Timeline for Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?
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| May 8, 2022 at 13:19 | comment | added | Max Lonysa Muller | @Wolfgang Ah yes, such a tag might indeed be useful. If you'd like to collaborate on this, it might be wise to coordinate which integrals may be worthwhile to consider, and exchange ideas. You can contact me via maxmuller100 [at] gmail [dot] com. I am for instance interested in how you obtain conjectures for closed forms of integrals - I now try them via wolfram alpha and then I put the digits in OEIS, but I suspect there's a faster method through some kind of symbolic computation software. If we work together, we might be more efficient and we might be able to find the Stirling-zeta values | |
| May 8, 2022 at 11:07 | comment | added | Wolfgang | @MaxMuller Of course. As you may imagine, I have already been looking at this kind of integrals, and I am even mulling about introducing a tag like "umbral-closed-form-zeta" to easily identify them. Also on math.stackexchange.com, which tends to have more integrals of this kind than mathoverflow.net. | |
| May 7, 2022 at 21:53 | comment | added | Max Lonysa Muller | @Wolfgang Very interesting, thank you. Is this perhaps something you'd be interested working on with me? For instance, we could start be listing some possible functions and closed form evaluations of integrals of integer powers of them, to try to find likely candidates that amound to the Stirling-zeta values | |
| May 7, 2022 at 10:19 | comment | added | Wolfgang | @PietroMajer No more mystery about the general form of $I_{n,m}$. It's all here. | |
| May 6, 2022 at 19:06 | comment | added | Wolfgang | ... are the Mittag-Leffler polynomials. BTW you can see that for those, the arguments of the zetas are able to "progress in both directions", comparing examples #1 and #4/5. So there should be chances that for the polynomials $\prod_{j=1}^k(x-j)$ (i.e. Stirling) there is also a positive answer. Maybe not with the zetas directly, but some multiplicator as for the $tanh^n/x^m$. | |
| May 6, 2022 at 18:57 | comment | added | Wolfgang | @MaxMuller That is a nice question, I wasn't aware of it. As you say, the $(\frac{\text{arcsinh}(x)}{x})^{k}$ comes close, but the polynomials for that are $\prod_{j=-k}^k(x-j)$ and $\prod_{j=1-k}^k(x-(j+\frac12))$ for odd and even degree respectively, while the Stirling numbers are the coefficients of $\prod_{j=1}^k(x-j)$. I would expect corresponding integral families (or sums involving generalized harmonic numbers) to exist, but haven't encountered yet. The only other polynomials with a similar function which I am aware of ... (cont'd) | |
| May 5, 2022 at 22:08 | comment | added | Max Lonysa Muller | @Wolfgang This is the question: mathoverflow.net/questions/409560/… | |
| May 5, 2022 at 21:54 | comment | added | Max Lonysa Muller | @Wolfgang It's tangentially related to this problem and it's more of a research proposal, though perhaps you have an answer at hand already. I'm looking for a family of integrals that amounts to $\sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1) $ for a given power $k$ of a function. Based on <a href="mathoverflow.net/questions/231964/…> question you asked earlier, it seems that $(\frac{\text{arcsinh}(x)}{x})^{k} $ might come close. Perhaps you could find a function that would actually answer the question | |
| May 5, 2022 at 20:31 | comment | added | Wolfgang | @MaxMuller sure but why not here? | |
| May 5, 2022 at 15:41 | comment | added | Max Lonysa Muller | @Wolfgang I have some questions about this problem I'd like to ask you, is there a way I could contact you? | |
| Apr 28, 2022 at 8:47 | comment | added | Wolfgang | for the record: some great answers for $I_{2,2}$ are here | |
| Nov 18, 2021 at 15:45 | history | edited | Wolfgang | CC BY-SA 4.0 |
simplified the expressions I(m,n)
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| Jul 6, 2017 at 18:23 | comment | added | Wolfgang | @user75829 Interesting. Never heard of gd(x) before! But as far as I see, it is essentially just notation? | |
| Jun 8, 2017 at 16:57 | comment | added | Wolfgang | Let us continue this discussion in chat. | |
