17
$\begingroup$

The following problem is proposed by Cornel Valean:$$\operatorname{P.V.} \int_0^1 \frac{1}{x (1-x)}\arctan \left(\frac{8 x^2-4 x^3+14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16}\right) \textrm{d}x=\log \left(\frac{5}{4}\right) \arctan\left(\frac{1}{2}\right).$$

The question is how can we prove this equality without using the main result in this paper and without the use of complex numbers?

$\endgroup$
21
  • 1
    $\begingroup$ Probably the arctan is a sum of arctan simpler. $\endgroup$ – FDP Feb 1 at 12:36
  • 2
    $\begingroup$ @Ali Shadhar: Of course not (for now) ;). Still searching ! $\endgroup$ – FDP Feb 2 at 3:56
  • 4
    $\begingroup$ The arctangent term splits into $\arctan\left(\frac {2}{x^2-x-4}\right)+ \arctan\left(\frac {-4x}{2x^2-x-4}\right)$. I don’t think that helps though. $\endgroup$ – Sophie Feb 4 at 12:58
  • 4
    $\begingroup$ @Ali Shadar: Sorry but here in my country this integral doesn't exist. To be correct the question needs to mention that this is the principal value that is required. $\endgroup$ – FDP Feb 7 at 13:58
  • 2
    $\begingroup$ The question does not really fulfill the conditions of a question on this site. We have indeed the origin of the question, but there is a lot of context missing. As it is stated, it is rather a puzzle, some integral which somebody could compute has some value, the puzzle being to reconstruct this value constrained to the prohibition of solving the integral without the simplest methods used to solve such integrals. Well, i will drop an answer, but please show the own effort to solve the integral. In this way, we will clearly see what is allowed. $\endgroup$ – dan_fulea Feb 8 at 11:59
12
+100
$\begingroup$

Too long for a comment.

Since $$T(x)=\dfrac{-4x^3 +8x^2+14x-8}{2x^4-3x^3-11x^2+16x+16} =\dfrac{R(x)-S(x)}{1+R(x)S(x)},$$ where $$R(x)=\dfrac2{x^2-x-4},\quad S(x)=\dfrac{4x}{2x^2-x-4},\quad x\in[0,1],\tag1$$ (pointed by Sophie), then $$\arctan T(x) = \arctan R(x) - \arctan S(x),$$ (see also WA plot)

Decompozition of arctangent

$$I=\int\limits_0^1\arctan T(x)\; \dfrac{\text dx}{x(1-x)} =\int\limits_0^1\arctan T(x)\;\left(\dfrac1x+\dfrac1{1-x}\right)\,\text dx, $$$$ I=\int\limits_0^1\big(2\arctan R(x)-\arctan S(x)-\arctan S(1-x)\big)\; \dfrac{\text dx}{x},\tag2$$ wherein $$S(1-x)=\dfrac{4-4x}{2x^2-3x-3}.\tag{1a}$$ This transformation makes the integral convergent, and the numerical integration gives $$I\approx 0.10345,99740\,30782\,04062\,03377\,83301\,52783\,80081\,8.$$ Obtained result exactly corresponds to the OP value.

However, analytical result does not obtained.

We can additionally decompose arctangents of $\;R(x),S(x),S(1-x)\;$ to the elementary terms.

Assuming $$R(x) = \dfrac{(a+2cx)+(b-2cx)}{1-(a+2cx)(b-2cx)},$$ and taking in account coefficients proportionality, one can get $$ \begin{cases} a+b=8c^2\\ 2c(b-a)=4c^2\\ ab-1=16c^2\\ c>0 \end{cases}\Rightarrow \begin{cases} a=4c^2-c\\ b=4c^2+c\\ 16c^4-17c^2-1=0\\ c>0 \end{cases} $$$$ c=\sqrt{\dfrac{\sqrt{353}+17}{32}}\approx1.05753\,68526,\tag3 $$$$ a=4c^2-c,\quad b=4c^2+c,$$ $$\arctan R(x)=\arctan(a+2cx)+\arctan(b-2cx),$$ $$\arctan R(x)=\arctan(4c^2+c(2x-1))+\arctan(4c^2-c(2x-1)).\tag4$$

