Conjecture $Re,operatorname{Li}_2left(frac12+frac...












17












$begingroup$


I numerically discovered the following conjecture:
$$Re,operatorname{Li}_2left(frac12+frac i6right)stackrel{color{gray}?}=frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).$$
It holds numerically with a precision of more than $30000$ decimal digits.



Could you suggest any ideas how to prove it?



Can we find a closed form for $Im,operatorname{Li}_2left(frac12+frac i6right)$?



Is there a general method to find closed forms of expressions of the form $Re,operatorname{Li}_2(p+iq)$, $Im,operatorname{Li}_2(p+iq)$ for $p,qinmathbb Q$?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I love your conjecture. Maybe better than Ramanujan ! :)
    $endgroup$
    – Hexacoordinate-C
    Sep 6 '15 at 21:49
















17












$begingroup$


I numerically discovered the following conjecture:
$$Re,operatorname{Li}_2left(frac12+frac i6right)stackrel{color{gray}?}=frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).$$
It holds numerically with a precision of more than $30000$ decimal digits.



Could you suggest any ideas how to prove it?



Can we find a closed form for $Im,operatorname{Li}_2left(frac12+frac i6right)$?



Is there a general method to find closed forms of expressions of the form $Re,operatorname{Li}_2(p+iq)$, $Im,operatorname{Li}_2(p+iq)$ for $p,qinmathbb Q$?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I love your conjecture. Maybe better than Ramanujan ! :)
    $endgroup$
    – Hexacoordinate-C
    Sep 6 '15 at 21:49














17












17








17


12



$begingroup$


I numerically discovered the following conjecture:
$$Re,operatorname{Li}_2left(frac12+frac i6right)stackrel{color{gray}?}=frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).$$
It holds numerically with a precision of more than $30000$ decimal digits.



Could you suggest any ideas how to prove it?



Can we find a closed form for $Im,operatorname{Li}_2left(frac12+frac i6right)$?



Is there a general method to find closed forms of expressions of the form $Re,operatorname{Li}_2(p+iq)$, $Im,operatorname{Li}_2(p+iq)$ for $p,qinmathbb Q$?










share|cite|improve this question









$endgroup$




I numerically discovered the following conjecture:
$$Re,operatorname{Li}_2left(frac12+frac i6right)stackrel{color{gray}?}=frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).$$
It holds numerically with a precision of more than $30000$ decimal digits.



Could you suggest any ideas how to prove it?



Can we find a closed form for $Im,operatorname{Li}_2left(frac12+frac i6right)$?



Is there a general method to find closed forms of expressions of the form $Re,operatorname{Li}_2(p+iq)$, $Im,operatorname{Li}_2(p+iq)$ for $p,qinmathbb Q$?







calculus complex-analysis closed-form conjectures polylogarithm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 6 '15 at 20:27









Vladimir ReshetnikovVladimir Reshetnikov

24.2k4119231




24.2k4119231








  • 1




    $begingroup$
    I love your conjecture. Maybe better than Ramanujan ! :)
    $endgroup$
    – Hexacoordinate-C
    Sep 6 '15 at 21:49














  • 1




    $begingroup$
    I love your conjecture. Maybe better than Ramanujan ! :)
    $endgroup$
    – Hexacoordinate-C
    Sep 6 '15 at 21:49








1




1




$begingroup$
I love your conjecture. Maybe better than Ramanujan ! :)
$endgroup$
– Hexacoordinate-C
Sep 6 '15 at 21:49




$begingroup$
I love your conjecture. Maybe better than Ramanujan ! :)
$endgroup$
– Hexacoordinate-C
Sep 6 '15 at 21:49










3 Answers
3






active

oldest

votes


















8












$begingroup$

I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.



