Evaluate this laplace transform of 1−e−tt and the integral exists according to wolfram

Name: Evaluate this laplace transform of 1−e−tt and the integral exists according to wolfram
Uploaded: 2022-02-23T17:22:57+00:00
Description: Evaluate this laplace transform of 1−e−tt and the integral exists according to wolfram

WriteI(s)=∫0∞1−e−tte−stdtfor Re s&gt;0. Compute I′(s) (by differentiating under the integral sign), and from that obtain I(s) by noting that limRes⇒+∞I(s)=0 by the dominated convergence theorem.I(s)=log⁡(1+1s)I am gettingI′(s)=−1s+1integrating both sides givesI(s)=ln⁡(1s+1)You don't get a factor of s down by differentiating.∂∂se−st=−t⋅e−stso it`sI′(s)=∫0∞(e−t−1)⋅e−stdt

"Evaluate this laplace transform of 1−e−tt and the..." solved and explained

Marla Payton

Answered question

2021-12-11

Evaluate this laplace transform of

\frac{1 - e^{- t}}{t}

and the integral exists according to wolfram

Answer & Explanation

Mary Goodson

Beginner2021-12-12Added 37 answers

Write
$I (s) = \int_{0}^{\infty} \frac{1 - e^{- t}}{t} e^{- s t} d t$
for Re s>0. Compute I′(s) (by differentiating under the integral sign), and from that obtain I(s) by noting that $lim_{R e s \Rightarrow + \infty} I (s) = 0$ by the dominated convergence theorem.
$I (s) = \log (1 + \frac{1}{s})$
I am getting
$I^{'} (s) = - \frac{1}{s + 1}$
integrating both sides gives
$I (s) = \ln (\frac{1}{s + 1})$
You don't get a factor of s down by differentiating.
$\frac{\partial}{\partial s} e^{- s t} = - t \cdot e^{- s t}$
so it`s
$I^{'} (s) = \int_{0}^{\infty} (e^{- t} - 1) \cdot e^{- s t} d t$

Lindsey Gamble

Beginner2021-12-13Added 38 answers

Use the fact that

\frac{1 - e^{- t}}{t} = \int_{0}^{1} d u e^{- t u}

So the integral in question is

\int_{0}^{\infty} d t \int_{0}^{1} d u e^{- t (u + s)}

Because all integrals here converge absolutely, we may switch the order of integration and get

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

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