Derive Inverse Laplace transform by differentiating F(s) and Integrating f(t) (5.5-21)

Question

Request:

Please check my work. I cannot duplicate the answer in text although it is very close. I believe the problem lies in how I take the derivative. Is there a better way to calculate the inverse transform other than the present method given?

Given:

Find the inverse Laplace transform $\mathcal{L}^{-1}\{tan^{-1}(\frac{s}{\omega})\}$ by first differentiating the Laplace transform and then applying the integral.

Solution:

$$-F'(s)=-\frac{1}{\omega}\cdot \frac{1}{({\frac{s}{\omega}})^2+1}\cdot \frac{\omega}{\omega}$$

$$=-\frac{\omega}{s^2+\omega^2}\cdot$$

$$\mathcal{L}^{-1}\{-F'(s)\}=-\omega sin(\omega t)$$

$$\mathcal{L}^{-1}\{tan^{-1}(\frac{s}{\omega})\}=\frac{f(t)}{t}=-\frac{\omega}{t}\cdot sin(\omega t)$$

Answer In Text:

$$\mathcal{L}^{-1}\{tan^{-1}(\frac{s}{\omega})\}=\frac{1}{t}\cdot sin(\omega t)$$

Question:

What went wrong?

Answer 1

First there is no minus sign for $F'(s)$, i.e.

$F'(s)= {\dfrac{ω}{s^2 + ω^2}}$, not $F'(s)= -{\dfrac{ω}{s^2 + ω^2}}$

Second

$L^{-1}({\dfrac{ω}{s^2 + ω^2}}) = \sin(ωt)$, not $ω\sin(ωt)$