What is the integral from 2 to infinity of te^{-st} ?

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

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integral from 2 to infinity of te^{-st}

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

- \frac{- 2 e ^{- 2 s} s - e ^{- 2 s}}{s ^{2}}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - \frac{- 2 e ^{- 2 s} s - e ^{- 2 s}}{s ^{2}}

Simplify

= - \frac{- 2 e ^{- 2 s} s - e ^{- 2 s}}{s ^{2}}

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