Laplace transform of the derivative of a function with jump discontinuity at t=0 While trying to find the inverse laplace transform of $F(s)=s\log\frac{s-1}{s+1}$, I first calculated at the inverse laplace transform of $$G(s)=\log\frac{s-1}{s+1}$$ which came out to be $$g(t)=\frac{e^{-t}-e^t}{t}$$ (ignoring the heaviside function, I don't think it would make a difference here.)
Now, to find the inverse laplace transform of $F(s)=sG(s)$ I would usually use the theorem $$\mathcal{L}\{g'(t)\}=sG(s)-g(0)$$ But here $g$ has a jump discontinuity at $t=0$ $$g(0^+)=2\neq g(0^-)=0$$ Initially I just used $g(0^+)=-2$ in place of $g(0)$ which resulted in $$f(t)=g'(t)-2\delta(t)$$Besides the fact that $2\delta(t)$ term is not in the answer given in the textbook, it doesn't make much sense to me either, we can only speak of $g'(t)$ for $t>0$. I assume I was wrong to use $g(0^+)=-2$ in the equation, so what would be the correct method to find $\mathcal{L}^{-1}\{sG(s)\}$ when $g(t)$ is discontinuous at $t=0$? Could the issue here be $g(t)$ itself, that $\mathcal{L}^{-1}\{\log\frac{s-1}{s+1}\}$ is not in fact $\frac{e^{-t}-e^t}{t}$? I ask this because I am not entirely sure whether $g(t)$ would satisfy this condition for the existence of a laplace transform:

$f(t)$ is piecewise continuous on $0\leq t\leq A$ for every $A>0$

I will edit this question to add my derivation of $g(t)$ if needed, any help would be appreciated here.
 A: I don't recommend forcing derivative's theorem in these cases because it really gets messy, specially when considering generalized functions and their initial conditions.
I would go this way using convolution:
$$F(s)=s \log \frac{s-1}{s+1}$$
$$f(t)=\mathcal{L}^{-1} \{s\} * \mathcal{L}^{-1} \left\{\log \frac{s-1}{s+1} \right\} = \delta'(t) * \frac{e^{-t}-e^t}{t} = \int_0^t \delta'(t-x) \frac{e^{-x}-e^x}{x} dx$$
and to solve the integral, I'll use the filter property of dirac's delta function and its derivatives
$$\int_0^t \frac{d^1}{dx^1}\big( \delta(t-x) \big) \frac{e^{-x}-e^x}{x} dx = (-1)^1 \frac{d^1}{dx^1} \left. \left( \frac{e^{-x}-e^x}{x} \right) \right|_{x=t} =  \frac{e^{-t}(t+1)+e^t(t-1)}{t^2}$$
mind that $\delta'(t-x)=\delta'(x-t)$ for the integral purposes and because $x$ and $t$ take positive values.
my answer is slightly diferent form what Wolfram Alpha computes. But maybe this result is right, at least for a part of it. Tell me if this answer is the one given to you. And please anyone let me know if I made any mistakes.
