Question

laplace transform of (1-cost)÷t
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Answer 1

Step-by-step explanation:

L.T ( 1/t -cos t/t) =

1/s^2- tan-1 s

Answer 2

Answer:

Laplace transform of $\frac{1-Cost}{t} = \ln \left(\frac{\sqrt{s^{2}+1}}{s}\right)$

Step-by-step explanation:

Here function f(t) ,$f(t)=\frac{1-\cos (t)}{t}$

                            $f(t)=\frac{1}{t}-\frac{Cost}{t}$

laplace transform of f(t) given by $F(s)=\int_{0}^{+\infty} f(t) \cdot e^{-s \cdot t} \cdot d t$

so f(t) becomes                               $f(t)=\frac{1}{t}-\frac{Cost}{t}$  

                                                               $=\int_{s}^{\infty} \frac{1}{u} d u+\int_{s}^{\infty} \frac{u}{u^{2}+1} d u$

            $\int_{s}^{\infty} \frac{1}{u} d u=\ln (\infty)-\ln (s) . \int_{s}^{\infty} \frac{u}{u^{2}+1} d u=\frac{1}{2} \ln (\infty)-\frac{1}{2} \ln \left(s^{2}+1\right)$

$\begin{aligned}&\text F(s)=\ln (\infty)-\ln (s)-\frac{1}{2} \ln (\infty)+\frac{1}{2} \ln \left(s^{2}+1\right) \\&F(s)=\frac{1}{2} \ln \left(s^{2}+1\right)-\ln (s)=\ln \left(\frac{\sqrt{s^{2}+1}}{s}\right)\end{aligned}$