A set of divergent integrals that I think, equal to $-\gamma$

Question

So, we take $\frac{\text{sgn}(x-1)}{x}$ and apply $\mathcal{L}_t[t f(t)](x)$ four times. The transform is known to keep area under the curve. These integrals, I think, are equal to minus Euler-Mascheroni constant. Since they all have infinite parts that cancel each other, their values are finite. I have already applied Laplace transforms to regularize divergent integrals in a similar way.

$$\int_0^\infty \frac{\text{sgn}(x-1)}{x}dx=\int_0^\infty\frac{2 e^{-x}-1}{x}dx=\int_0^\infty\frac{x-1}{x (x+1)}dx=\int_0^\infty \left(2 e^x \text{Ei}(-x)+\frac{1}{x}\right) dx=$$ $$\int_0^\infty \frac{x^2-2 x \log (x)-1}{(x-1)^2 x} dx=-\gamma$$

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?

Answer 1

There might be a mistake here, but formally I'm getting

\begin{align} \int_{x=0}^{x=\infty} \frac{\mathrm{sgn}(x-1)}{x} \mathrm{d}x &= \int_{x=1}^{x=\infty} \frac{1}{x} \mathrm{d}{x} - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= \int_{y^{-1}=1}^{y^{-1}=\infty} \frac{1}{y^{-1}} \mathrm{d}y^{-1} - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= \int_{y=1}^{y=0} y \mathrm{d}y^{-1} - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= -\int_{y=1}^{y=0} y y^{-2} \mathrm{d}y - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= -\int_{y=1}^{y=0} \frac{1}{y} \mathrm{d}y - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= \int_{y=0}^{y=1} \frac{1}{y} \mathrm{d}y - \int_{x=0}^{x=1} \frac{1}{x} \mathrm{d}x \\ &= 0 \end{align}