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

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

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

Laplace of (1-et)/t

Solution:

Let us recall the division by t formula:

L$\Big[ \dfrac{f(t)}{t} \Big]$ = $\int_s^\infty F(s) ds$ where F(s) = L{f(t)}, the Laplace transform of f(t).

Put f(t) = 1-et in the formula, so that we have:

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

= L{1} – L{et}

= $\dfrac{1}{s}$ – $\dfrac{1}{s-1}$ as we know L{eat} = 1/(s-a).

Now, from the above formula, we get that

L$\Big\{\dfrac{1-e^t}{t} \Big\}$ = $\int_s^\infty \Big[ \dfrac{1}{s} – \dfrac{1}{s-1} \Big] ds$

= $\Big[ \log s – \log (s-1) \Big]_s^\infty$

= $\Big[ \log \dfrac{s}{s-1} \Big]_s^\infty$

= lims→∞ log $\dfrac{s}{s-a}$ – $\log \dfrac{s}{s-1}$

= log 1 – $\log \dfrac{s}{s-1}$

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

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

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FAQs

Q1: What is the Laplace of (1-et)/t?

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

Q2: Find L{(1-et)/t}.