Computing the Laplace transform of $\tan(pt)$

Question

I've been thinking of using complex number approach , what's your view guys ?.

Answer 1

Since: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{1} $$ by considering the logarithmic derivative of both sides we have: $$ \tan(x) = \sum_{n\geq 0}\frac{8x}{(2n+1)^2\pi^2-4x^2}=\sum_{n\geq 0}\left(\frac{-1}{x-\frac{2n+1}{2}\pi}+\frac{-1}{x+\frac{2n+1}{2}\pi}\right)\tag{2}$$ so $\tan(x)$ does not have a Laplace transform, but, at least formally, has a nice inverse Laplace transform: $$\mathcal{L}^{-1}(\tan(x))=-2\sum_{n\geq 0}\cosh\left(\frac{2n+1}{2}\pi s\right).\tag{3} $$ For practical purposes, we may consider partial sums for the RHS of $(2)$ or the RHS of $(3)$.