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I'm interested in integrals of the form $$I(a,b)=\int_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx,\color{#808080}{\text{ for }a>0,\,b>0}\tag1$$ It's known$\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that $$I(a,0)=\frac{\pi^2}4\left[\ln\left(1+\frac1a\right)+\frac{\ln(1+a)}a\right].\tag2$$ Maple and Mathematica are also able to evaluate $$I(1,1)=\frac{3\pi^2}4\ln2-\frac{21}8\zeta(3).\tag3$$


Is it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?

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    $\begingroup$ When is the book coming out? ;) $\endgroup$ – nbubis May 12 '15 at 18:39
  • $\begingroup$ i would try the following: introduce an additional parameter $c$ in the first $\text{arccot}$ and then differentiate w.r.t to all three paramaters. The resulting integral should be doable. Then integrating back and see how far you get staying as general eas possible. I'm somehow expecting that we end up with a quite ugly combinations of dilogarithms. $\endgroup$ – tired May 12 '15 at 19:25
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By the substitution $x\mapsto x^{-1}$ it is not hard to see that $$I(\nu^{-1},1)=\int^\infty_0\frac{\arctan^2{x}\arctan(\nu x)}{x^2}{\rm d}x$$

First, start off by re-expressing the integral \begin{align} \int^\pi_0x^2\cos(nx)\cot\left(\frac{x}{2}\right)\ {\rm d}x &=-{\rm Re}\int_C\frac{z+1}{z-1}z^{n-1}\ln^2{z}\ {\rm d}z\\ &=(-1)^{n}\int^1_0\frac{1-x}{1+x}x^{n-1}(\ln^2{x}-\pi^2)\ {\rm d}x-\int^1_0\frac{1+x}{1-x}x^{n-1}\ln^2{x}\ {\rm d}x\\ \end{align} where $C$ is the arc joining $z=1$ to $z=-1$. (The first equality follows from $z=e^{ix}$, whereas the second follows from the fact that the integral along $[-1,\epsilon]\cup\epsilon\exp(i[\pi,0])\cup[\epsilon,1]\cup C$ is $0$ since $z=1$ is a removable singularity and the indent around $z=0$ vanishes.)

Next, note that \begin{align} I_\nu(\nu^{-1},1) &=\int^\frac{\pi}{2}_0\frac{x^2}{\tan{x}(\cos^2{x}+\nu^2\sin^2{x})}{\rm d}x\\ &=\frac{1}{4}\int^\pi_0\frac{x^2}{\tan\left(\frac{x}{2}\right)(1+(1-\nu^2)\cos{x}+\nu^2)}{\rm d}x\\ &=\frac{1}{8\nu}\int^\pi_0\left(x^2\cot\left(\frac{x}{2}\right)+2\sum^\infty_{n=1}\left(\frac{\nu-1}{\nu+1}\right)^nx^2\cos(nx)\cot\left(\frac{x}{2}\right)\right){\rm d}x\\ &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\int^1_0\frac{1-x}{(1+x)(1+\xi x)}(\ln^2{x}-\pi^2)+\frac{1+x}{(1-x)(1-\xi x)}\ln^2{x}\ {\rm d}x\right) \end{align} Here $\xi=\dfrac{\nu-1}{\nu+1}$. Utilising the partial fraction decompositions $$\frac{1-x}{(1+x)(1+\xi x)}=\frac{\nu+1}{1+x}-\frac{\nu}{1+\xi x}$$ $$\frac{1+x}{(1-x)(1-\xi x)}=\frac{\nu+1}{1-x}-\frac{\nu}{1-\xi x}$$ in tandem with the easily verifiable fact $$\int^1_0\frac{\ln^2{x}}{1+\lambda x}{\rm d}x=-\frac{2{\rm Li}_3(-\lambda)}{\lambda}$$ yields \begin{align} I_\nu(\nu^{-1},1) &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\frac{\pi^2\nu}{\xi}\ln(1+\xi)-\pi^2(\nu+1)\ln{2}+\frac{7\zeta(3)}{2}(\nu+1)-\frac{4\nu}{\xi}\chi_3(\xi)\right)\\ &=\chi_3\left(\frac{\nu-1}{\nu+1}\right)-\frac{7\zeta(3)}{8}+\frac{\pi^2}{4}\ln\left(1+\frac{1}{\nu}\right) \end{align} where $\displaystyle\chi_s(z)=\sum_{n\ \text{odd}}\frac{z^n}{n^s}=\frac{1}{2}\left({\rm Li}_s(z)-{\rm Li}_s(-z)\right)$ is the Legendre-chi function.