Similarly, assuming $$S(x) = \dfrac{-4x}{-2x^2+x+4} = \dfrac{(gx+d)+(hx-d)}{1-(gx+d)(hx-d)},$$ one can get $$ \begin{cases} g+h=-(1+d^2)\\ 4d(g-h)=1+d^2\\ 2gh=1+d^2\\ d>0 \end{cases}\Rightarrow \begin{cases} (2g+1)+(2h+1)=-2d^2\\ (2g+1)(2h+1)=1\\ 2d((2g+1)-(2h+1))=1+d^2\\ d>0, \end{cases} $$$$ \begin{cases} (2g+1),(2h+1)=-d^2\pm\sqrt{d^4-1}\\ 4d\sqrt{d^4-1}=1+d^2\\ d>0, \end{cases} \begin{cases} 16d^4-17d^2-1=0\\ g,h=\dfrac{-(1+d^2)\pm\sqrt{d^4-1}}2\\ d>0, \end{cases} $$$$ g,h=\dfrac{1+d^2}{-(1+d^2)\mp\sqrt{d^4-1}} =\dfrac{4d}{-4d\mp1}, $$$$ d=\sqrt{\dfrac{\sqrt{353}+17}{32}}=c,\quad g = -\dfrac{4c}{4c+1},\quad h = -\dfrac{4c}{4c-1}, $$$$ \arctan S(x) = \arctan(gx+d)+\arctan(hx-d) $$$$ \arctan S(x) = \arctan\dfrac{4c^2+c-4cx}{4c+1}-\arctan\dfrac{4c^2-c+4cx}{4c-1} \tag5 $$$$ \arctan S(1-x) = \arctan\dfrac{4c^2-3c+4cx}{4c+1}-\arctan\dfrac{4c^2+3c-4cx}{4c-1} .\tag6 $$ Obtained integrals allow analytical calculation via integral logarithmic function of complex argument.

Analytpcal integration

Taking the real parts, one can get $$J(p,q,x) = \int \arctan(px+q)\,\dfrac{\text dx}{x}=J_1(p,q,x)+J_2(p,q,x)+J_3(p,q,x),\tag7$$ where $$J_1(p,q,x) = (\ln p+\ln x)\arctan(px+q),$$ $$J_2(p,q,x)=\dfrac12\Im\left(\text{Li}_2\left(1+\dfrac{px}{q-i}\right)-\text{Li}_2\left(1+\dfrac{px}{q+i}\right)\right), $$$$ J_3(p,q,x)=\dfrac12\Im\big(\ln(q+i)\ln(1-i(px+q))-\ln(q-i)\ln(1+i(px+q))\big).$$ From $(4),(5),(6)$ should that the term with $\;\ln x\;$ does not influent to the goal definite integral. Therefore, we can assume $$J_1(p,q,1)-J_1(p,q,0) = \ln p\arctan\,\dfrac p{1+(p+q)q}.\tag8$$ Also, $$J_2(p,q,1)-J_2(p,q,0) = \dfrac12\Im\left(\text{Li}_2\left(1+\dfrac{p}{q-i}\right)-\text{Li}_2\left(1+\dfrac{p}{q+i}\right)\right),$$ $$J_3(p,q,x) = \dfrac12\Im\bigg(\ln(q+i)\ln(1-i(px+q))+\ln(q+i)\ln(1+i(px+q)) $$$$ -\ln(q+i)\ln(1+i(px+q))-\ln(q-i)\ln(1+i(px+q))\bigg)$$ $$= \dfrac12\Im\bigg(\ln(q+i)\ln(1+(px+q)^2)-\ln(q^2+1)\ln(1+i(px+q))\bigg),$$ $$J_3(p,q,1)-J(p,q,0) = \dfrac12\Im\bigg(\ln(q+i)\ln\,\dfrac{1+(p+q)^2}{1+q^2}-\ln(q^2+1)\ln\dfrac{1+i(p+q)}{1+iq}\bigg) $$$$ = \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln\dfrac{(1+i(p+q))(q+i)}{1+iq}\bigg) $$$$ = \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln\dfrac{(1+i(p+q))(q+i)(1-iq)}{1+q^2}\bigg) $$$$ = \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln((q+i)^2(p+q-i)) \bigg) $$

But this approach contradicts to the OP requirements.

$\endgroup$
15
  • 4
    $\begingroup$ Thanks for the efforts Yuri . +1 $\endgroup$ – Ali Shadhar Feb 8 at 13:46
  • 1
    $\begingroup$ $+100$ for the hard work (leaving aside the requirements). Besides, maybe the MSE community should seriously think to consider creating, say, MSE Extreme, where questions of this type and with similar requirements could be asked without being concerned that they would possibly be too tough, too difficult, and wouldn't fulfill some requirements of the site. $\endgroup$ – user97357329 Feb 9 at 20:53
  • 2
    $\begingroup$ @user97357329 One would have to be a prophet to know in advance how hard an answer turns out to be. Also, maybe there is still an easy solution out there (just need to have a good idea). I don't see anything on this question which would not fulfill MSE requirements. On contrary, I would hate to see MSE to turn into a site solving just homeworks and similar "easy" stuff. $\endgroup$ – pisoir Feb 9 at 21:50
  • 1
    $\begingroup$ There’s an easier way to split the arctangent into the four terms (it’s how I found the one in my comment) by associating it with the argument of a complex number $\arctan(\dots)=\arg(2x^4-3x^3-11x^2+16x+16+i(-4x^3+8x^2+14x-8))$ and then factoring that as a polynomial with coefficients in $\mathbb Z[i]$. There’s a bit of trouble after that to split the square root of a complex number into real and imaginary parts but I still think it’s overall less computation. $\endgroup$ – Sophie Feb 10 at 0:15
  • 1
    $\begingroup$ great answer Yuri!! Congrats. $\endgroup$ – Ali Shadhar Feb 10 at 13:16
5
$\begingroup$

