Find the Laplace transform of (1-e^t)/t

The Laplace transform of (1-e^t)/t is equal to log[(s-1)/s]. In this post, we find Laplace of (1-e^t)/t. The Laplace transform formula of (1-e^t)/t is given by

L{(1-e^t)/t} = log[(s-1)/s].

Laplace of (1-e^t)/t

Answer: The Laplace transform of (1-et)/t is log[(s-1)/s].

Solution:

Let us recall the division by t formula:

L$[{\displaystyle \frac{f(t)}{t}}]$ = ${\int }_s^{\mathrm{\infty }}F(s)ds$

where F(s) = L{f(t)}, the Laplace transform of f(t). Put f(t) = 1-e^t in this formula, so that we have:

F(s) = L{f(t)} = L{1-e^t}

⇒ F(s) = L{1} – L{e^t}

⇒ F(s) = ${\displaystyle \frac{1}{s}}$ – ${\displaystyle \frac{1}{s-1}}$ as we know L{e^at} = 1/(s-a).

Now, from the above formula, we get that

L$\{{\displaystyle \frac{1-e^t}{t}}\}$ = ${\int }_s^{\mathrm{\infty }}[{\displaystyle \frac{1}{s}}–{\displaystyle \frac{1}{s-1}}]ds$

= $[\log s–\log (s-1){]}_s^{\mathrm{\infty }}$

= $[\log {\displaystyle \frac{s}{s-1}}{]}_s^{\mathrm{\infty }}$

= lim_s→∞ log ${\displaystyle \frac{s}{s-a}}$ – $\log {\displaystyle \frac{s}{s-1}}$

= log 1 – $\log {\displaystyle \frac{s}{s-1}}$

= $\log {\displaystyle \frac{s-1}{s}}$ since log 1 = 0.

So the Laplace transform of (1-e^t)/t is equal to log[(s-1)/s] which is proved using the division by t formula.

More Laplace Transforms:

Main Article: Laplace Transform: Definition, Table, Formulas, Properties

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