Integrating back, \begin{align} I_\nu(\nu^{-1},1) &=\underbrace{\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)-\frac{7\zeta(3)}{8}\nu}_{\text{Let this be C}}+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_3(v)}{(1+v)^2}{\rm d}v\\ &=C-\left.\frac{2\chi_3(v)}{1+v}\right|^\frac{1-\nu}{1+\nu}_1+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_2(v)}{v(1+v)}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}-\left.\color{white}{\frac{1}{1}}2\chi_2(v)\ln(1+v)\right|^\frac{1-\nu}{1+\nu}_1+\int^\frac{1-\nu}{1+\nu}_1\frac{\ln(1+v)\ln\left(\frac{1+v}{1-v}\right)}{v}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln{2}\\ &\ \ \ \ +\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)-\ln^2(1-v)+\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v \end{align} Repeatedly integrating by parts, it is not hard to derive that \begin{align} \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)}{v}{\rm d}v =&\ -\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)+\frac{7\zeta(3)}{8}-{\rm Li}_3\left(\frac{1+\nu}{2}\right)+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)\\ &\ +\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)\\ -\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1-v)}{v}{\rm d}v =&\ {\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)\\ \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v =&\ -2\chi_3(\nu)+2\chi_2(\nu)\ln{\nu}+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right) \end{align} Therefore, we get, for $I(\nu^{-1},1)$, \begin{align} I(\nu^{-1},1) =&\color{brown}{\ (1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-2\chi_3(\nu)+{\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_3\left(\frac{1+\nu}{2}\right)-\frac{7\zeta(3)}{8}\nu}\\ &\ \color{brown}{+2\chi_2(\nu)\ln{\nu}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)}\\ &\ \color{brown}{+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)+\frac{\pi^2}{4}\ln{2}}\\ &\ \color{brown}{+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right)+\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)-\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)}\\ &\ \color{brown}{-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)} \end{align} If no mistakes were made, this formula should hold for $0<\nu\le1$. Further simplifications to the formula may be possible through some polylogarithm identities.

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    $\begingroup$ nice work (+1). what exactly is the reason for the reformulation in terms of the two auxillary integrals? i would have guessed that one can dig in the orignal integral by more or less the same techniques. $\endgroup$ – tired May 16 '15 at 18:01
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    $\begingroup$ Bravo!! Very good work! $\endgroup$ – Iaroslav Blagouchine May 16 '15 at 23:45
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    $\begingroup$ Very nice result! +1. By the way, maybe I'm wrong, but in the last expression there is a condition $0 < a \leq 1$, therefore this is a closed-form for $I(c,1)$ for $c \geq 1$. $\endgroup$ – user153012 May 17 '15 at 1:39
  • $\begingroup$ @tired Thanks. I have followed your suggestion and simplified the answer. $\endgroup$ – M.N.C.E. May 17 '15 at 8:58
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    $\begingroup$ @user153012 Thank you. Yes, this should hold for $c\ge 1$. I have not checked if the formula is valid for $0<c<1$ though. $\endgroup$ – M.N.C.E. May 17 '15 at 9:01
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$$\begin{align}I(2,1)&=\frac{\pi^2}3\ln2-\frac{\pi^2}6\ln3+2\ln^22\cdot\ln3-3\ln2\cdot\ln^23+\frac{29}{24}\ln^33\\&+\frac{73}{16}\zeta(3)-2\ln2\cdot\operatorname{Li}_2\left(\tfrac13\right)-\frac{13}4\operatorname{Li}_3\left(\tfrac13\right)-4\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$


$$\begin{align}I(3,1)&=\frac{13\,\pi^2}{12}\ln2-\frac{4\,\pi^2}9\ln3-\frac13\ln2\cdot\ln^23+\frac7{18}\ln^33\\&-\frac{13}8\zeta(3)+\ln3\cdot\operatorname{Li}_2\left(\tfrac13\right)+\frac43\operatorname{Li}_3\left(\tfrac13\right)-\frac23\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$


Update (in response to a comment): $$\begin{align}&I(\phi,1)=\frac32\ln^32+\frac{\pi^2}{12}\Big[\left(6-3\sqrt5\right)\ln2+\left(3\sqrt5-4\right)\ln\left(1+\sqrt5\right)\Big]+\frac{51-21\sqrt5}{48}\zeta(3)\\&-\frac{\ln\left(1+\sqrt5\right)}2\Bigg[15\ln^22-15\ln\left(1+\sqrt5\right)\ln2+4\ln^2\left(1+\sqrt5\right)+2\operatorname{Li}_2\left(\frac{1-\sqrt5}4\right)\Bigg]\\&-\ln\left(3+\sqrt5\right)\operatorname{Li}_2\left(\sqrt5-2\right)+\frac{11+3\sqrt5}{48}\operatorname{Li}_3\left(9-4\sqrt5\right)-\frac{13+3\sqrt5}6\operatorname{Li}_3\left(\sqrt5-2\right)\end{align}$$

