Inverse Laplace Transform of $\ln[\frac{s^2+a^2}{s^2+b^2}]$

Question

How does one find $\mathcal{L}^{-1}\{\ln[\frac{s^2+a^2}{s^2+b^2}]\}$?

I've tried splitting it up into $\mathcal{L}^{-1}\{\ln(s^2+a^2)\}-\mathcal{L}^{-1}\{\ln(s^2+b^2)\}$. However, I can't think of any way to actually take the inverse transform of $\mathcal{L}^{-1}\{\ln(s^2+a^2)\}$.

Answer 1

$\mathcal{L}^{-1}\left\{\ln\dfrac{s^2+a^2}{s^2+b^2}\right\}$

$=\mathcal{L}^{-1}\left\{\int_s^\infty\left(\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right)ds\right\}$

$=\dfrac{1}{t}\mathcal{L}^{-1}\left\{\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right\}$

$=\dfrac{2\cos at-2\cos bt}{t}$

Answer 2

You can write $$ \begin{align} F^{\prime}(s)&=[\ln(s^{2}+a^{2})-\ln(s^{2}+b^{2})]^\prime\\ &=\frac{2s+2a}{s^{2}+a^{2}}-\frac{2s+2b}{s^{2}+b^{2}}\\ &=\frac{2s}{s^{2}+a^{2}}+\frac{2a}{s^{2}+a^{2}}-\frac{2s}{s^{2}+b^{2}}-\frac{2b}{s^{2}+b^{2}}\\ &\to 2(\cos at)+2(\sin at)-2(\cos bt)-2(\sin bt) \end{align} $$