Here is a full detailed answer respecting the rules of the OP as seen in the revision #1:

https://math.stackexchange.com/posts/4007626/revisions

The peculiar added condition of "not using complex numbers" came after some hours.

Some differences between revisions

It is easy to add those $39$ characters without suggesting why or how this should work.

But well, even so, there is certainly a way to "avoid $\Bbb C$" below, by using a messy language. All i need is a close form for a primitive function for $\int\frac 1x\arctan F(x)\; dx$ with $F$ rational. One can write this primitive function as if there would be no complex numbers on this world. The reader really willing to do this can do it, just write formulas for the real and for the imaginary part of the logarithm and dilogarithm. (Then work with them.) For instance, for $z=r(\cos\theta+i\sin \theta)\in\Bbb C$ we have explicit real functions $z\to|z|$ and $z\operatorname{arg}z$ and the formula $\log (r(\cos\theta+i\sin \theta))=\log r +i\theta$ can be written without using complex numbers, when we need only $\log z$ and $\log^2 z$, but where is the point in doing so?!

So i will write these primitives using complex numbers, since my concern is rather on concise ideas, not on typing to avoid $i$, and since my hand has to type all the stuff, in a magnitude highly contrasting the OP in detail and precision. The OP does not even mention any try, and - by the way - does not satisfy (m)any of the current common guide lines of this site.

The accepted answer also uses $\Im$, so i will do it too. We see numbers and prove the result as stated. Well, with complex numbers.


Outline:

  • $(1)$ Use Sophie's splitting, which deserves its own line: $$ \tag{$1$} \arctan h(x) = \arctan 2{x^2-x-4}-\arctan\frac{4x}{2^2-x-4}\ . $$

  • $(2)$ Get rid of the P.V. by resolving the singularities. For this write $\frac 1{x(1-x)}$ as $\frac 1x+\frac1{1-x}$, for the expressions in $\frac 1{1-x}$ use the substitution $y=1-y$, and resolve the uniqe singularity that comes from $\frac 1x$. This leads to an expression of the given integral $J$ in the form $$ \tag{$2$} J = 2J_1-J_2-J_3\ , $$ where $J_1$ is an integral of $\displaystyle\frac 1x\arctan F(x)$ on $[0,1]$, and $J_2,J_3$, are similar, replacing the rational function $F$ by other rational functions $G,H$.

  • $(3)$ Use $$ \tag{$3$} \arctan t = \frac 1{2i}\log\frac{1+it}{1-it} $$ to pass from $\arctan$ to $\log$, which can be further splitted.

  • $(4)$ Write $$ \tag{4} \frac {1+iF(x)}{1-iF(x)}=\frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\ , $$ and split the logarithm. Similarly for $G,H$, where the values $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ appear. Then we integrate the pieces using the formula:

  • $(5)$ $$ \tag{$5$} \int\frac 1x\log(x-a)\; dx % = \log(x-a)\log \frac xa % +\operatorname{Li}_2\left(1-\frac xa\right) + C'' % = % \log(x-a)\log \frac xa % -\operatorname{Li}_2\left(\frac xa\right) + C' % -\log\left(1-\frac xa\right)\log\frac xa % = % \log(-a)\log\frac xa -\operatorname{Li}_2\left(\frac xa\right) + C' = \log(-a)\log x -\operatorname{Li}_2\left(\frac xa\right) + C \ . $$

  • $(6)$ We have explicit real numbers for the integrals $J_1,J_2,J_3$ so far. We work rather in $i\Bbb R$, seen as $\Bbb C\cong \Bbb R\oplus i\Bbb R$ taken modulo the first $\oplus$ component $\Bbb R$. These integrals are sum of terms involving dilogarithms and products of two logarithms, calculated in points which are expressions in $a_1,a_2,a_3,a_4$ for $J_1$, and similarly for $J_2,J_3$. We use dilogarithmic identities to get rid of all dilogaritm values. The properties of the dilogarithm, allowing to express $\operatorname{Li}_2(1-z)$ and $\operatorname{Li}_2(1/z)$ in terms of $\operatorname{Li}_2 z$, are useful:

  • $(7)$ We exhibit a simple hidden relation between the "$a,b,c$-values", which immediately leads to the simplification of the logarithm products.