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    $\begingroup$ Obviously. Are you sure about that $\frac{7}{18}$? $\endgroup$ – marty cohen May 13 '15 at 0:57
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    $\begingroup$ @Winther I checked to 1000 decimal places - it matches. $\endgroup$ – Vladimir Reshetnikov May 14 '15 at 22:24
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    $\begingroup$ @Cleo Do you get your answers using integer relation algorithms? Could you post a closed form for something more tricky, like $I(\phi,1)$ where $\phi$ is the golden ratio? $\endgroup$ – Vladimir Reshetnikov May 15 '15 at 19:46
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    $\begingroup$ @martycohen Yes, absolutely. $\endgroup$ – Cleo May 15 '15 at 23:48
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    $\begingroup$ @VladimirReshetnikov I updated my answer. $\endgroup$ – Cleo May 15 '15 at 23:49
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I finally managed to find a general solution to this problem. Although I put a lot of effort into simplification of the result, it is still not as pretty and symmetric as I would like it to be. I hope to improve it later. Sorry for poor typesetting.

Assuming $0<a<b<c,$ $$\int_0^\infty\operatorname{arccot}(ax)\,\operatorname{arccot}(bx)\,\operatorname{arccot}(cx)\,dx=\\ \frac1{24 a b c}\left(2 a (b+2 c) \ln ^3(a)-3 \left(2 b (a+c) \ln (b)+(4 a c-2 b c) \ln (c)+\\ b \left(-2 c \operatorname{artanh}\left(\frac{b}{c}\right)+(c-a) \ln (c-a)-(a+c) \ln (a+c)+a \ln \left(c^2-b^2\right)\right)\right) \ln ^2(a)\\ +3 \left(b c \left(\ln ^2(b)+2 \left(2 \ln (a+c)+\ln \left(-\frac{c (b+c)}{(a-b) (a-c) (b-c)}\right)\right) \ln (b)+\ln ^2(b-a)\\ -3 \ln ^2(c)+\ln ^2(c-a)-\ln ^2(c-b)+\ln ^2(b+c)-4 \operatorname{artanh}\left(\frac{a}{c}\right) \ln (a+b)+\\ 2 \ln (b-a) \ln \left(\frac{c}{c-a}\right)-6 \ln (c) \ln (a+c)+2 \ln (a+c) \ln (c-b)\\ +4 \ln (c) (\ln (c-a)+\ln (c-b))-4 \ln (c) \ln (b+c)-2 \ln (c-a) \ln (b+c)\right)\\ +a \left(b \left(2 \ln ^2(b)-2 \ln \left(\frac{a^2-c^2}{b^2-c^2}\right) \ln (b)+\ln ^2(a+b)-\ln ^2(c-a)-\ln ^2(a+c)\\ -3 \ln ^2(b+c)-2 \ln (b-a) \ln (c-b)+2 \ln (c-a) (\ln (b-a)+\ln (b+c))\\ +2 \ln (a+b) \ln \left(\frac{b+c}{c^2-a^2}\right)+2 \ln (a+c) \ln \left(c^2-b^2\right)\right)-c \left(\ln ^2(b)-2 (\ln ((b-a) c)-\ln (a+c)) \ln (b)+\ln ^2(b-a)+\ln ^2(a+b)\\ -3 \ln ^2(c)+\ln ^2(a+c)+\ln ^2(c-b)+2 \ln ^2(b+c)+2 \ln (b-a) \ln \left(\frac{c}{c-b}\right)\\ -2 \ln (a+b) \ln (b+c)-2 \ln (a+c) \ln (c (b+c))\right)\right)\right) \ln (a)\\ -2 (a (b-c)+b c) \ln ^3(b)+a c \ln ^3(b-a)-b c \ln ^3(b-a)-3 a c \ln ^3(c)+5 b c \ln ^3(c)\\ +2 a b \ln ^3(c-a)-2 b c \ln ^3(c-a)+2 a c \ln ^3(a+c)-2 b c \ln ^3(a+c)-a b \ln ^3(c-b)\\ -b c \ln ^3(c-b)-a b \ln ^3(b+c)-3 b c \ln ^3(b+c)-3 b c \ln (b-a) \ln ^2(c)\\ -3 a b \ln ((a-b) (b-c)) \ln ^2(c-a)+3 b c \ln ((a-b) (b-c)) \ln ^2(c-a)\\ +3 b c \ln \left(\left(b^2-a^2\right) c\right) \ln ^2(a+c)-3 a c \ln (a+b) \ln ^2(c-b)+3 a c \ln ((b-a) c) \ln ^2(c-b)\\ -3 a b (\ln (b-a)-\ln ((a+b) (a+c))) \ln ^2(c-b)+3 b c \ln (c (a+c)) \ln ^2(c-b)\\ -3 a c \ln \left(\frac{b-a}{a+b}\right) \ln ^2(b+c)+6 a c \ln (c) \ln ^2(b+c)-12 b c \ln (c) \ln ^2(b+c)\\ +3 a b \ln (a+c) \ln ^2(b+c)+3 b c \left(\ln \left(b^2-a^2\right)+\ln (a+c)\right) \ln ^2(b+c)\\ -3 a c \ln ^2(b-a) \ln \left(1-\frac{b}{c}\right)-b c \pi ^2 \left(\ln (b-a)+\ln \left(\frac{b-c}{a-c}\right)\right)-5 a c \pi ^2 \ln (c)-5 b c \pi ^2 \ln (c)\\ -3 b c \ln ^2(b-a) \ln \left(-\frac{c}{b-c}\right)-12 b c \ln ^2(c) \ln \left(\frac{c-a}{a+c}\right)-12 b c \ln (a+b) \ln (c) \ln \left(\frac{c-a}{a+c}\right)\\ +\pi ^2 a c (\ln (b-a)-\ln (a+c))+3 a b \pi ^2 \ln (a+c)+4 b c \pi ^2 \ln (a+c)-6 b c \ln ^2(c) \ln (c-b)\\ +6 a b \ln (b-a) \ln (c-a) \ln (c-b)-6 b c \ln (b-a) \ln (c-a) \ln (c-b)\\ -6 a b \ln (a+b) \ln (a+c) \ln (c-b)-6 b c \ln (a+b) \ln (a+c) \ln (c-b)\\ -12 