The Laplace transform of (1-sint)/t does not exist. That is, the formula for the Laplace of 1-sint divided by t is given as follows:
L{$\frac{1-\sin t}{t}$} = Undefined.
Let us now find the Laplace of (1-sint)/t.
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What is the Laplace transform of (1-sint)/t?
| Answer: The Laplace of (1-sint)/t is not defined |
Solution:
To find the Laplace of (1-sint)/t, we will use the division by t formula. This formula states that if the Laplace transform of f(t) is F(s), then we have
L$\Big[ \dfrac{f(t)}{t} \Big]$ = $\int_s^\infty F(s) ds$ …(I)
Put f(t) = 1-sint.
So F(s) = L{f(t)} = L{1-sint} = L{1} – L{sint}. As the Laplace of sin(at) is a/(s2+a2), we get that
F(s) = $\dfrac{1}{s}$ – $\dfrac{1}{s^2+1}$.
Now, using the formula (I),
L{$\frac{1-\sin t}{t}$} = $\int_s^\infty \Big[ \dfrac{1}{s} – \dfrac{1}{s^2+1} \Big] ds$
= $\Big[ \log s – \tan^{-1}s \Big]_s^\infty$
= (lims→∞ log s) -π/2 – log s + tan-1s
As logs when s→∞ is undefined, this does not exist.
So the Laplace transform of (1-sint)/t does NOT exist.
More Laplace Transforms:
Main Article: Laplace Transform: Definition, Table, Formulas, Properties
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FAQs
Answer: The Laplace of (1-sint)/t does NOT exist.