| Jun 8, 2017 at 14:37 | comment | added | Agno | @Wolfgang. I believe I have found a clue. When you expand all formulae (i.e. no brackets left) and then divide each coefficient of $e_k$ by $2n$, then you exactly get the Sloane's series (to be read as a triangle) oeis.org/A141904 (numerator) and oeis.org/A142048 (denominator). These are related to the Taylor expansion of $\arctan(1)$, but unfortunately no easy closed form is mentioned there. I keep searching! | |
| Jun 8, 2017 at 14:03 | comment | added | Wolfgang | @Agno You are welcome. I like your optimism, and I already start sharing it :) Hoping you'll find something. Big prime factors are no threat indeed, and BTW the coefs from the right seem to be polynomial: $[e_{2n-1}]I_{2n,2}=\frac43n(n-1)$, $[e_{2n-3}]I_{2n,2}=-\frac2{45}n(n-2)(10n-7)$, $[e_{2n-5}]I_{2n,2}=\frac2{2835}n(n-3)(70n^2-147n+62)$ etc. | |
| Jun 8, 2017 at 12:37 | comment | added | Agno | Thanks for your swift response! It is really nice playing with these formulae. | |
| Jun 8, 2017 at 12:30 | comment | added | Wolfgang | and $I_{16,2}=\frac1{ 70945875}( 286876800 e_3- 1721992128 e_5+ 5375115200 e_7- 10526195680 e_9+ 13477464000 e_{11} -11048637600 e_{13} + 5297292000 e_{15}-1135134000 e_{17} )$ $ \qquad \ $ It didn't occur to me to look up the numerators in the oeis :) | |
| Jun 8, 2017 at 12:30 | comment | added | Wolfgang | @Agno $I_{8,2} =\frac1{ 105}( 352 e_3- 1232 e_5+ 1680 e_7- 840 e_9) $ $I_{10,2} =\frac1{ 945}( 3378 e_3- 14360 e_5+ 27090 e_7- 25200 e_9+ 9450 e_{11}) $ $I_{12,2}=\frac1{ 51975}( 195240 e_3- 957308 e_5+ 2244000 e_7- 2938320 e_9+ 2079000 e_{11}- 623700 e_{13} $ $I_{14,2}=\frac1{ 675675}( 2642070 e_3- 14482832 e_5+ 39859820 e_7- 65745680 e_9+ 66216150 e_{11}- 37837800 e_{13}+9459450 e_{15} )$ | |
| Jun 8, 2017 at 11:43 | comment | added | Agno | @Wolfgang The coefficient of $e_3$ for each $I_{2n,2}$ seems to be $\sum_{k=1}^{n} \frac{2}{2k-1}$. See also oeis.org/A074599. I am optimistic similar forms could be found for the other $e_k$ coefficients. Would you be so kind to share in a comment the $e_k$ sequences for $I_{8,2}$,$I_{10,2}$ and $I_{12,2}$ like in your OP? | |
| Jun 7, 2017 at 20:02 | vote | accept | Wolfgang | ||
| Jun 7, 2017 at 18:27 | answer | added | Igor Khavkine | timeline score: 43 | |
| Jun 7, 2017 at 8:37 | comment | added | Wolfgang | @PietroMajer No... it ends with $(-1)^{n+1}\,2n\,e_{2n+1}$, but the coefficients of $e_3$ for $n=3,...,8$ are $46/15,352/105,1126/315=3378/9!!,13016/3465=39048/11!!,...$ having in their numerators biggest prime factors $23,11,563,1627,88069,1423$. This does not look very promising. | |
| Jun 7, 2017 at 7:39 | comment | added | Pietro Majer | Do you have a general formula for $I_{2n,2}$? | |
| Jun 7, 2017 at 7:34 | comment | added | Igor Khavkine | @Wolfgang, the trick is to start with that contour (which obviously gives the correct answer) and then deform it so that it encircles only poles. You can see a related example in this answer. | |
| Jun 7, 2017 at 7:08 | comment | added | Wolfgang | @IgorKhavkine But as far as I see, there are no poles inside that contour. | |
| Jun 7, 2017 at 6:56 | comment | added | Igor Khavkine | You can apply the contour integral method to the equivalent integral $\frac{1}{2\pi i} \oint \frac{\operatorname{tahn}^3 z}{z^2} \log(-z) \, dz$, where the contour clockwise tightly encircles the positive real axis. | |
| Jun 7, 2017 at 6:43 | comment | added | Wolfgang | @GeraldEdgar Yes I know but wow, that "really means nothing". Note that the ISC doesn't even find $1.1274284420316\approx I_{3,3}$ :-( | |
| Jun 7, 2017 at 6:00 | answer | added | T. Amdeberhan | timeline score: 20 | |
| Jun 7, 2017 at 0:38 | comment | added | Gerald Edgar | $I_{3,2} \approx 1.154785313323$ is not recognized by the ISC ... isc.carma.newcatle.edu.au | |
| Jun 6, 2017 at 22:56 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
edited title
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| Jun 6, 2017 at 20:21 | comment | added | Sylvain JULIEN | A worth noticing feature of your expressions of the considered integrals in terms of the $e_{i}$ is that they're linear, so maybe using a wise mix of integration techniques and linear algebra could lead to something interesting. | |
| Jun 6, 2017 at 19:56 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 1 character in body
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| Jun 6, 2017 at 17:04 | comment | added | LSpice | I don't know whether to hope that we do, or that we don't, find an integration technique called "nospoon's method" in future calculus textbooks. | |
| Jun 6, 2017 at 16:29 | history | asked | Wolfgang | CC BY-SA 3.0 |