b c \ln (c) \ln (a+c) \ln (c-b)-a b \pi ^2 (\ln (c-a)+\ln (c-b))\\ +3 a c \ln ^2(a+b) \ln \left(\frac{c (c-b)}{b+c}\right)+3 a c \ln ^2(c) (\ln (b-a)-\ln (b+c))+9 b c \ln ^2(c) \ln (b+c)\\ -6 a c \ln (a+b) \ln (c) \ln (b+c)+6 b c \ln (a+b) \ln (c) \ln (b+c)+12 b c \ln (c) \ln (c-a) \ln (b+c)\\ +5 \pi ^2 a b \ln (b+c)+6 a c \pi ^2 \ln (b+c)+3 b c \pi ^2 \ln (b+c)-3 a b \ln ^2(a+c) \ln \left(\frac{b+c}{a+b}\right)\\ +6 a c \ln (b-a) \ln (a+c) \ln (c (b+c))-6 b c \ln (b-a) \ln (a+c) \ln (c (b+c))\\ -3 a c \ln ^2(a+c) (\ln (b-a)+\ln (c (b+c)))-3 b c \ln ^2(a+b) \ln \left(-\frac{c (b+c)}{b-c}\right)\\ -3 \ln ^2(b) \left(a \left(c (\ln (b-a)+\ln (c)-2 \ln (a+c)+\ln (b+c))\\ +b \ln \left(\frac{b^2-c^2}{a^2-c^2}\right)\right)-b c \ln \left(\frac{(a-b) c (c-b)}{a^2-c^2}\right)\right)-6 a c \ln (b-a) \ln (c) \ln \left(c^2-b^2\right)\\ +6 b c \ln (b-a) \ln (c) \ln \left(c^2-b^2\right)\\ -\ln (b) \left(b c \left(3 \left(-\ln ^2(a+b)+2 (\ln ((a+c) (b+c))-\ln (c-a)) \ln (a+b)\\ +2 \ln ^2(c)+\ln ^2(c-a)+\ln ^2(a+c)-3 \ln ^2(b+c)-2 \ln (a+c) \ln (c-b)\\ +2 \ln (c-a) \ln (b+c)+2 \ln (c) (\ln ((a+c) (b+c))-2 \ln (c-a))\\ -2 \ln (b-a) \ln \left(c^2-a^2\right)+2 \ln (b-a) \ln \left(c^2-b^2\right)\right)+\pi ^2\right)\\ +a \left(c \left(6 (\ln (b-a)+\ln (c)-\ln (a+c)) \ln \left(\frac{a+c}{b+c}\right)+\pi ^2\right)\\ +3 b \left(\ln ^2(a+b)+2 \ln \left(\frac{b+c}{c^2-a^2}\right) \ln (a+b)-\ln ^2(c-a)-\ln ^2(a+c)-3 \ln ^2(b+c)\\ +2 \ln (b-a) \ln \left(\frac{a-c}{b-c}\right)+2 \ln (c-a) \ln (b+c)+2 \ln (a+c) \ln \left(c^2-b^2\right)+2 \pi ^2\right)\right)\right)\\ +6 \left(2 \left(a \ln \left(\frac{a}{c}\right)+b \ln \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(\frac{a}{c}\right) c+2 \left(a \ln \left(\frac{a}{c}\right)+b \ln \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(-\frac{c}{a}\right) c-2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a+c)}{a (b-c)}\right) c+2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a-c)}{a (b+c)}\right) c-2 a \operatorname{Li}_3\left(\frac{a}{c}\right) c+2 b \operatorname{Li}_3\left(-\frac{b}{c}\right) c-2 b \operatorname{Li}_3\left(\frac{b}{c}\right) c+2 a \operatorname{Li}_3\left(-\frac{c}{a}\right) c-(a-b) \operatorname{Li}_3\left(\frac{(b-a) c}{b (a+c)}\right) c-(a-b) \operatorname{Li}_3\left(\frac{a (c-b)}{(a-b) c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a (c-b)}{(a+b) c}\right) c-(a-b) \operatorname{Li}_3\left(\frac{b-a}{b+c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a+b}{b+c}\right) c-(a+b) \operatorname{Li}_3\left(\frac{a (b+c)}{(a+b) c}\right) c+(a-3 b) \zeta (3) c+2 a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(\frac{a}{b}\right)+2 a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(-\frac{b}{a}\right)+\left(a b \ln \left(\frac{a}{b}\right)+b c \ln \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right) \operatorname{Li}_2\left(\frac{b-a}{b-c}\right)+\left(a b \ln \left(\frac{a}{b}\right)+b c \ln \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right) \operatorname{Li}_2\left(\frac{b-c}{a+b}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+b}{b+c}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+c}{b+c}\right)-2 a b \operatorname{Li}_3\left(\frac{a}{b}\right)+2 a b \operatorname{Li}_3\left(-\frac{b}{a}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a-b}{a-c}\right)-a (b-c) \operatorname{Li}_3\left(\frac{b-a}{b-c}\right)+a (b-c) \operatorname{Li}_3\left(\frac{a+b}{b-c}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a (b-c)}{b (a-c)}\right)+b (a+c) \operatorname{Li}_3\left(\frac{a+b}{a+c}\right)+b (a+c) \operatorname{Li}_3\left(\frac{b (a+c)}{a (b-c)}\right)+b (a-c) \operatorname{Li}_3\left(\frac{b (a-c)}{a (b+c)}\right)-b (a-c) \operatorname{Li}_3\left(\frac{c-a}{b+c}\right)+b (a+c) \operatorname{Li}_3\left(\frac{a+c}{b+c}\right)-a (b+c) \operatorname{Li}_3\left(\frac{a (b+c)}{b (a+c)}\right)\right)\right)$$