Let's go.


  • $(1)$ and $(2)$

Let $J$ be the given expression $$ J = \lim_{\epsilon\searrow 0}\int_{0+\epsilon}^{1-\epsilon} \frac 1 {x(1-x)}\arctan \underbrace{\frac{-4x^3 +8 x^2 +14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16} }_{h(x)} \;dx \ . $$ I hate P.V. - so we split $\displaystyle \frac 1{x(1-x)}=\frac 1x+\frac 1{1-x}$ and accordingly the integral into two pieces. The special values of $h$ in zero and one are $h(0) = -8/16=-1/2$, $h(1) = 10/20=1/2$. So $h(0)+h(1)=0$, giving $$ \begin{aligned} J &= \lim_{\epsilon\searrow 0} \int_{0+\epsilon}^{1-\epsilon} \frac 1x(\arctan h(x)-\arctan h(0))\; dx \\ &\qquad\qquad+ \lim_{\epsilon\searrow 0} \int_{0+\epsilon}^{1-\epsilon} \frac 1{1-x}(\arctan h(x)- \arctan h(1))\; dx \\ &= \int_0^1 \frac 1x(\arctan h(x)-\arctan h(0))\; dx + \int_0^1 \frac 1{1-x}(\arctan h(x)-\arctan h(1))\; dx \\ &= \int_0^1 \frac 1x(\arctan h(x)-\arctan h(0))\; dx + \int_0^1 \frac 1x(\arctan h(1-x)-\arctan h(1))\; dx \\ &= 2 \underbrace{\int_0^1\frac 1x\arctan \frac {x^2-x}{2(x^2-x-5)}\; dx}_{J_1} \\ &\qquad\qquad- \underbrace{\int_0^1\frac 1x\arctan \frac {4x}{2x^2-x-4}\; dx}_{J_2} \\ &\qquad\qquad\qquad\qquad- \underbrace{\int_0^1\frac 1x\arctan \frac {8x^2-24x}{6x^2+7x-25}\; dx}_{J_3} \ . \end{aligned} $$ At the last step we have used first Sophie's splitting, which deserves again its own line: $$ \arctan h(x) = \arctan 2{x^2-x-4}-\arctan\frac{4x}{2^2-x-4}\ , $$ to undo the chicanery in the expression of the integral.

(My interest in this problem came from the explicit computation of the separated integrals $J_1$, $J_2$, $J_3$. Yes, this is possible.)


  • $(3)$

Now we use $\displaystyle\arctan t = \frac 1{2i}\log\frac{1+it}{1-it}$, replace each $\arctan$ by $\log$, split every $\log$ by computing roots of equations of second degree for the three functions $$ \begin{aligned} F(x) &=\frac {x^2-x}{2(x^2-x-5)}\ ,\\ G(x) &=\frac {4x}{2x^2-x-4}\ , \\ H(x) &= \frac {8x^2-24x}{6x^2+7x-25}\ . \end{aligned} $$ ("Periods" may be introduced, but will not, final numerical checks are enough.)


  • $(4)$:

We have explicitly: $$ \begin{aligned} \frac {1+iF(x)}{1-iF(x)} &= \frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\ ,\\ \frac {1+iF(x)}{1-iF(x)} &= \frac{b_3b_4}{b_1b_2}\cdot\frac{(x-b_1)(x-b_2)}{(x-b_3)(x-b_4)}\ ,\\ \frac {1+iF(x)}{1-iF(x)} &= \frac{c_3c_4}{c_1c_2}\cdot\frac{(x-c_1)(x-c_2)}{(x-c_3)(x-c_4)}\ , \end{aligned} $$ where $$ \begin{aligned} a_1 &= \frac 12(1 - \sqrt{17-8i})\ , & a_3 &= \frac 12(1-\sqrt{17+8i})=\bar a_1\\ a_2 &= \frac 12(1 + \sqrt{17-8i})\ , & a_4 &= \frac 12(1+\sqrt{17+8i})=\bar a_2\ , \\[2mm] b_1 &= \frac 14(1-4i - \sqrt{17-8i})\ , & b_3 &= \frac 14(1+4i - \sqrt{17+8i})=\bar b_1\ ,\\ b_2 &= \frac 14(1-4i + \sqrt{17-8i})\ , & b_4 &= \frac 14(1+4i + \sqrt{17+8i})=\bar b_2\ , \\[2mm] c_1 &= \frac 14(3+4i - \sqrt{17-8i})\ , & c_3 &= \frac 14(3-4i - \sqrt{17+8i})=\bar c_1\ ,\\ c_2 &= \frac 14(3+4i + \sqrt{17-8i})\ , & c_4 &= \frac 14(3-4i + \sqrt{17+8i})=\bar c_2\ . \end{aligned} $$ One can replace the square root $\sqrt{17-8i}$ by its longer version, using: $$ \begin{aligned} 2\Re \sqrt{17-8i} & = \sqrt{17-8i}+\sqrt{17+8i} =\sqrt{\ \left(\sqrt{17-8i} + \sqrt{17+8i}\right)^2\ }\\ &=\sqrt{34+2\sqrt{17^2+8^2}}=\sqrt{34+2\sqrt{353}}\ , \\[3mm] 2i\Im \sqrt{17-8i} & = \sqrt{17-8i}-\sqrt{17+8i} =\frac{(17-8i)-(17+8i)}{\sqrt{34+2\sqrt{353}}}\\ &=-i\sqrt{-34+2\sqrt{353}}\ . \end{aligned} $$


