How to find the inverse Laplace $\mathcal{L}^{-1}[\ln(1+\frac{a^2}{s^2})]$

Question

I am trying to calculate the inverse Laplace $\mathcal{L}^{-1}[\ln(1+\frac{a^2}{s^2})]$

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**My attempt:**

We know from a basic Laplace property that

$$\mathcal{L}^{-1}[F'(s)]=(-1)tf(t) = -t\mathcal{L}[F(s)] \quad (*)$$

Therefore $$ F(s) = \ln(1+\frac{a^2}{s^2}) = \ln(\frac{s^2 + a^2}{s^2}) = \ln(s^2 + a^2) - \ln(s^2) \quad (1)$$

Therefore $$ F'(s) = \frac{2s}{s^2+a^2} - \frac{2}{s} \quad (2)$$

So from (1), (2), (*) we have that

$$ \mathcal{L}[F(s)] = \frac{2}{t}\cos(at)-2$$

And now I am stuck. It feels I solved the difficult part and I stuck on something trivial but still, I don't know how to go on in order to find the inverse Laplace.

Answer 1

You should have:

$$\mathcal{L}^{-1}[F'(s)]=(-1)tf(t) = \color {red}{-t\mathcal{L^{-1}}}[F(s)]$$ Therefore you have that: $$\color {red} {\mathcal{L^{-1}}}[F'(s)] = 2\cos(at)-2$$ $$-t {\mathcal{L^{-1}}}[F(s)] = 2\cos(at)-2$$ $${\mathcal{L^{-1}}}[F(s)] = -\dfrac 1t(2\cos(at)-2)$$ Finally: $$f(t) =-\frac{2}{t}(\cos(at)-1)$$