What is the integral of e^{-st}ln(t) ?

The integral of e^{-st}ln(t) is -e^{-st} 1/s ln(t)+1/s \Ei(-st)+C

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integral of e^{-st}ln(t)

\int e^{- s t} ln (t) d t

- e^{- s t} \frac{1}{s} ln (t) + \frac{1}{s} Ei (- s t) + C

Solution steps

\int e^{- s t} ln (t) d t

Apply Integration By Parts

: - e^{- s t} \frac{1}{s} ln (t) - \int - \frac{e ^{- s t}}{s t} d t

= - e^{- s t} \frac{1}{s} ln (t) - \int - \frac{e ^{- s t}}{s t} d t

\int - \frac{e ^{- s t}}{s t} d t = - \frac{1}{s} Ei (- s t)

= - e^{- s t} \frac{1}{s} ln (t) - (- \frac{1}{s} Ei (- s t))

Simplify

= - e^{- s t} \frac{1}{s} ln (t) + \frac{1}{s} Ei (- s t)

Add a constant to the solution

= - e^{- s t} \frac{1}{s} ln (t) + \frac{1}{s} Ei (- s t) + C

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What is the integral of e^{-st}ln(t) ?
The integral of e^{-st}ln(t) is -e^{-st} 1/s ln(t)+1/s \Ei(-st)+C