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#### Alexmahone

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- Jan 26, 2012

- 268

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- Jan 26, 2012

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- Jan 26, 2012

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Straight forward application of the definition:

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

CB

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- Jan 26, 2012

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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

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- Jan 26, 2012

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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?

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- #5

- Jan 26, 2012

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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.

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- Jan 26, 2012

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