  • $(5)+(6)$:

We compute $J_1$, $J_2$, $J_3$ to some extent. In fact, i wanted to computed $J_1$ explicitly, and this is possible, we have a close formula for it involving the functions $\log$ and $\arctan$ at special places. For $J_2$, $J_3$ individually, we have to add the dilogarithm.

  • $(5)+(6)$ The computation of $J_1$:

We have collected all ingrediends. The following computations are considered in $i\Bbb R$ modulo the real part $\Bbb R$, and we will write $\equiv$ for this modulo computation.

Let $a$ be a number in the list $a_1,a_2;a_3,a_4$. Then $$ \int\frac 1x\log(x-a)\; dx % = \log(x-a)\log \frac xa % +\operatorname{Li}_2\left(1-\frac xa\right) + C'' % = % \log(x-a)\log \frac xa % -\operatorname{Li}_2\left(\frac xa\right) + C' % -\log\left(1-\frac xa\right)\log\frac xa % = % \log(-a)\log\frac xa -\operatorname{Li}_2\left(\frac xa\right) + C' = \log(-a)\log x -\operatorname{Li}_2\left(\frac xa\right) + C \ . $$ The above primitive makes sense in $x=1$. For $x\to 0$ we may have problems with $\log x$, but there are none after we consider all contributions from $\log\left(\frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\right)$ to the factor $\log x$:

$$ \lim_{\epsilon\to 0}\int_\epsilon^1\frac1x\log\frac{x-a}{-a}\; dx =-\operatorname{Li}_2\left(\frac 1a\right)\ , $$ since the two $\log\epsilon$ terms finally cancel.

Note that $(a_1,a_3)$ and $(a_2,a_4)$ are pairs of conjugated complex numbers, $a_1=\bar a_3$, $a_2=\bar a_4$. The result of the integration is thus in $i\Bbb R$: $$ \begin{aligned} 2i\; J_1 &= 2i\;\int_0^1\frac 1x\arctan F(x)\; dx \\ &= 2i\;\int_0^1\frac 1x\log\left( \frac{a_3a_4}{a_1a_2}\cdot\frac{(x-a_1)(x-a_2)}{(x-a_3)(x-a_4)}\right)\; dx \\ &= -\operatorname{Li}_2\left(\frac 1{a_1}\right) -\operatorname{Li}_2\left(\frac 1{a_2}\right) +\operatorname{Li}_2\left(\frac 1{a_3}\right) +\operatorname{Li}_2\left(\frac 1{a_4}\right) \\ &\equiv -2\operatorname{Li}_2\left(\frac 1{a_1}\right) -2\operatorname{Li}_2\left(\frac 1{a_2}\right) \text{ since $a_3=\bar a_1$ and $a_4=\bar a_2$} \\ &= 2\underbrace{ (\operatorname{Li}_2(a_1) +\operatorname{Li}_2(1-a_1))}_{\pi^2/6 -\log a_1\log a_2} \\&\qquad\qquad +\log^2 (-a_1)+\log^2(-a_2) \\ &\qquad\qquad\text{ and one may STOP HERE, but optionally let's get $J_1$ explicitly...} \\ &\equiv -2\log a_1\log a_2 \\ &\qquad\qquad + \log^2 a_1+\log^2a_2 +2i\pi\log\frac{a_2}{a_1} \\ &\equiv \log^2\frac {a_1}{a_2} +2i\pi\log\frac {a_2a_4}{a_1a_3} \\ &= \log^2\frac {a_1^2}{a_1a_2} -2\pi i\log\frac {a_1^2a_3^2}{a_1a_2a_3a_4} \\ &= \log^2\frac {a_1^2}{-4+2i} -2\pi i\log\frac {a_1^2a_3^2}{20} \\ &\equiv \left( \frac 12 \log\frac {(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2} {4^2\cdot 20} + i\left(2\operatorname{arg}(a_1)+\arctan\frac 12-\pi\right)\right)^2 \\ &\qquad -i\pi\; \log\frac{(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2}{4^2\cdot 20} \\ &\equiv i \log\frac {(1 - \sqrt{34+2\sqrt{353}} + \sqrt{353})^2} {4^2\cdot 20} \cdot \left(2\operatorname{arg}(a_1)+\arctan\frac 12-2\pi\right) \ . \end{aligned} $$ We have used $a_1+a_2=1=a_3+a_4$, $a_1a_2=-4+2i$, $a_3a_4=-4-2i$, $a_1a_2a_3a_4=20$.