Here is the equivalent Mathematica expression. The formula can be proved using differentiation by parameters $a,b,c.$

In fact, the integrand even has a closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms, but it is too large to put it here.

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    $\begingroup$ How much paper did you use to simplify this?? $\endgroup$ – user85798 Dec 5 '15 at 21:52
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    $\begingroup$ You may be interested in this paper (especially appendix B): homeweb.unifr.ch/kellerha/pub/gafa.pdf . There are a handful of auxiliary functions defined there to denote various sums of polylog terms, which would probably go a long way towards grouping the terms of your ghastly result in a more organized manner. There are also a few general closed forms for certain integrals that may also come in handy. And congratulations on your progress! $\endgroup$ – David H Dec 7 '15 at 3:15
  • $\begingroup$ I made some progress on simplification. A draft is here. $\endgroup$ – Vladimir Reshetnikov Dec 10 '15 at 19:09
  • $\begingroup$ It starts looking that the antiderivative itself might be not as scary as it seemed in the beginning. But it needs some more work... $\endgroup$ – Vladimir Reshetnikov Dec 13 '15 at 4:28
  • $\begingroup$ @VladimirReshetnikov are there any values $0<a<b<c$ such that this integral can be expressed in terms of elementary functions only, without $\text{Li}_2$ and $\text{Li}_3$? $\endgroup$ – user294724 Dec 13 '15 at 20:07
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Some further raw results (largely inspired by Cleo's answer :+1) for the impatient :