We know that :



$$
{Li}_{2}(bar{z})=bar{{Li}_{2}(z)}
$$



So :



$$
Re{{Li}_{2}(z)}=frac{bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
$$



So:



$$
Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(frac{1}{2}-frac{i}{6})}{2}\
=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(1-(frac{1}{2}+frac{i}{6}))}{2}
$$



Now let's use:



$$
{Li}_{2}(z)+{Li}_{2}(1-z)=frac{{pi}^{2}}{6}-ln{z}ln{(1-z)}
$$



So we get:



$$
Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(1-(frac{1}{2}+frac{i}{6}))}}{2}\
=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}}{2}
$$



Let's compute those logarithms:



$$
ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
=ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
=frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
$$



Taking their product:
$$
ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}=frac{1}{4}{ln{(frac{5}{18}})}^{2}+{(arctan{frac{1}{3}})}^{2}
$$



Finally:



$$
Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{18}{5}})}^{2}-frac{1}{2}{(arctan{1/3})}^{2}\
=frac{7(pi)^2}{48}-frac{{(arctan{2})}^{2}}{3}-frac{{(arctan{3})}^{2}}{6}-frac{1}{8} {ln{frac{18}{5}}}^{2}
$$



There's a general formula using the same method. But the imaginary part doesn't have a known closed form:



$$
Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
    $endgroup$
    – Jack D'Aurizio
    Sep 6 '15 at 21:16








  • 1




    $begingroup$
    Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
    $endgroup$
    – Oussama Boussif
    Sep 6 '15 at 21:17






  • 1




    $begingroup$
    @OussamaBoussif: You can say that again.
    $endgroup$
    – Lucian
    Sep 6 '15 at 23:49






  • 2




    $begingroup$
    @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
    $endgroup$
    – Vladimir Reshetnikov
    Sep 7 '15 at 17:42






  • 2




    $begingroup$
    There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
    $endgroup$
    – Vladimir Reshetnikov
    Sep 7 '15 at 18:42





















3












$begingroup$

First of all we know that:




$$
operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
$$




Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:




$$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$




where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
$$begin{align}
operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
end{align}$$
Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that




$$
operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
$$






Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
$$
e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
$$
has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.



Because of $(diamondsuit)$ and $(heartsuit)$ we have
$$
operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
$$
for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.



For the logarithm term we get
$$
Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
$$
and
$$
Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
$$



We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
$$
color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
$$



Now we could obtain your conjectured closed-form:
$$
Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
$$



For the imaginary part we have
$$begin{align}
Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
end{align}$$



By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

    First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
    $$begin{aligned}
    Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
    &=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
    &=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
    &=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
    &=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
    end{aligned}$$

    Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
    Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1424600%2fconjecture-re-operatornameli-2-left-frac12-frac-i6-right-frac7-pi2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8












      $begingroup$

      I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.



      We know that :



      $$
      {Li}_{2}(bar{z})=bar{{Li}_{2}(z)}
      $$



      So :



      $$
      Re{{Li}_{2}(z)}=frac{bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
      $$



      So:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(frac{1}{2}-frac{i}{6})}{2}\
      =frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(1-(frac{1}{2}+frac{i}{6}))}{2}
      $$



      Now let's use:



      $$
      {Li}_{2}(z)+{Li}_{2}(1-z)=frac{{pi}^{2}}{6}-ln{z}ln{(1-z)}
      $$



      So we get:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(1-(frac{1}{2}+frac{i}{6}))}}{2}\
      =frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}}{2}
      $$



      Let's compute those logarithms:



      $$
      ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
      =ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
      =frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
      $$



      Taking their product:
      $$
      ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}=frac{1}{4}{ln{(frac{5}{18}})}^{2}+{(arctan{frac{1}{3}})}^{2}
      $$



      Finally:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{18}{5}})}^{2}-frac{1}{2}{(arctan{1/3})}^{2}\
      =frac{7(pi)^2}{48}-frac{{(arctan{2})}^{2}}{3}-frac{{(arctan{3})}^{2}}{6}-frac{1}{8} {ln{frac{18}{5}}}^{2}
      $$



      There's a general formula using the same method. But the imaginary part doesn't have a known closed form:



      $$
      Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
      $$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
        $endgroup$
        – Jack D'Aurizio
        Sep 6 '15 at 21:16








      • 1




        $begingroup$
        Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
        $endgroup$
        – Oussama Boussif
        Sep 6 '15 at 21:17






      • 1




        $begingroup$
        @OussamaBoussif: You can say that again.
        $endgroup$
        – Lucian
        Sep 6 '15 at 23:49






      • 2




        $begingroup$
        @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 17:42






      • 2




        $begingroup$
        There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 18:42


















      8












      $begingroup$

      I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.



