Evaluate this laplace transform of PSK\frac{1-e^{-t}}{t}ZSK and the integral exists

Marla Payton 2021-12-11 Answered
Evaluate this laplace transform of \(\displaystyle{\frac{{{1}-{e}^{{-{t}}}}}{{{t}}}}\) and the integral exists according to wolfram

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Expert Answer

Mary Goodson
Answered 2021-12-12 Author has 1742 answers

Write
\(I(s)=\int_0^{\infty} \frac{1-e^{-t}}{t}e^{-st}dt\)
for Re s>0. Compute I′(s) (by differentiating under the integral sign), and from that obtain I(s) by noting that \(\displaystyle\lim_{{{R}{e}{s}\Rightarrow+\infty}}{I}{\left({s}\right)}={0}\) by the dominated convergence theorem.
\(\displaystyle{I}{\left({s}\right)}={\log{{\left({1}+{\frac{{{1}}}{{{s}}}}\right)}}}\)
I am getting
\(\displaystyle{I}'{\left({s}\right)}=-{\frac{{{1}}}{{{s}+{1}}}}\)
integrating both sides gives
\(\displaystyle{I}{\left({s}\right)}={\ln{{\left({\frac{{{1}}}{{{s}+{1}}}}\right)}}}\)
You don't get a factor of s down by differentiating.
\(\displaystyle{\frac{{\partial}}{{\partial{s}}}}{e}^{{-{s}{t}}}=-{t}\cdot{e}^{{-{s}{t}}}\)
so it`s
\(I'(s)=\int_0^{\infty} (e^{-t}-1) \cdot e^{-st} dt\)

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Lindsey Gamble
Answered 2021-12-13 Author has 4878 answers
Use the fact that
\(\displaystyle{\frac{{{1}-{e}^{{-{t}}}}}{{{t}}}}={\int_{{0}}^{{{1}}}}{d}{u}{e}^{{-{t}{u}}}\)
So the integral in question is
\(\displaystyle{\int_{{0}}^{{\infty}}}{\left.{d}{t}\right.}{\int_{{0}}^{{1}}}{d}{u}{e}^{{-{t}{\left({u}+{s}\right)}}}\)
Because all integrals here converge absolutely, we may switch the order of integration and get
\(\displaystyle{\int_{{0}}^{{1}}}{d}{u}{\int_{{0}}^{{\infty}}}{\left.{d}{t}\right.}{e}^{{-{t}{\left({u}+{s}\right)}}}={\int_{{0}}^{{1}}}{\frac{{{d}{u}}}{{{u}+{s}}}}={\log{{\left({1}+{\frac{{{1}}}{{{s}}}}\right)}}}\)
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