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 $\displaystyle t>4$.

Straight forward application of the definition:

\[ F(s)=\int_0^{\infty}f(t)e^{-st}\; dt=\int_1^4e^{-st}\;dt \]

CB

 

[Alexmahone]

Straight forward application of the definition:

\[ F(s)=\int_0^{\infty}f(t)e^{-st}\; dt=\int_1^4e^{-st}\;dt \]

CB

Thanks. How would the answer differ if one of the endpoints 1 or 4 (or both) were excluded?

 

[Ackbach]

Thanks. How would the answer differ if one of the endpoints 1 or 4 (or both) were excluded?

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]

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.

That's exactly what I meant. Thanks.

 

[Ackbach]

That's exactly what I meant. Thanks.

You're welcome, as always!