The Laplace transform of (1-cost)/t is equal to 1/2 log[(s^{2}+1)/s^{2}]. Here we find Laplace of (1-cost)/t.

The Laplace transform formula of (1-cost)/t is given by

L{(1-cost)/t} = 1/2 log[(s^{2}+1)/s^{2}].

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## Find the Laplace transform of (1-cost)/t

Answer: 1/2 log[(1+s^{2})/s^{2}] is the Laplace transform of (1-cost)/t. |

**Solution:**

Here we will use the following formula:

L$\Big[ \dfrac{f(t)}{t} \Big]$ = $\int_s^\infty F(s) ds$ **…(I)**

where F(s) = L{f(t)}, the Laplace transform of f(t).

To find the Laplace transform of (1-cost)/t, let us put f(t) = 1-cost in the above formula (I).

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

= L{1} – L{cost}

= $\dfrac{1}{s}$ – $\dfrac{s}{s^2+1}$ as the Laplace of cos(at) is s/(s^{2}+a^{2}).

Now, from the formula (I), we have that

L{(1-cost)/t} = $\int_s^\infty \Big[ \dfrac{1}{s} – \dfrac{s}{s^2+1} \Big] ds$

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

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

= $\dfrac{1}{2}\Big[ \log \dfrac{s^2}{1+s^2} \Big]_s^\infty$

= $\dfrac{1}{2}$ lim_{s→∞} log $\dfrac{s^2}{1+s^2}$ – $\dfrac{1}{2} \log \dfrac{s^2}{1+s^2}$

= $\dfrac{1}{2}$ log 1 + $\dfrac{1}{2} \log \dfrac{1+s^2}{s^2}$

= $\dfrac{1}{2} \log \dfrac{1+s^2}{s^2}$ as the value of log 1 equals 0.

So the Laplace transform of (1-cost)/t is equal to 1/2 log[(1+s^{2})/s^{2}]. and this is obtained by the division by t formula of Laplace transforms.

More Laplace Transforms:

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

Laplace tansform of a constant

Find Laplace transform of cos^{2}t

## FAQs

**What is the Laplace of (1-cost)/t?**

Answer: The Laplace of (1-cost)/t is 1/2 log[(1+s^{2})/s^{2}].

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.