What is the integral from 3 to infinity of e^{-st}(2t-3) ?

The integral from 3 to infinity of e^{-st}(2t-3) is -(-(3e^{-3s})/s-(2e^{-3s})/(s^2))

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integral from 3 to infinity of e^{-st}(2t-3)

\int_{3}^{\infty} e^{- s t} (2 t - 3) d t

- (- \frac{3 e ^{- 3 s}}{s} - \frac{2 e ^{- 3 s}}{s ^{2}})

Solution steps

\int_{3}^{\infty} e^{- s t} (2 t - 3) d t

Compute the indefinite integral:

\int e^{- s t} (2 t - 3) d t = - e^{- s t} \frac{1}{s} (- 3 + 2 t) - e^{- s t} \frac{2}{s ^{2}} + C

Compute the boundaries:

\int_{3}^{\infty} e^{- s t} (2 t - 3) d t = 0 - (- \frac{3 e ^{- 3 s}}{s} - \frac{2 e ^{- 3 s}}{s ^{2}})

= 0 - (- \frac{3 e ^{- 3 s}}{s} - \frac{2 e ^{- 3 s}}{s ^{2}})

Simplify

= - (- \frac{3 e ^{- 3 s}}{s} - \frac{2 e ^{- 3 s}}{s ^{2}})

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What is the integral from 3 to infinity of e^{-st}(2t-3) ?
The integral from 3 to infinity of e^{-st}(2t-3) is -(-(3e^{-3s})/s-(2e^{-3s})/(s^2))