      We know that :



      $$
      {Li}_{2}(bar{z})=bar{{Li}_{2}(z)}
      $$



      So :



      $$
      Re{{Li}_{2}(z)}=frac{bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
      $$



      So:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(frac{1}{2}-frac{i}{6})}{2}\
      =frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(1-(frac{1}{2}+frac{i}{6}))}{2}
      $$



      Now let's use:



      $$
      {Li}_{2}(z)+{Li}_{2}(1-z)=frac{{pi}^{2}}{6}-ln{z}ln{(1-z)}
      $$



      So we get:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(1-(frac{1}{2}+frac{i}{6}))}}{2}\
      =frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}}{2}
      $$



      Let's compute those logarithms:



      $$
      ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
      =ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
      =frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
      $$



      Taking their product:
      $$
      ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}=frac{1}{4}{ln{(frac{5}{18}})}^{2}+{(arctan{frac{1}{3}})}^{2}
      $$



      Finally:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{18}{5}})}^{2}-frac{1}{2}{(arctan{1/3})}^{2}\
      =frac{7(pi)^2}{48}-frac{{(arctan{2})}^{2}}{3}-frac{{(arctan{3})}^{2}}{6}-frac{1}{8} {ln{frac{18}{5}}}^{2}
      $$



      There's a general formula using the same method. But the imaginary part doesn't have a known closed form:



      $$
      Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
      $$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
        $endgroup$
        – Jack D'Aurizio
        Sep 6 '15 at 21:16








      • 1




        $begingroup$
        Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
        $endgroup$
        – Oussama Boussif
        Sep 6 '15 at 21:17






      • 1




        $begingroup$
        @OussamaBoussif: You can say that again.
        $endgroup$
        – Lucian
        Sep 6 '15 at 23:49






      • 2




        $begingroup$
        @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 17:42






      • 2




        $begingroup$
        There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 18:42
















      8












      8








      8





      $begingroup$

      I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.



      We know that :



      $$
      {Li}_{2}(bar{z})=bar{{Li}_{2}(z)}
      $$



      So :



      $$
      Re{{Li}_{2}(z)}=frac{bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
      $$



      So:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(frac{1}{2}-frac{i}{6})}{2}\
      =frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(1-(frac{1}{2}+frac{i}{6}))}{2}
      $$



      Now let's use:



      $$
      {Li}_{2}(z)+{Li}_{2}(1-z)=frac{{pi}^{2}}{6}-ln{z}ln{(1-z)}
      $$



      So we get:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(1-(frac{1}{2}+frac{i}{6}))}}{2}\
      =frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}}{2}
      $$



      Let's compute those logarithms:



      $$
      ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
      =ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
      =frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
      $$



      Taking their product:
      $$
      ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}=frac{1}{4}{ln{(frac{5}{18}})}^{2}+{(arctan{frac{1}{3}})}^{2}
      $$



      Finally:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{18}{5}})}^{2}-frac{1}{2}{(arctan{1/3})}^{2}\
      =frac{7(pi)^2}{48}-frac{{(arctan{2})}^{2}}{3}-frac{{(arctan{3})}^{2}}{6}-frac{1}{8} {ln{frac{18}{5}}}^{2}
      $$



      There's a general formula using the same method. But the imaginary part doesn't have a known closed form:



      $$
      Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
      $$






      share|cite|improve this answer











      $endgroup$



      I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.



      We know that :



      $$
      {Li}_{2}(bar{z})=bar{{Li}_{2}(z)}
      $$



      So :



      $$
      Re{{Li}_{2}(z)}=frac{bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
      $$



      So:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(frac{1}{2}-frac{i}{6})}{2}\
      =frac{{Li}_{2}(frac{1}{2}+frac{i}{6})+{Li}_{2}(1-(frac{1}{2}+frac{i}{6}))}{2}
      $$



      Now let's use:



      $$
      {Li}_{2}(z)+{Li}_{2}(1-z)=frac{{pi}^{2}}{6}-ln{z}ln{(1-z)}
      $$



      So we get:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(1-(frac{1}{2}+frac{i}{6}))}}{2}\
      =frac{frac{{pi}^{2}}{6}-ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}}{2}
      $$



