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I have to evaluate this definite integral: $$Z=\int_0^\infty\operatorname{arccot}(x)\,\operatorname{arccot}(2x)\,\operatorname{arccot}(5x)\,dx$$ My CAS was only able to find its approximate numeric value: $$Z\approx0.796300956669079523165601562454031588576893734085453548868394...$$ Is there an approach that would allow to evaluate it in a closed form?

I looked up this integral in Gradshteyn-Ryzhyk, but the closest one I found was formula 4.511: $$\int_0^\infty\operatorname{arccot}(px)\,\operatorname{arccot}(qx)\,dx=\frac\pi2\left[\frac1p\,\ln\left(1+\frac p q\right)+\frac1q\,\ln\left(1+\frac q p\right)\right]$$ Is there a way to generalize it to a product of 3 arccotangents? Any help is appreciated.

Piotr Shatalin
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    i suspect (!) that it might be possible to solve this integral by differentiation $\partial_a\partial_b\partial_cZ(a,b,c)=-\int_0^{\infty}\frac{x^3}{(1+(a x)^2)(1+(b x)^2)(1+(c x)^2)}$. things will get a little bit tedious but it might be worth a shot. – tired Dec 01 '15 at 01:32
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    u might enjoy this http://math.stackexchange.com/questions/1279165/integrals-of-the-form-large-int-0-infty-operatornamearccotx-cdot-operat/1285149#1285149 – tired Dec 01 '15 at 01:42
  • This is a great question! I feel there should be a closed form for this integral, possibly involving polylogarithms. – Vladimir Reshetnikov Dec 01 '15 at 05:17
  • @VladimirReshetnikov https://gist.github.com/anonymous/1546aa11a1d3803b6b71 – Start wearing purple Dec 01 '15 at 13:49
  • @Startwearingpurple 1. Where did this come from? 2. Each logarithmic term is a polynomial of degree at most $3$ in terms $\log n$, $n\in{2,3,5,7}$, so substantial simplifications are possible – Julian Rosen Dec 02 '15 at 17:55
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    @JulianRosen I explained the idea in an answer that was immediately downvoted (so I decided to delete it). It is actually very simple: $$\int\ln(x-a)\ln(x-b)\ln(x-c),dx$$ has an expression in terms of polylogs and elementary functions. Rewriting arccot's in the logarithmic form and carefully taking the limits yields the result even if we replace $2$ and $5$ by arbitrary parameters. – Start wearing purple Dec 02 '15 at 21:47

1 Answers1

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Using a formula from my another answer, we can get: $$Z=\frac1{960}\Big[96\operatorname{Li}_3\left(\tfrac13\right)+744\operatorname{Li}_3\left(\tfrac23\right)-780\operatorname{Li}_3\left(\tfrac15\right)-1152\operatorname{Li}_3\left(\tfrac25\right)\\ +408\operatorname{Li}_3\left(\tfrac35\right)-60\operatorname{Li}_3\left(\tfrac45\right)-720\operatorname{Li}_3\left(\tfrac16\right)-840\operatorname{Li}_3\left(\tfrac56\right)\\ -48\operatorname{Li}_3\left(\tfrac17\right)-1032\operatorname{Li}_3\left(\tfrac27\right)-192\operatorname{Li}_3\left(\tfrac37\right)-192\operatorname{Li}_3\left(\tfrac47\right)\\ -1200\operatorname{Li}_3\left(\tfrac57\right)+120\operatorname{Li}_3\left(\tfrac67\right)-112\operatorname{Li}_3\left(\tfrac18\right)-168\operatorname{Li}_3\left(\tfrac78\right)\\ -192\operatorname{Li}_3\left(\tfrac3{10}\right)+168\operatorname{Li}_3\left(\tfrac7{10}\right)+120\operatorname{Li}_3\left(\tfrac5{12}\right)\\ -120\ln5\cdot\left[4\operatorname{Li}_2\left(\tfrac13\right)+2\operatorname{Li}_2\left(\tfrac25\right)-\operatorname{Li}_2\left(\tfrac17\right)+4\operatorname{Li}_2\left(\tfrac27\right) -3\operatorname{Li}_2\left(\tfrac37\right)\right]\\ +24\ln2\cdot\left[12\operatorname{Li}_2\left(\tfrac13\right)+20\operatorname{Li}_2\left(\tfrac25\right)-12\operatorname{Li}_2\left(\tfrac17\right)+20\operatorname{Li}_2\left(\tfrac27\right)-8\operatorname{Li}_2\left(\tfrac37\right)\right]\\ \\ +1364\ln^32+100\ln^33+148\ln^35+424\ln^37\\ -228\ln3\cdot\ln^22-168\ln5\cdot\ln^22+1176\ln^23\cdot\ln2-624\ln^25\cdot\ln2-648\ln^27\cdot\ln2\\+108\ln3\cdot\ln^25-36\ln3\cdot\ln^27-600\ln^23\cdot\ln5+564\ln^25\cdot\ln7-600\ln^27\cdot\ln5\\+504\ln3\cdot\ln7\cdot\ln2+48\ln5\cdot\ln7\cdot\ln2-288\ln3\cdot\ln5\cdot\ln7\\ -2\pi^2\cdot(3\ln2-76\ln3+37\ln5+36\ln7)+3151\,\zeta(3)\Big]$$

Here is equivalent Mathematica expression.

Community
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Vladimir Reshetnikov
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    Omg! In Italy we would say "sei fortissimo" which roughly translate in "you are damn powerful". +1 – AlienRem Dec 04 '15 at 19:47
  • It seems like this is a pure Mathematica result? If this is that case why not add the code used [or the steps to simplify first] to getting it so people acctually can learn something from this answer? (a naive Mathematica 9.0 evaluation does not find the answer) – Winther Dec 05 '15 at 18:54
  • @Winther Indeed, I use Mathematica as my working environment (rather than plain paper) to apply rules and to check results numerically. But all simplification rules for polylogarithms were hand-written based on known identities (I use Lewin's book as a reference), because Mathematica is very poor at polylogarithm simplifications on its own. – Vladimir Reshetnikov Dec 05 '15 at 19:25
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    My point was not with you using Mathematica, but rather that it would be good to add some steps / code so that other people can reproduce it / learn some tricks on using software like this to evaluate integrals - otherwise I don't see much value in answers like this. As I said Mathematica 9.0 does not evaluate it directly so one must use some trick first. I suspect you had to use something like arccot$(z) \equiv \frac{i}{2}\left[\log(1-i/z) - \log(1+i/z)\right]$ and split into $8$ $\log() \log() \log() $-integrals to get it started?! – Winther Dec 07 '15 at 00:43
  • @Winther The trick to make Mathematica to find an antiderivative is to rewrite the integral as Integrate[(π/2 - (Log[1 - I x] - Log[1 + I x]) I/2) (π/2 - (Log[1 - I a x] - Log[1 + I a x]) I/2) (π/2 - (Log[1 - I b x] - Log[1 + I b x]) I/2), x]. This evaluates to an expression with LeafCount exceeding $20000$, and then all you need is to simplify it :) – Vladimir Reshetnikov Dec 12 '15 at 00:10