Here, $\operatorname{arg}(a_1)$ is explicit, just use the real part and the absolute value (or the imaginary part instead) to write this as a value of the $\arcsin$ or $\arccos$ or $\arctan$: $$ \begin{aligned} \Re a_1 &=a_1+a_3=1-\frac 12\sqrt{34+2\sqrt{353}})\ ,\\ |a_1|^2 &= a_1\bar a_1 =a_1a_3 = \frac 14(1-\sqrt{34+2\sqrt{353}}+\sqrt{353})\ . \end{aligned} $$ Note that we need to take care which sign is taken in each of the cases $\log(-a_j)=\log(a_j)\pm\log(-1)$. The sign is plus for $a_2,a_3$ and minus for $a_1,a_4$.

The result is an element in $i\Bbb R$. The $\arctan\frac 12$ comes when getting the argument for $\displaystyle \frac{-4-2i}{-4+2i}= \frac{-2-i}{-2+i}= \frac{(2+i)^2}{2^2+1^2}$, which is twice the argument of $2+i$.


For a numerical check, look in the previous version of this edited answer. (Or in the commented block of this version.) A lot of work. $$ % Numerical check for $J_1$. We use sage. % % F = 1/2*(x^2 - x)/(x^2 - x - 5) % a1, a2 = ((1+i*F)/(1-i*F)).numerator() .roots(multiplicities=False, ring=QQbar) % a3, a4 = ((1+i*F)/(1-i*F)).denominator().roots(multiplicities=False, ring=QQbar) % % A1, A2, A3, A4 = a1.n(200), a2.n(200), a3.n(200), a4.n(200) # approximations % % J1 = integral( atan(F(x)) / x, x, 0, 1, hold=True ).n(200) % print(f'Numerical value of J1 is:\n{J1}') % % def T(a): return -dilog(1/a) % J1_v1 = 1/(2*i) * ( T(A1) + T(A2) - T(A3) - T(A4) ) % print(f'Numerical value of J1 rewritten as signed sum of dilog values is:\n{J1_v1}') % % J1_v2 = 1/(2*i) * ( dilog(A1) + dilog(A2) - dilog(A3) - dilog(A4) % + 1/2 * ( log(-A1)^2 + log(-A2)^2 - log(-A3)^2 - log(-A4)^2 ) ) % print(f'Numerical value of J1 rewritten as signed sum of dilog and log values is:\n{J1_v2.n(200)}') % % J1_v3 = 1/(2*i) * ( -log(A1)*log(A2) + log(A3)*log(A4) % + 1/2 * ( log(-A1)^2 + log(-A2)^2 - log(-A3)^2 - log(-A4)^2 ) ) % print(f'Numerical value of J1 rewritten using only log values is:\n{J1_v3.n(200)}') % % J1_v4 = 1/(2*i) * 1/2 * ( log(A1/A2)^2 - log(A3/A4)^2 + 2*i*pi*log(A2*A4/A1/A3)) % print(f'Numerical value of J1 rewritten as signed sum of log values is:\n{J1_v4.n(200)}') % % J1_v5 = 1/(2*i) * ( 1/2*log( A1^2/(-4 + 2*i) )^2 - 1/2*log( A3^2/(-4 - 2*i) )^2 % - i*pi * log(A1^2*A3^2 / 20 ) ) % print(f'Numerical value of J1 rewritten is:\n{J1_v5.n(200)}') % % s = sqrt(353) % S = sqrt(34 + 2*s) % E = (1 - S + s)^2/16/20 % J1_v6 = 1/2 * ( log(E) * ( 2*A1.argument() + atan(1/2) - 2*pi ) ).n(200) % print(f'Numerical value of J1 from explicit formula is:\n{J1_v6.n(200)}') % % This gives the confirmations: % % Numerical value of J1 is: % 0.048392365330461616 % Numerical value of J1 rewritten as signed sum of dilog values is: % 0.048392365330461619161981331382180296201063587714971948595743 % Numerical value of J1 rewritten as signed sum of dilog and log values is: % 0.048392365330461619161981331382180296201063587714971948595740 % Numerical value of J1 rewritten using only log values is: % 0.048392365330461619161981331382180296201063587714971948595742 % Numerical value of J1 rewritten as signed sum of log values is: % 0.048392365330461619161981331382180296201063587714971948595755 % Numerical value of J1 finally rewritten is: % 0.048392365330461619161981331382180296201063587714971948595755 % Numerical value of J1 from explicit formula is: % 0.048392365330461619161981331382180296201063587714971948595743 % % Since i am interested in the K-theory of number fields, here the number field involved being the field % $F$ associated to the polynomial $x^4 - 34x^2 + 353$, the above collapsing of the dilog terms % is such a pitty... (Well the collapse is an instance of K-theory, we use the symbolic relation in % $K_2(F)$ written as $[a]+[1-a]=0$, in ad-hoc terms meaning that $\operatorname{Li}_2(a)+\operatorname{Li}_2(1-a)$ % is a sum of "simpler polylogarithms", so i am not so highly disappointed.) $$