      Let's compute those logarithms:



      $$
      ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
      =ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
      =frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
      $$



      Taking their product:
      $$
      ln{(frac{1}{2}+frac{i}{6})}ln{(frac{1}{2}-frac{i}{6})}=frac{1}{4}{ln{(frac{5}{18}})}^{2}+{(arctan{frac{1}{3}})}^{2}
      $$



      Finally:



      $$
      Re{{Li}_{2}(frac{1}{2}+frac{i}{6})}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{18}{5}})}^{2}-frac{1}{2}{(arctan{1/3})}^{2}\
      =frac{7(pi)^2}{48}-frac{{(arctan{2})}^{2}}{3}-frac{{(arctan{3})}^{2}}{6}-frac{1}{8} {ln{frac{18}{5}}}^{2}
      $$



      There's a general formula using the same method. But the imaginary part doesn't have a known closed form:



      $$
      Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Sep 11 '15 at 7:56

























      answered Sep 6 '15 at 20:59









      Oussama BoussifOussama Boussif

      2,897722




      2,897722








      • 1




        $begingroup$
        (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
        $endgroup$
        – Jack D'Aurizio
        Sep 6 '15 at 21:16








      • 1




        $begingroup$
        Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
        $endgroup$
        – Oussama Boussif
        Sep 6 '15 at 21:17






      • 1




        $begingroup$
        @OussamaBoussif: You can say that again.
        $endgroup$
        – Lucian
        Sep 6 '15 at 23:49






      • 2




        $begingroup$
        @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 17:42






      • 2




        $begingroup$
        There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 18:42
















      • 1




        $begingroup$
        (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
        $endgroup$
        – Jack D'Aurizio
        Sep 6 '15 at 21:16








      • 1




        $begingroup$
        Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
        $endgroup$
        – Oussama Boussif
        Sep 6 '15 at 21:17






      • 1




        $begingroup$
        @OussamaBoussif: You can say that again.
        $endgroup$
        – Lucian
        Sep 6 '15 at 23:49






      • 2




        $begingroup$
        @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 17:42






      • 2




        $begingroup$
        There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
        $endgroup$
        – Vladimir Reshetnikov
        Sep 7 '15 at 18:42










      1




      1




      $begingroup$
      (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
      $endgroup$
      – Jack D'Aurizio
      Sep 6 '15 at 21:16






      $begingroup$
      (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $text{Re},text{Li}_2left(frac{1}{2}+iqright)$ for any $qinmathbb{R}$.
      $endgroup$
      – Jack D'Aurizio
      Sep 6 '15 at 21:16






      1




      1




      $begingroup$
      Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
      $endgroup$
      – Oussama Boussif
      Sep 6 '15 at 21:17




      $begingroup$
      Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough.
      $endgroup$
      – Oussama Boussif
      Sep 6 '15 at 21:17




      1




      1




      $begingroup$
      @OussamaBoussif: You can say that again.
      $endgroup$
      – Lucian
      Sep 6 '15 at 23:49




      $begingroup$
      @OussamaBoussif: You can say that again.
      $endgroup$
      – Lucian
      Sep 6 '15 at 23:49




      2




      2




      $begingroup$
      @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
      $endgroup$
      – Vladimir Reshetnikov
      Sep 7 '15 at 17:42




      $begingroup$
      @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part.
      $endgroup$
      – Vladimir Reshetnikov
      Sep 7 '15 at 17:42




      2




      2




      $begingroup$
      There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
      $endgroup$
      – Vladimir Reshetnikov
      Sep 7 '15 at 18:42






      $begingroup$
      There are some know results: $$Im,operatorname{Li}_2(i)=G,$$ $$Im,operatorname{Li}_2(1+i)=G+fracpi4ln2,$$ $$Im,operatorname{Li}_2!left(frac{1+i}2right)=G-fracpi8ln2,$$ $$Im,operatorname{Li}_2left(ileft(2+sqrt3right)right)=frac23G-frac{5 pi}{12}lnleft(2-sqrt3right),$$ $$Im,operatorname{Li}_2!left(frac{1+i}{sqrt2}right)=frac{1-sqrt8}4G- frac{1+sqrt2}{16} pi^2+ frac{sqrt2}{32}psi^{(1)}!left(tfrac18right).$$
      $endgroup$
      – Vladimir Reshetnikov
      Sep 7 '15 at 18:42













      3












      $begingroup$

      First of all we know that:




      $$
      operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
      $$




      Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:




      $$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$




      where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
      $$begin{align}
      operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
      operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
      end{align}$$
      Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that




      $$
      operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
      $$






      Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
      $$
      e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
      $$
      has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.



