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Working on an algebra of divergent integrals I came to the following relation: If $\tau=\int_0^\infty dx$ then

$$\ln (\tau+a)=\int_{0}^\infty \psi'(x+1/2+a)dx$$

and this directly gives the following relation (for finite $z$):

$$\frac1\pi\ln \left(\frac{\tau +\frac{z}{\pi }}{\tau -\frac{z}{\pi }}\right)=\tan z$$

I wonder what interesting consequences there could be of such connection between trigonometric and inverse trigonometric (logarithmic) functions?

P.S. If I did all the manipulations correctly, this gives the following relation:

$$\tan z=\frac2\pi\operatorname{arctanh} \frac{z}{\tau\pi}$$

Anixx
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1 Answers1

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[This could be a very long comment, if they were allowed]

I think your manipulations may have gone awry somewhere. So your $\tau$ is an infinity that grows like $\Omega(x)$ as $x\to\infty$. One way to approach things would be to replace $\tau$ with $\lim_{x\to\infty}x$ everywhere, argue 'continuity', and pull it out front.

You ought to obtain the exact same results with using an infinitesimal $\epsilon$ for $\frac{1}{\tau}$. But that doesn't seem to work either.

You might be better off starting with simpler identities, based on some written-up formalism that you can point to. There is quite a lot of mathematics on infinities and infinitesimals, done in a variety of formalisms, that works. You just have not given us enough information to help you.

Jacques Carette
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  • Of couse there can (and most probably are) the mistakes, but this is not one. $\tau$ is not "infinity", it is a value of a divergent integral. There are many infinite values there in this approach. Yes, one can write $\operatorname{gen} \lim_{x\to\infty}x=\tau$, and this would be correct given that generalized limit is not normal limit but rather the integral of derivative. Writing this like a normal limit is wrong. – Anixx Nov 08 '19 at 22:34
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    I did say "an infinity", not "infinity". I've done work (unpublished) on formalisms for divergent integrals too. Without details, it's really hard to give you a better answer. In my paragraph "I think ..." I made a guess, and it could be wrong. We won't be able to help unless you tell us more about what $\tau$ really means. – Jacques Carette Nov 08 '19 at 22:37
  • Oh, I just realized that you did not see the link to the post where this is explained. Here is it: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 Also look at the table of those extended numbers representing the integrals in various forms: https://extended.fandom.com/wiki/Extended_Wiki#Some_extended_numbers – Anixx Nov 08 '19 at 22:41
  • Ok, I'll that carefully and come back. But maybe not be immediately. I might also delete my answer if it becomes clear that it (my answer) is useless. – Jacques Carette Nov 08 '19 at 22:59
  • You opinion and criticism is especially interesting given you are an expert in computer algebra software. – Anixx Nov 09 '19 at 12:41
  • I've read what you've linked to (thanks). Seems very nice - but there are too many 'leaps' in between steps. I was trying to implement things [in Maple], and couldn't, because too many details were skipped. I would need more formal definitions (or a guide to them) instead of trying to sort them out from the 'intuition' and the results. A FULL derivation of int(x^n,x=0..infinity) for example would be very useful. – Jacques Carette Nov 10 '19 at 14:55
  • Indeed, implementation of this system is complicated in a CAS. For instance it should be able to somehow automatically find regularized values of divergent integrals (which Mathematica cannot do, it can do it only for series, and even that not perfect and with limited set of methods). – Anixx Nov 10 '19 at 15:13
  • I would be very happy to answer all your questions that I can, maybe we should contact via e-mail. – Anixx Nov 10 '19 at 15:15
  • The majority of the relations come from the Faulhaber's formula. For instance, for finding the regular part of $f(\omega_-+a)$ I use the formula $f'(a)=\operatorname{reg} (f(\omega_++a)-f(\omega_-+a))$. So I do the following in Mathematica: Sum[f'(x),x]. Of the possible resulting functions (that differ by an arbitrary 1-periodic one) I chose the one which is Newton-analytic (equal to its Newton series), which is usually what Mathematica gives anyway, and then chose the arbitrary additive constant based on regularization of the series at one arbitrary chosen point. – Anixx Nov 10 '19 at 15:18
  • One of the rules is that any Cesaro or Abel (but not Ramanujan/Zeta) summable sum or integral should be automatically converted to its regularized value (they are equivalent as one of the postulates). This should apply to all summation methiods based on "averaging" and "means". – Anixx Nov 10 '19 at 15:32
  • One aspect I omitted there is that this system greatly simplifies many physical formulas. For instance, the mean energy of quantum harmonic oscillator which is usually taken as $\epsilon ={\frac {h\nu }{2}}+{\frac {h\nu }{e^{h\nu /kT}-1}}$ can be rewritten as $\epsilon =\operatorname{reg}\left(h\nu\omega_++kT e^{\frac{h\nu\omega_-}{kT}}\right)$ Moreover, this appears in all physical fields that use regularization, such as calculation of the Casimir effect force or quantum field theory. Maybe even more, the use of these xtended numbers in these fields would better reflect the underlying nature. – Anixx Nov 10 '19 at 16:00
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    Do indeed contact me by email - I'm a quick Google away. I think implementing some of this in a CAS would be a lot of fun. – Jacques Carette Nov 10 '19 at 22:03