$$I\left(4,1\right)=- \frac{1915}{128}\zeta(3)+\left(\frac {359}{192}\ln(2) + 2\ln(3) + \frac{175}{96}\ln(5)\right)\pi^2 +\frac{513}{32}\ln(2)^2\ln(5) - \frac{1535}{96}\ln(2)^3 - \frac {15}4\ln(3)^3 - \frac{1031}{96}\ln(5)^3 - \frac{31}8\ln(2)\ln(5)^2 + \frac{93}8\ln(3)\ln(5)^2 + \frac {45}4\operatorname{Li}_3(1/3) +\frac{93}8\operatorname{Li}_3(2/3) - \frac{81}{32}\operatorname{Li}_3(1/5) - \operatorname{Li}_3(2/5) - \frac 38\operatorname{Li}_3(3/5) - \frac {91}{32}\operatorname{Li}_3(4/5)- \frac 38\operatorname{Li}_3(1/6) + \frac 5{16}\operatorname{Li}_3(1/10) \\- \ln(2)\left(12\operatorname{Li}_2(1/3) + 8\operatorname{Li}_2(1/4) + 2\operatorname{Li}_2(1/5)+\frac {163}{16}\operatorname{Li}_2(2/5) + \frac {195}{16}\operatorname{Li}_2(3/5)\right) \\- \frac{189}{16}\,\ln(5)\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))$$


$$I\left(5,1\right)=- \frac{577}{160}\zeta(3)+\left(\frac {11}{15}\ln(3) + \frac{191}{480}\ln(4) - \frac{11}{60}\ln(5)\right)\pi^2 + \frac{17}{10}\ln(3)\ln(5)^2 - \frac{17}{30}\ln(3)^3 - \frac{173}{960}\ln(4)^3 - \frac{181}{120}\ln(5)^3 - \frac 1{10}\ln(4)\ln(5)^2 + \frac{51}{160}\ln(4)^2\ln(5) + 2\operatorname{Li}_3(1/3) + \frac 75\operatorname{Li}_3(2/3) - \frac 38\operatorname{Li}_3(1/5) + \frac 9{10}\operatorname{Li}_3(2/5) - \frac 35\operatorname{Li}_3(3/5) -\frac 18\operatorname{Li}_3(4/5) + \frac 14\operatorname{Li}_3(1/10) - \frac 75\ln(2)\,\left(\operatorname{Li}_2(1/3)+\frac{\operatorname{Li}_2(1/4)}2 +\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5)\right) - \ln(5)\,\left(\operatorname{Li}_2(2/5)+2\operatorname{Li}_2(3/5)\right)$$


$$I\left(\frac 12,1\right)=-\frac{39}8\zeta(3)+\left(\ln(2)+5\ln(3)+2\ln(5)\right)\zeta(2)-\frac 5{12}\ln(3)^3+\ln(2)^2\ln(25/8)-\ln(25/9)\ln(5)^2 -\ln(4)\left(\operatorname{Li}_2(1/3)+\frac 34 \operatorname{Li}_2(1/4)\right)-(\ln(4)+2\ln(5))\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))+\frac 12\operatorname{Li}_3(1/3)+2\operatorname{Li}_3(2/3)$$


$$I\left(\frac 13,1\right)=-\frac{139}{24}\zeta(3)+\left(\frac 94\ln(2)-\frac 5{12}\ln(3)+\frac 13\ln(5)\right)\pi^2 + 2\ln(2)^2\ln(5) - 2\ln(2)^3 -\frac 23\ln(3)^3 - 2\ln(5)^3 + 2\ln(3)\ln(5)^2 + 2\operatorname{Li}_3(1/3) + 2\operatorname{Li}_3(2/3) - 2\ln(2)\,\left(\operatorname{Li}_2(1/3)+\frac 12\operatorname{Li}_2(1/4)+\operatorname{Li}_2(2/5) +\operatorname{Li}_2(3/5)\right) - 2\ln(5)\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))$$


