# Laplace transforms A powerful tool in solving problems inengineering and physics is the Laplace transform.

tinfoQ 2020-10-18 Answered

Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
where we assume s is a positive real number. For example, to find the Laplace transform of $f\left(t\right)={e}^{-t}$, the following improper integral is evaluated using integration by parts:
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{e}^{-t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+1\right)t}dt=\frac{1}{\left(s+1\right)}$
Verify the following Laplace transforms, where u is a real number.
$f\left(t\right)=1\to F\left(s\right)=\frac{1}{s}$

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## Expert Answer

stuth1
Answered 2020-10-19 Author has 97 answers
Step 1
$f\left(t\right)=1\to F\left(s\right)=\frac{1}{s}$
Step 2
Given a function f(t), the Laplace transform is a new function F(s) is defined by
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$ where we assume s is a positive real number.
For f(t)=1, the Laplace transform is given by
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot 1dt,$

Therefore,
$F\left(s\right)=-{\int }_{0}^{-\mathrm{\infty }}\frac{\left({e}^{u}\right)}{s}du$
$={\int }_{-\mathrm{\infty }}^{0}\frac{\left({e}^{u}\right)}{s}du\left[\because {\int }_{a}^{b}f\left(x\right)dx=-{\int }_{b}^{a}f\left(x\right)dx\right]$
$=\frac{1}{s}{\int }_{-\mathrm{\infty }}^{0}\left({e}^{u}\right)du$
$=\frac{1}{s}\left[{e}^{u}{\right]}_{-\mathrm{\infty }}^{0}$
$=\frac{1}{s}\left[{e}^{0}-{e}^{-\mathrm{\infty }}\right]$
$=\frac{1}{s}\left[1-0\right]$
$=\frac{1}{s}$
Hence, verified.
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