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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?

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

Right?

Anixx
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1 Answers1

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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}

user76284
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  • @Anixx How do you get that change of variable? – user76284 Mar 28 '21 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. – Anixx Mar 28 '21 at 20:53
  • On the other hand, the transforming with Laplace transform as indicated in the question usually works well. – Anixx Mar 28 '21 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. – user76284 Mar 28 '21 at 20:57
  • No, $I\ne 0$ of course. Its regularized value is $0$, yes, but as a divergent integral it is definitely not zero. – Anixx Mar 28 '21 at 20:58
  • @Anixx Are we not working in the context of regularization and divergent integrals? – user76284 Mar 28 '21 at 20:59
  • @Anixx The second example makes more sense, though. – user76284 Mar 28 '21 at 21:01
  • Not exactly. I treat divergent integrals as (infinite) entities. The change of variables is not applicable to divergent (to infinity) integrals with improper bounds (and the regularized value also can change). This is not surprising, because they do not satisfy the requirements of change-of-variable theorems. – Anixx Mar 28 '21 at 21:02
  • Look also here: https://math.stackexchange.com/questions/4075766/why-is-ln-0-ne-ln-infty Here I separately transformed $\int_0^1 1/x dx$ and $\int_1^\infty 1/x dx$ and found that the difference between them is $\gamma$. From this it directly follows that their infinite parts cancel out and $\int_0^\infty \frac{\text{sign} (x-1)}x dx=-\gamma$ – Anixx Mar 28 '21 at 21:08
  • @Anixx I'll check that out. – user76284 Mar 28 '21 at 21:15
  • The question of substitution of variables in divergent integrals has been raised by me here: https://math.stackexchange.com/questions/3892992/is-there-a-solid-reason-why-some-people-assume-the-fundamental-theorem-of-calcul – Anixx Mar 28 '21 at 21:16
  • Here is another post regarding regularization of divergent integrals using Laplace transform: https://mathoverflow.net/questions/385813/can-we-meaningfully-ascribe-values-to-these-divergent-integrals and this result is actually supported by the theory of hyperfunctions, for instance. – Anixx Mar 28 '21 at 21:18
  • @Anixx I'll check that out as well. I'd delete the second equation of that question though, since $I = 0$ in the context of regularized integrals, so $I = 2I$ is true. – user76284 Mar 28 '21 at 21:20
  • No, I am not talking in the context of regularized values (even if finding the regularized value is the aim). – Anixx Mar 28 '21 at 21:21
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    @Anixx If you're not looking for regularizations, then I don't know what to tell you. The integrals are simply divergent, then. – user76284 Mar 28 '21 at 21:22
  • And regularization of an integral that has its divergent parts cancel out is different than regularization of an integral that diverges to infinity. For instance, it is okay for me to say $\int_0^\infty (x-\frac2{x^3})dx=0$, but totally not okay to say $\int_0^\infty1dx=0$ or $\int_0^\infty 1/x^2dx=0$ (the later two integrals are equal though). – Anixx Mar 28 '21 at 21:24
  • You can look here for more on the framework I work in: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Mar 28 '21 at 21:26
  • @Anixx It's perfectly okay to say $\int_0^\infty dx = 0$, if you're willing to extend the integral beyond its usual domain. The reason is that $\int_0^\infty dx = \int_0^\infty \lim_{\varepsilon \downarrow 0} e^{-x \varepsilon} (1 - x \varepsilon) dx \stackrel{!}{=} \lim_{\varepsilon \downarrow 0} \int_0^\infty e^{-x \varepsilon} (1 - x \varepsilon) dx = 0$. See here for details. – user76284 Mar 28 '21 at 21:31
  • I do not contest that the regularized value of this integral is zero. But it is infinite, hence, not zero. In my framework this integral is an important constant. I use a special symbol for it. $\int_0^\infty dx=\int_0^\infty \frac1{x^2}dx=\tau$ – Anixx Mar 28 '21 at 21:35
  • If you look at the linked page you will see that there are divergent integrals equal to $2\tau, \tau/2, \tau+1/2, \tau-1/2, \tau^2, \tau^3$ and so on. – Anixx Mar 28 '21 at 21:38
  • It seems, there is a fundamental difference between various regularization methods: https://mathoverflow.net/questions/391387/did-anyone-ever-propose-the-distinction-between-divergent-to-infinity-as-oppos – Anixx Apr 28 '21 at 11:40