$$I\left(\frac 23,1\right)=\frac{573}{128}\zeta(3)+\left(\frac{31}{64}\ln(2)-\frac 23\ln(3)-\frac 5{48}\ln(5)\right)\pi^2 + \frac {47}{32}\ln(2)^3 + \frac 34\ln(3)^3 + \frac {79}{32}\ln(5)^3 - 3\ln(2)\ln(5)^2 + \frac{45}{32}\ln(2)^2\ln(5) - \frac 74\ln(3)\ln(5)^2 - \frac 54\operatorname{Li}_3(1/3) - \frac 94\operatorname{Li}_3(2/3) - \frac {17}{32}\operatorname{Li}_3(1/5) - \operatorname{Li}_3(1/6) - \frac {13}8\operatorname{Li}_3(2/5) - \operatorname{Li}_3(3/5) - \frac {27}{32}\operatorname{Li}_3(4/5) - \frac 5{16}\operatorname{Li}_3(1/10) + \ln(2)\,\left(\frac 54\operatorname{Li}_2(1/3) + \frac 58\operatorname{Li}_2(1/4)\right) + (\ln(3)-\ln(2))\operatorname{Li}_2(1/5) + \frac{3\ln(2)+7\ln(5)}4\,(\operatorname{Li}_2(2/5) + \operatorname{Li}_2(3/5))$$


$$I\left(\frac 14,1\right)=\frac{335}{32}\zeta(3)-\left(\frac{95}{48}\ln(2)+\frac 23\ln(3)+\frac 1{24}\ln(5)\right)\pi^2 + \frac {107}{24}\ln(2)^3 + \frac 73\ln(3)^3 + \frac {41}8\ln(5)^3 - \ln(2)\ln(5)^2 - \frac{21}8\ln(2)^2\ln(5) - \frac {11}2\ln(3)\ln(5)^2 - 7\operatorname{Li}_3(1/3) - \frac{11}2\operatorname{Li}_3(2/3) - \frac 78\operatorname{Li}_3(1/5) + \operatorname{Li}_3(2/5) - \frac 32\operatorname{Li}_3(3/5) + \frac 38\operatorname{Li}_3(4/5) - \frac 32\operatorname{Li}_3(1/6)- \frac 54\operatorname{Li}_3(1/10) + \ln(2)\;\left(4\operatorname{Li}_2(1/3) + 2\operatorname{Li}_2(1/4) + \frac 54\operatorname{Li}_2(2/5) + \frac {13}4\operatorname{Li}_2(3/5)\right) + \frac {19}4\ln(5)\;\left(\operatorname{Li}_2(2/5) + \operatorname{Li}_2(3/5)\right)$$


Raw expressions for (the much needed) simplifications and manipulations
($I(4,1),I(5,1),I(1/2,1),I(1/3,1),I(2/3,1),I(1/4,1)$ respectively)

- 1915/128*zeta(3)+(359/192*ln(2) + 2*ln(3) + 175/96*ln(5))*PI^2 +(513/32*ln(2)^2*ln(5) - 1535/96*ln(2)^3 - 15/4*ln(3)^3 - 1031/96*ln(5)^3 - 31/8*ln(2)*ln(5)^2  + 93/8*ln(3)*ln(5)^2 + 45/4*polylog(3, 1/3) +93/8*polylog(3, 2/3) - 81/32*polylog(3, 1/5) - 3/8*polylog(3, 1/6) - polylog(3, 2/5) - 3/8*polylog(3, 3/5) - 91/32*polylog(3, 4/5) + 5/16*polylog(3, 1/10) - 12*ln(2)*polylog(2, 1/3) - 8*ln(2)*polylog(2, 1/4) - 2*ln(2)*polylog(2, 1/5) - 163/16*ln(2)*polylog(2, 2/5) - 195/16*ln(2)*polylog(2, 3/5) - 189/16*ln(5)*polylog(2, 2/5) - 189/16*ln(5)*polylog(2, 3/5))

- 577/160*zeta(3) +(191/240*ln(2)+11/15*ln(3)-11/60*ln(5))*PI^2 + 17/10*ln(3)*ln(5)^2 - 17/30*ln(3)^3 - 173/960*ln(4)^3 - 181/120*ln(5)^3  - 1/10*ln(4)*ln(5)^2 + 51/160*ln(4)^2*ln(5) + 2*polylog(3, 1/3) + 7/5*polylog(3, 2/3) - 3/8*polylog(3, 1/5) + 9/10*polylog(3, 2/5) - 3/5*polylog(3, 3/5) - 1/8*polylog(3, 4/5) + 1/4*polylog(3, 1/10) - 7/10*ln(4)*(polylog(2,1/3)+polylog(2,1/4)/2+polylog(2,2/5)+polylog(2,3/5)) - ln(5)*polylog(2, 2/5) - 2*ln(5)*polylog(2, 3/5)

