# What is the Laplace transform of (e^at-cosbt)/t [Solved]?

The Laplace transform of (e^at-cosbt)/t is equal to log[(s2+b2)1/2/(s-a)]. Here we will find the Laplace of (eat-cosbt)/t using the division by t formula of Laplace transforms.

Note that

## Laplace of (eat-cosbt)/t

To find the Laplace transform of (eat-cosbt)/t, we will follow the steps discussed below.

Step 1: At first, we will find the Laplace of eat-cosbt. We know that

1. L{eat} = 1/(s-a)
2. L{cosbt} = s/(s2+b2)

So L{eat-cosbt} = L{eat} – L{cosbt} = $\dfrac{1}{s-a} – \dfrac{s}{s^2+b^2}$.

Step 2: Now apply the division by t formula which says that if L{f(t)}=F(s), then

L$\Big( \dfrac{f(t)}{t} \Big)$ = $\int_s^\infty F(s) ds$ …(∗)

Step 3: So the Laplace of (eat-cosbt)/t is

L{(eat-cosbt)/t} = $\int_s^\infty \Big( \dfrac{1}{s-a} – \dfrac{s}{s^2+b^2} \Big) ds$

= $\Big[ \log (s-a) – \dfrac{1}{2} \log (s^2+b^2) \Big]_s^\infty$

= $\dfrac{1}{2} \log \Big[\dfrac{(s-a)^2}{s^2+b^2} \Big]_s^\infty$

= $\dfrac{1}{2}$ lims→∞ $\log \dfrac{(s-a)^2}{s^2+b^2}$ – $\dfrac{1}{2} \log \dfrac{(s-a)^2}{s^2+b^2}$

= $\dfrac{1}{2}$ lims→∞ $\log \dfrac{s^2(1-\frac{a}{s})^2}{s^2(1+\frac{b^2}{s^2})}$ – $\dfrac{1}{2} \log \dfrac{(s-a)^2}{s^2+b^2}$

= $\dfrac{1}{2} \cdot$ log 1 – $\dfrac{1}{2} \log \dfrac{(s-a)^2}{s^2+b^2}$

= $-\dfrac{1}{2} \log \dfrac{(s-a)^2}{s^2+b^2}$ as log1 =0

= $\dfrac{1}{2} \log \dfrac{s^2+b^2}{(s-a)^2}$.

So the Laplace transform of (eat-cosbt)/t is equal to log[(s2+b2)1/2/(s-a)], and this is proved by the division by t formula of Laplace transforms.

More Laplace Transforms:

Laplace Transform: Definition, Table, Formulas, Properties

Laplace transform of sint/t

Laplace transform of cost/t

Laplace transform of (1-cost)/t

Laplace transform of (1-et)/t

## FAQs

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

Answer: The Laplace transform of (et-cost)/t is equal to log[(s2+1)1/2/(s-1)].

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