- Thread starter
- #1

#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

- Thread starter Alexmahone
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 268

- Jan 26, 2012

- 890

Straight forward application of the definition:

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

CB

- Thread starter
- #3

- Jan 26, 2012

- 268

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

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

CB

- Admin
- #4

- Jan 26, 2012

- 4,198

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

- Thread starter
- #5

- Jan 26, 2012

- 268

That's exactly what I meant. Thanks.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.

- Admin
- #6

- Jan 26, 2012

- 4,198

You're welcome, as always!That's exactly what I meant. Thanks.