-39/8*zeta(3)+(ln(2)+5*ln(3)+2*ln(5))*zeta(2)-5/12*ln(3)^3+ln(2)^2*ln(25/8)-ln(25/9)*ln(5)^2 -ln(4)*(polylog(2,1/3)+3/4*polylog(2,1/4))- (ln(4)+2*ln(5))*(polylog(2,2/5)+polylog(2,3/5))+polylog(3,1/3)/2+2*polylog(3,2/3)

-139/24*zeta(3)+(9/4*ln(2)-5/12*ln(3)+1/3*ln(5))*PI^2 + 2*ln(2)^2*ln(5) - 2*ln(2)^3 -2/3*ln(3)^3 - 2*ln(5)^3  + 2*ln(3)*ln(5)^2 + 2*polylog(3,1/3) + 2*polylog(3, 2/3) - 2*ln(2)*(polylog(2, 1/3)+polylog(2, 1/4)/2+polylog(2, 2/5) +polylog(2, 3/5)) - 2*ln(5)*(polylog(2,2/5)+polylog(2, 3/5))

573/128*zeta(3)+(31/64*ln(2)-2/3*ln(3)-5/48*ln(5))*PI^2 + 47/32*ln(2)^3 + 3/4*ln(3)^3 + 79/32*ln(5)^3 - 3*ln(2)*ln(5)^2 + 45/32*ln(2)^2*ln(5) - 7/4*ln(3)*ln(5)^2 - 5/4*polylog(3, 1/3) - 9/4*polylog(3, 2/3) - 17/32*polylog(3, 1/5) - polylog(3, 1/6) - 13/8*polylog(3, 2/5) - polylog(3, 3/5) - 27/32*polylog(3, 4/5) - 5/16*polylog(3, 1/10) + ln(2)*(5/4*polylog(2,1/3) + 5/8*polylog(2, 1/4)) + (ln(3)-ln(2))*polylog(2, 1/5) + (3*ln(2)+7*ln(5))/4*(polylog(2, 2/5) + polylog(2, 3/5))

335/32*zeta(3)-(95/48*ln(2)+2/3*ln(3)+1/24*ln(5))*PI^2 + 107/24*ln(2)^3 + 7/3*ln(3)^3 + 41/8*ln(5)^3 - ln(2)*ln(5)^2 - 21/8*ln(2)^2*ln(5) - 11/2*ln(3)*ln(5)^2 - 7*polylog(3, 1/3) - 11/2*polylog(3, 2/3) - 7/8*polylog(3, 1/5) + polylog(3, 2/5) - 3/2*polylog(3, 3/5) + 3/8*polylog(3, 4/5) - 3/2*polylog(3, 1/6)- 5/4*polylog(3, 1/10) + ln(2)*(4*polylog(2, 1/3) + 2*polylog(2, 1/4) + 5/4*polylog(2, 2/5) + 13/4*polylog(2,3/5)) + 19/4*ln(5)*(polylog(2, 2/5) + polylog(2, 3/5))
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  • $\begingroup$ Can you do $I\left(\frac23,1\right)$? $\endgroup$ – Vladimir Reshetnikov May 15 '15 at 17:22
  • $\begingroup$ @VladimirReshetnikov: yes it works (I'll update my answer later). $\endgroup$ – Raymond Manzoni May 15 '15 at 17:34
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    $\begingroup$ @Ant: that's pslq (or LLL) at work! Compute a value to high precision (say $400$ digits), conjecture a vector of expressions like $[\ln(3)^3, \ln(3)\ln(2)^2,\zeta(3)]$ and so on and the software will use an integer relation algorithm to return you the wished integer coefficients like here. $\endgroup$ – Raymond Manzoni May 15 '15 at 18:15
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    $\begingroup$ That $\dfrac14$ looks so lonely there... He's searching for his brother, $\dfrac34$ , and he's all sad, and crying... :-$)$ $\endgroup$ – Lucian May 15 '15 at 18:15
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    $\begingroup$ @Ant: Of course without a proof you can't be sure but I could challenge you to disprove them! :-) This should be easy by increasing the precision but the neat thing about the integer relation algorithms is that you will get the same constants by increasing the precision once you crossed the minimal threshold while in my link the constants will change rather randomly with the number of digits (the total count of digits staying near the fixed precision). Of course this could be an illusion with a second threshold much higher and so on (like when playing with $\theta$ functions). $\endgroup$ – Raymond Manzoni May 16 '15 at 17:22

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