      Because of $(diamondsuit)$ and $(heartsuit)$ we have
      $$
      operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
      $$
      for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.



      For the logarithm term we get
      $$
      Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
      $$
      and
      $$
      Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
      $$



      We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
      $$
      color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
      $$



      Now we could obtain your conjectured closed-form:
      $$
      Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
      $$



      For the imaginary part we have
      $$begin{align}
      Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
      end{align}$$



      By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        First of all we know that:




        $$
        operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
        $$




        Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:




        $$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$




        where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
        $$begin{align}
        operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
        operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
        end{align}$$
        Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that




        $$
        operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
        $$






        Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
        $$
        e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
        $$
        has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.



        Because of $(diamondsuit)$ and $(heartsuit)$ we have
        $$
        operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
        $$
        for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.



        For the logarithm term we get
        $$
        Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
        $$
        and
        $$
        Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
        $$



        We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
        $$
        color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
        $$



        Now we could obtain your conjectured closed-form:
        $$
        Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
        $$



        For the imaginary part we have
        $$begin{align}
        Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
        end{align}$$



        By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          First of all we know that:




          $$
          operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
          $$




          Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:




          $$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$




          where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
          $$begin{align}
          operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
          operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
          end{align}$$
          Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that




          $$
          operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
          $$






          Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
          $$
          e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
          $$
          has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.



          Because of $(diamondsuit)$ and $(heartsuit)$ we have
          $$
          operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
          $$
          for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.



          For the logarithm term we get
          $$
          Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
          $$
          and
          $$
          Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
          $$



          We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
          $$
          color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
          $$



          Now we could obtain your conjectured closed-form:
          $$
          Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
          $$



          For the imaginary part we have
          $$begin{align}
          Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
          end{align}$$



          By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.






          share|cite|improve this answer











          $endgroup$



          First of all we know that:




          $$
          operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
          $$




          Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:




          $$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$




          where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
          $$begin{align}
          operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
          operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
          end{align}$$
          Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that




          $$
          operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
          $$






          Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
          $$
          e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
          $$
          has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.



          Because of $(diamondsuit)$ and $(heartsuit)$ we have
          $$
          operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
          $$
          for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.



          For the logarithm term we get
          $$
          Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
          $$
          and
          $$
          Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
          $$



          We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
          $$
          color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
          $$



          Now we could obtain your conjectured closed-form:
          $$
          Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
          $$



          For the imaginary part we have
          $$begin{align}
          Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
          end{align}$$



          By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 11 '15 at 0:43

























          answered Sep 11 '15 at 0:09









          user153012user153012

          6,30822277




          6,30822277























              1












              $begingroup$

              This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

              First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
              $$begin{aligned}
              Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
              &=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
              &=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
              &=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
              &=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
              end{aligned}$$

              Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
              Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

                First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
                $$begin{aligned}
                Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
                &=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
                &=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
                &=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
                &=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
                end{aligned}$$

                Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
                Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

                  First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
                  $$begin{aligned}
                  Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
                  &=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
                  &=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
                  &=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
                  &=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
                  end{aligned}$$

                  Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
                  Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$






                  share|cite|improve this answer









                  $endgroup$



                  This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

                  First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
                  $$begin{aligned}
                  Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
                  &=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
                  &=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
                  &=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
                  &=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
                  end{aligned}$$

                  Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
                  Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 3:31









                  Kemono ChenKemono Chen

                  2,9181739




                  2,9181739






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1424600%2fconjecture-re-operatornameli-2-left-frac12-frac-i6-right-frac7-pi2%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Ellipse (mathématiques)

                      Quarter-circle Tiles

                      Mont Emei