Inverse Laplace transform of $\ln \left( 1 + \frac{4 a}{s^2} \right)$

Question

I would like to find the following inverse Laplace transform.

$$\mathcal L^{-1} \left[ \ln \left( 1 + \frac{4 a}{s^2} \right) \right]$$

I don't know because of $4a$.

Answer 1

Formally we have $$\log\left(1+\frac{4a}{s^2}\right) = \sum_{n\geq 1}\frac{(-1)^{n+1}(4a)^n}{n s^{2n}} $$ and since $\mathcal{L}^{-1}\left(\frac{1}{s^{2n}}\right)=\frac{x^{2n-1}}{(2n-1)!} $ we have

$$\mathcal{L}^{-1}\log\left(1+\frac{4a}{s^2}\right) = \sum_{n\geq 1}\frac{(-1)^{n+1}(4a)^n}{n(2n-1)!}x^{2n-1}=\frac{2}{x}\sum_{n\geq 1}\frac{(-1)^{n+1}(4a)^n}{(2n)!}x^{2n}=\frac{2}{x}\left(1-\cos(2x\sqrt{a})\right). $$ In the opposite direction, the Laplace transform of $\frac{1-\cos(Ax)}{x}$ is given by the complex version of Frullani's theorem.