# Solve no.4 inverse laplace L^{-1}left{s lnleft(frac{s}{sqrt{s^2+1}}right)+cot^{-1s}right}

Question
Laplace transform
Solve no.4 inverse laplace
$$L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1s}\right\}$$

2021-02-26
Step 1
Given:
$$L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1s}\right\}$$ Step 2
Use the linear property of inverse Laplace transformation, we get
$$L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s\right\}=L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}+L^{-1}\left\{\cot^{-1}s\right\}$$
Now first find the inverse Laplace transformation of the first term
We know that the Laplace transformation formula:
$$\text{If } L\left\{f(t)\right\}=F(s) \text{ then } L\left\{t f(t)\right\}=\frac{-d}{ds}(F(s))$$
$$\Rightarrow f(t)=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}F(s)\right\}$$
$$L^{-1}\left\{s \ln \left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=-\frac{1}{t}L^{-1}\left\{\frac{d}{s}\left(s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right)\right\}$$
Step 3
compute the derivative, we get
$$\Rightarrow -\frac{1}{t}L^{-1}{\ln\frac{s}{\sqrt{s^2+1}+\frac{1}{s^2+1}}}$$
$$\Rightarrow -\frac{1}{t}L^{-1}\left\{\ln\frac{s}{\sqrt{s^2+1}}\right\}-\frac{1}{t}L^{-1}\left\{\frac{1}{s^2+1}\right\}$$
Now we know that $$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)$$
$$\Rightarrow \frac{-1}{t}L^{-1}\left\{\ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\} -\frac{\sin(t)}{t}$$
Use the quotient rule of logarithmic, we get
$$\Rightarrow -\frac{1}{t}L^{-1}\left\{\ln(s)-\frac{1}{2}\ln(s^2+1)\right\} -\frac{\sin(t)}{t}$$
Step 4
Use the inverse Laplace transformation formula in the first term:
$$f(t)=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}F(s)\right\}$$
$$\Rightarrow \frac{1}{t^2}L^{-1}\left\{\frac{d}{ds}\left[\ln(s)-\frac{1}{2}ln(s^2+1)\right]\right\}-\frac{sin(t)}{t}$$
$$\Rightarrow \frac{1}{t^2}L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{t^2}L^{-1}\left\{\frac{s}{s^2+1}\right\}-\frac{sin(t)}{t}$$
$$\Rightarrow \frac{1}{t^2}\left(L^{-1}\left\{{\frac{1}{s}}\right\}L^{-1}\left\{{\frac{s}{s^2+1}}\right\}\right)-\frac{\sin(t)}{t}$$
$$L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}$$
Step 5
Now find the inverse Laplace transformation of $$\cot^{-1}(s)$$
We know that $$\cot^{-1}(s)=\tan^{-1}(\frac{1}{s})$$
$$\Rightarrow L^{-1}\left\{\cot^{-1}s\right\}=L^{-1}\left\{\tan^{-1}\left(\frac{1}{s}\right)\right\}$$
$$\Rightarrow L^{-1}\left\{\tan^{-1}\left(\frac{1}{s}\right)\right\}=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}\tan^{-1}\left(\frac{1}{s}\right)\right\}$$
$$=-\frac{1}{t}L^{-1}\left\{\frac{1}{1+\left(\frac{1}{s}\right)^2} \cdot\left(\frac{-1}{s^2}\right)\right\}$$
$$=-\frac{1}{t}L^{-1}\left\{\frac{-1}{s^2+1}\right\}$$
$$\Rightarrow L^{-1}\left\{\cot^{-1}s\right\}=\frac{(\sin t)}{t}$$
Step 6
Now add the inverse Laplace transformation of both terms
$$L^{-1}\left\{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}$$
$$L^{-1}\left\{\cot^{-1}s\right\}=\frac{\sin(t)}{t}$$
$$\Rightarrow L^{-1}\left\{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}+\frac{\sin(t)}{t}$$
$$\text{Therefore, }$$
$$L^{-1}{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s}=\frac{(1-\cos t)}{t^2}$$

### Relevant Questions

Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
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Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
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Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
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The inverse Laplace transform for
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ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}$$
$$y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0$$
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