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

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?

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}

• @Anixx How do you get that change of variable? Mar 28 at 20:49
• Sorry, typo. Here are some examples why change of variables does not work with divergent integrals with improper bounds: $$I=\int_0^\infty1dx=2\int_0^\infty1du=2I,$$ $$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$ (with substitution $u=2x$). In the second case even the regularized value changes. Mar 28 at 20:53
• On the other hand, the transforming with Laplace transform as indicated in the question usually works well. Mar 28 at 20:54
• @Anixx The first line is true though, since $I = 0$. I don't know what's the justification for the second equality in the first line though, unless you already know that the integrals are 0. Mar 28 at 20:57
• @Anixx If you're not looking for regularizations, then I don't know what to tell you. The integrals are simply divergent, then. Mar 28 at 21:22