 # Solve no.4 inverse laplace L^{-1}left{s lnleft(frac{s}{sqrt{s^2+1}}right)+cot^{-1s}right} babeeb0oL 2021-02-25 Answered
Solve no.4 inverse laplace
${L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)+{\mathrm{cot}}^{-1s}\right\}$
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Step 1
Given:
${L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)+{\mathrm{cot}}^{-1s}\right\}$ Step 2
Use the linear property of inverse Laplace transformation, we get
${L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)+{\mathrm{cot}}^{-1}s\right\}={L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)\right\}+{L}^{-1}\left\{{\mathrm{cot}}^{-1}s\right\}$
Now first find the inverse Laplace transformation of the first term
We know that the Laplace transformation formula:

$⇒f\left(t\right)=-\frac{1}{t}{L}^{-1}\left\{\frac{d}{ds}F\left(s\right)\right\}$
${L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)\right\}=-\frac{1}{t}{L}^{-1}\left\{\frac{d}{s}\left(s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)\right)\right\}$
Step 3
compute the derivative, we get
$⇒-\frac{1}{t}{L}^{-1}\mathrm{ln}\frac{s}{\sqrt{{s}^{2}+1}+\frac{1}{{s}^{2}+1}}$
$⇒-\frac{1}{t}{L}^{-1}\left\{\mathrm{ln}\frac{s}{\sqrt{{s}^{2}+1}}\right\}-\frac{1}{t}{L}^{-1}\left\{\frac{1}{{s}^{2}+1}\right\}$
Now we know that ${L}^{-1}\left\{\frac{a}{{s}^{2}+{a}^{2}}\right\}=\mathrm{sin}\left(at\right)$
$⇒\frac{-1}{t}{L}^{-1}\left\{\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)\right\}-\frac{\mathrm{sin}\left(t\right)}{t}$
Use the quotient rule of logarithmic, we get
$⇒-\frac{1}{t}{L}^{-1}\left\{\mathrm{ln}\left(s\right)-\frac{1}{2}\mathrm{ln}\left({s}^{2}+1\right)\right\}-\frac{\mathrm{sin}\left(t\right)}{t}$
Step 4
Use the inverse Laplace transformation formula in the first term:
$f\left(t\right)=-\frac{1}{t}{L}^{-1}\left\{\frac{d}{ds}F\left(s\right)\right\}$
$⇒\frac{1}{{t}^{2}}{L}^{-1}\left\{\frac{d}{ds}\left[\mathrm{ln}\left(s\right)-\frac{1}{2}ln\left({s}^{2}+1\right)\right]\right\}-\frac{sin\left(t\right)}{t}$
$⇒\frac{1}{{t}^{2}}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{{t}^{2}}{L}^{-1}\left\{\frac{s}{{s}^{2}+1}\right\}-\frac{sin\left(t\right)}{t}$
$⇒\frac{1}{{t}^{2}}\left({L}^{-1}\left\{\frac{1}{s}\right\}{L}^{-1}\left\{\frac{s}{{s}^{2}+1}\right\}\right)-\frac{\mathrm{sin}\left(t\right)}{t}$
${L}^{-1}\left\{s\mathrm{ln}\left(\frac{s}{\sqrt{{s}^{2}+1}}\right)\right\}=\frac{1}{{t}^{2}}\left(1-\mathrm{cos}\left(t\right)\right)-\frac{\mathrm{sin}\left(t\right)}{t}$
Step 5
Now find the inverse Laplace transformation of