  • $(5)+(6)$ The computation of $J_2$ and $J_3$:

In the same manner, we get: $$ \begin{aligned} 2i\; J_2 &= 2i\;\int_0^1\frac 1x\arctan G(x)\; dx \\ &= 2i\;\int_0^1\frac 1x\log\left( \frac{b_3b_4}{b_1b_2}\cdot\frac{(x-b_1)(x-b_2)}{(x-b_3)(x-b_4)}\right)\; dx \\ &= -\operatorname{Li}_2\left(\frac 1{b_1}\right) -\operatorname{Li}_2\left(\frac 1{b_2}\right) +\operatorname{Li}_2\left(\frac 1{b_3}\right) +\operatorname{Li}_2\left(\frac 1{b_4}\right) \\ &\equiv -2\operatorname{Li}_2\left(\frac 1{b_1}\right) -2\operatorname{Li}_2\left(\frac 1{b_2}\right) \\ &= 2\operatorname{Li}_2(b_1) +2\operatorname{Li}_2(b_2) \\&\qquad\qquad +\log^2 (-b_1)+\log^2(-b_2)\ , \\[3mm] 2i\; J_3 &= 2i\;\int_0^1\frac 1x\arctan H(x)\; dx \\ &= 2i\;\int_0^1\frac 1x\log\left( \frac{c_3c_4}{c_1c_2}\cdot\frac{(x-c_1)(x-c_2)}{(x-c_3)(x-c_4)}\right)\; dx \\ &= -\operatorname{Li}_2\left(\frac 1{c_1}\right) -\operatorname{Li}_2\left(\frac 1{c_2}\right) +\operatorname{Li}_2\left(\frac 1{c_3}\right) +\operatorname{Li}_2\left(\frac 1{c_4}\right) \\ &= -2\operatorname{Li}_2\left(\frac 1{c_1}\right) -2\operatorname{Li}_2\left(\frac 1{c_2}\right) \\ &= 2\operatorname{Li}_2(c_1) +2\operatorname{Li}_2(c_2) \\&\qquad\qquad +\log^2 (-c_1)+\log^2(-c_2)\ , \\[3mm] 2i(J_2+J_3) &\equiv 2\cdot \underbrace{( \operatorname{Li}_2(b_1) +\operatorname{Li}_2(c_2))}_{\pi^2/6 -\log b_1\log c_2} \\ &+2\cdot \underbrace{( \operatorname{Li}_2(c_1) +\operatorname{Li}_2(b_2))}_{\pi^2/6 -\log c_1\log b_2} \\&\qquad\qquad +\log^2 (-b_1)+\log^2(-b_2) +\log^2 (-c_1)+\log^2(-c_2) \\ &\qquad\qquad\qquad\qquad\text{ since $b_1+c_2=1=b2+c_1$} \\ &\equiv -2\log b_1\log c_2 - 2\log c_1\log b_2 \\&\qquad\qquad +\log^2 (-b_1)+\log^2(-b_2) +\log^2 (-c_1)+\log^2(-c_2) \\ &\equiv -2\log b_1\log b_2 -2\log b_1\log c_2 - 2\log c_1\log b_2 -2\log c_1\log c_2 \\&\qquad\qquad +\log^2 (-b_1)+\log^2(-b_2) + 2\log b_1\log b_2 \\&\qquad\qquad +\log^2 (-c_1)+\log^2(-c_2) + 2\log c_1\log c_2 \\ &\equiv -2(\log b_1 + \log c_1)(\log b_2 + \log c_2) \\&\qquad\qquad +\log^2 (-b_1)+\log^2(-b_2) + 2(\log(-b_1) - i\pi)(\log(-b_2) - i\pi) \\&\qquad\qquad +\log^2 (-c_1)+\log^2(-c_2) + 2(\log(-c_1) + i\pi)(\log(-c_2) + i\pi) \\ &\equiv -2(\log b_1 + \log c_1)(\log b_2 + \log c_2) \\&\qquad\qquad +(\log (-b_1)+\log(-b_2))^2 - 2\pi i(\log(-b_1) + \log(-b_2)) \\&\qquad\qquad +(\log (-c_1)+\log(-c_2))^2 + 2\pi i(\log(-c_1) + \log(-c_2)) \\ &= -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad +(\log (b_1b_2)+0\pi i)^2 - 2\pi i(\log(b_1b_2)+0\pi i) \\&\qquad\qquad +(\log (c_1c_2)-2\pi i)^2 + 2\pi i(\log(c_1c_2)-2\pi i) \\ &= -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad +\log^2 (-2) - 2\pi i\log(-2) \\&\qquad\qquad +\left(\log \frac {-3+4i}2 -2\pi i\right)^2 + 2\pi i\left(\log \frac {-3+4i}2 - 2\pi i\right) \\ &= -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad +(\log 2 +i\pi)^2 - 2\pi i(\log 2+i\pi) \\&\qquad\qquad + \left(\log \frac 52 +i\pi - i\arctan \frac 43 -2\pi i\right)^2 + 2\pi i\left(\log \frac 52 +i\pi - i\arctan \frac 43 -2\pi i\right) \\ &\equiv -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad +2\pi i \log 2 - 2\pi i\log 2 \\&\qquad\qquad - 2i \log \frac 52\left(\pi +\arctan \frac 43\right) + 2\pi i \log \frac 52 \\ &\equiv -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad - 2i \log \frac 52\arctan \frac 43 \\ &\equiv -2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad - 4i \log \frac 52\arctan \frac 12 \ . \end{aligned} $$ No intermediary step was omitted. The above shows how a piece of the final result comes in.

