Computing the Laplace transform of $\tan(pt)$ I've been  thinking of using  complex number approach , what's your view guys ?.
 A: 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 $(3)$.
A: Using CAS like Mathematica we have:
$$\mathcal{L}_t[\tan (p t)](s)=-\frac{i}{s}+\frac{\Phi \left(-1,1,1-\frac{i s}{2 p}\right)}{p}$$
where:
$\Phi \left(-1,1,1-\frac{i s}{2 p}\right)$ is HurwitzLerchPhi function
Using:
$$\tan (x)=-i+\sum _{k=1}^{\infty } -2 i (-1)^k e^{-2 i k x}$$
we have:
$$\mathcal{L}_t[\tan (p t)](s)=\sum _{k=1}^{\infty } \mathcal{L}_t\left[-i-2 i (-1)^k e^{-2 i k p t}\right](s)=-\frac{i}{s}-\sum
   _{k=1}^{\infty } \frac{2 i (-1)^k}{2 i k p+s}=-\frac{i}{s}+\frac{\Phi \left(-1,1,1-\frac{i s}{2 p}\right)}{p}$$
