Question

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right)$

Where $f={\mathcal{L}}^{-1}\left\{F\right\}$.Use this equation in Problems 33-36 to compute

${\mathcal{L}}^{-1}\left\{F\right\}$.role="math" localid="1664422930667" $F\left(s\right)=arctan\left(\frac{1}{s}\right)$

Short answer

$F\left(s\right)=arctan\left(\frac{1}{s}\right)$

Step-by-step solution

Simplify the function and find derivative 

Find the derivative of F with respect to s:

$$\begin{gathered} \frac{dF}{ds}=\frac{d}{ds}\left(arc\tan \frac{1}{s}\right) \\ =\frac{1}{\left(1+{\displaystyle \frac{1}{s^2}}\right)}\left(-\frac{1}{s^2}\right) \\ =-\frac{1}{s^2+1} \end{gathered}$$

Find the Laplace inverse

From the given condition, we have ${\mathcal{L}}^{-1}\left(F\right)=\frac{1}{{\left(-t\right)}^n}{\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}$,apply this to find the Laplace inverse as:

$$\begin{gathered} {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{{\left(-t\right)}^1}{\mathcal{L}}^{-1}\left\{\frac{dF}{ds}\right\} \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{\left(-t\right)}{\mathcal{L}}^{-1}\left(-\frac{1}{s^2+1}\right) \\ =-\left(-\frac{1}{t}\right)\left({\mathcal{L}}^{-1}\left(\frac{1}{s^2+1}\right)\right) \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\sin t \end{gathered}$$

Therefore,

${\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\sin t$