The above gives is a more or less explicit value for the sum $(J_2+J_3)$.


  • $(7)$:

We are in position to put all together. For this, the following simple relations connecting the $(a,b,c)$-values are freeing the door in the wall: $$ \boxed{\qquad b_1c_1=\frac 12 a_1^2\qquad\text{ and }\qquad b_2c_2=\frac 12a_2^2\ .\qquad} $$ It can be checked by hand or computer. This was in my initial answer the missing bridge to get to the end. (Having it, all the computations can be directed to the target, so the answer was dramatically changed in form, but not in ideas.)

Using it, and putting all together: $$ \begin{aligned} 2i J &= 2i(2J_1-J_2-J_3) \\ &=2i\cdot 2J_2 - 2i(J_2+J_3) \\ &\equiv - 4 \log a_1\log a_2 + 2\log^2 (-a_1)+2\log^2(-a_2) \\ &\qquad\qquad +2\log (b_1 c_1)\log (b_2 c_2) \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &= - 4 \log a_1\log a_2 + 2\log^2 (-a_1)+2\log^2(-a_2) \\&\qquad\qquad + 2\log \frac {a_1^2}2 \log \frac {a_2^2}2 \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &= - 4 \log a_1\log a_2 + 2(\log a_1 -i\pi)^2 + 2(\log a_2 + i\pi)^2 \\&\qquad\qquad + 2(2\log a_1 - 2\pi i - \log 2)(2 \log a_2 -\log 2) \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &\equiv + 4 \log a_1\log a_2 + 2\log^2 a_1 - 4\pi i\log a_1 + 2\log^2 a_2 + 4\pi i\log a_2 \\&\qquad\qquad -4\log 2(\log a_1 +\log a_2) -4\pi i(2 \log a_2 - \log 2) \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &= 2(\log a_1 + \log a_2)^2 -4\pi i(\log a_1 + \log a_2) \\&\qquad\qquad -4\log 2(\log a_1 +\log a_2) + 4\pi i \log 2 \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &= 2\log^2 (a_1a_2) -4\pi i\log (a_1a_2) -4\log 2\log (a_1a_2) + 4\pi i \log 2 \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &\qquad\qquad\text{ and with $\log(a_1a_2)=\log(-4+2i)=\frac 12\log 20+i(\pi-\arctan(1/2))$...} \\ &\equiv 2i\log 20\left(\pi-\arctan\frac 12\right) -2\pi i\log 20 -4i\log 2\left(\pi-\arctan\frac 12\right) + 4\pi i \log 2 \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &\equiv -2i\log 20\arctan\frac 12 +4i\log 2\arctan\frac 12 \\&\qquad\qquad +4i\log \frac 52\arctan \frac 12 \\ &=2i\log \frac 54\arctan \frac 12\ . \\ &\qquad\qquad\text{ So finally:} \\[5mm] J&=\log \frac 54\arctan \frac 12\ . \end{aligned} $$ $\square$

$\endgroup$
1
  • $\begingroup$ Thank you so much for your hard work +1 $\endgroup$ – Ali Shadhar Feb 15 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.