What is the integral from 0 to infinity of e^{-st}t ?

The integral from 0 to infinity of e^{-st}t is 1/(s^2)

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integral from 0 to infinity of e^{-st}t

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

\frac{1}{s ^{2}}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\infty} e^{- s t} t d t = 0 - (- \frac{1}{s ^{2}})

= 0 - (- \frac{1}{s ^{2}})

Simplify

= \frac{1}{s ^{2}}

\int (1 + x) x d x

f (x) = 9 e^{x} cos (x), (0, 9)

y^{^{''}} + y = x^{3}

\int_{0}^{y} y e^{- y} d y

\frac{d y}{d t} y = (2 t - 1) (3 t - 4)^{- 1}

What is the integral from 0 to infinity of e^{-st}t ?
The integral from 0 to infinity of e^{-st}t is 1/(s^2)