So, we take $\frac{\text{sgn}(x-1)}{x}$ and apply $\mathcal{L}_t[t f(t)](x)$ four times. The transform keeps area. These integrals are minus Euler-Mascheroni constant:
$$\int_0^\infty \frac{\text{sgn}(x-1)}{x}dx=\int_0^\infty\frac{2 e^{-x}-1}{x}dx=\int_0^\infty\frac{x-1}{x (x+1)}dx=\int_0^\infty \left(2 e^x \text{Ei}(-x)+\frac{1}{x}\right) dx=$$ $$\int_0^\infty \frac{x^2-2 x \log (x)-1}{(x-1)^2 x} dx=-\gamma$$
Yes?
Proof: take 2 of them and find average:
$$\int \frac12\left(\frac{x-1}{x (x+1)}+ \left(2 e^x \text{Ei}(-x)+\frac{1}{x}\right)\right)dx=-\gamma$$
Right?