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.

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


enter image description here

Proof: take 2 of them and find average:

$$\int \frac12\left(\frac{x-1}{x (x+1)}+ \left(2 e^x \text{Ei}(-x)+\frac{1}{x}\right)\right)dx=-\gamma$$

enter image description here



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}

  • $\begingroup$ @Anixx How do you get that change of variable? $\endgroup$ – user76284 Mar 28 at 20:49
  • $\begingroup$ 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. $\endgroup$ – Anixx Mar 28 at 20:53
  • $\begingroup$ On the other hand, the transforming with Laplace transform as indicated in the question usually works well. $\endgroup$ – Anixx Mar 28 at 20:54
  • $\begingroup$ @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. $\endgroup$ – user76284 Mar 28 at 20:57
  • 1
    $\begingroup$ @Anixx If you're not looking for regularizations, then I don't know what to tell you. The integrals are simply divergent, then. $\endgroup$ – user76284 Mar 28 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.