Polylog property $operatorname{Li}_n(1-i)+(-1)^noperatorname{Li}_n(frac{1+i}2)$












3












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How can we calculate $$L_n=operatorname{Li}_n(1-i)+(-1)^noperatorname{Li}_nleft(frac{1+i}2right)?$$




I have found the following result manually:
$$L_1=-frac12pi i+lnfrac{1-i}2=-frac34pi i-frac12ln2$$
$$L_2=frac1{16}pi^2-ileft(frac14piln2+Gright)+frac5{96}pi^2-frac18ln^22+ileft(G-frac18piln2right)\
=frac{11}{96}pi^2-frac18ln^22-ifrac38piln2$$
Something I have found by using Mathematica:
$$L_3=-frac{1}{48} ln^32+frac{11}{192} pi ^2 ln2+i left(-frac{7 pi ^3}{128}-frac{1}{32} 3 pi ln^22right)$$
$$L_4=frac{1313 pi ^4}{92160}-frac{1}{384} ln ^42+frac{11}{768} pi ^2 ln ^22+i left(-frac{1}{64} pi ln ^32-frac{7}{256} pi ^3 ln 2right)$$
It seems like there is a method that Mathematica knows to simplify $L_n$. What is the method?






calculus polylogarithm







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    3












    $begingroup$



    How can we calculate $$L_n=operatorname{Li}_n(1-i)+(-1)^noperatorname{Li}_nleft(frac{1+i}2right)?$$




    I have found the following result manually:
    $$L_1=-frac12pi i+lnfrac{1-i}2=-frac34pi i-frac12ln2$$
    $$L_2=frac1{16}pi^2-ileft(frac14piln2+Gright)+frac5{96}pi^2-frac18ln^22+ileft(G-frac18piln2right)\
    =frac{11}{96}pi^2-frac18ln^22-ifrac38piln2$$
    Something I have found by using Mathematica:
    $$L_3=-frac{1}{48} ln^32+frac{11}{192} pi ^2 ln2+i left(-frac{7 pi ^3}{128}-frac{1}{32} 3 pi ln^22right)$$
    $$L_4=frac{1313 pi ^4}{92160}-frac{1}{384} ln ^42+frac{11}{768} pi ^2 ln ^22+i left(-frac{1}{64} pi ln ^32-frac{7}{256} pi ^3 ln 2right)$$
    It seems like there is a method that Mathematica knows to simplify $L_n$. What is the method?






    calculus polylogarithm







    share|cite|improve this question









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      3












      3








      3





      $begingroup$



      How can we calculate $$L_n=operatorname{Li}_n(1-i)+(-1)^noperatorname{Li}_nleft(frac{1+i}2right)?$$




      I have found the following result manually:
      $$L_1=-frac12pi i+lnfrac{1-i}2=-frac34pi i-frac12ln2$$
      $$L_2=frac1{16}pi^2-ileft(frac14piln2+Gright)+frac5{96}pi^2-frac18ln^22+ileft(G-frac18piln2right)\
      =frac{11}{96}pi^2-frac18ln^22-ifrac38piln2$$
      Something I have found by using Mathematica:
      $$L_3=-frac{1}{48} ln^32+frac{11}{192} pi ^2 ln2+i left(-frac{7 pi ^3}{128}-frac{1}{32} 3 pi ln^22right)$$
      $$L_4=frac{1313 pi ^4}{92160}-frac{1}{384} ln ^42+frac{11}{768} pi ^2 ln ^22+i left(-frac{1}{64} pi ln ^32-frac{7}{256} pi ^3 ln 2right)$$
      It seems like there is a method that Mathematica knows to simplify $L_n$. What is the method?






      calculus polylogarithm







      share|cite|improve this question









      $endgroup$





      How can we calculate $$L_n=operatorname{Li}_n(1-i)+(-1)^noperatorname{Li}_nleft(frac{1+i}2right)?$$




      I have found the following result manually:
      $$L_1=-frac12pi i+lnfrac{1-i}2=-frac34pi i-frac12ln2$$
      $$L_2=frac1{16}pi^2-ileft(frac14piln2+Gright)+frac5{96}pi^2-frac18ln^22+ileft(G-frac18piln2right)\
      =frac{11}{96}pi^2-frac18ln^22-ifrac38piln2$$
      Something I have found by using Mathematica:
      $$L_3=-frac{1}{48} ln^32+frac{11}{192} pi ^2 ln2+i left(-frac{7 pi ^3}{128}-frac{1}{32} 3 pi ln^22right)$$
      $$L_4=frac{1313 pi ^4}{92160}-frac{1}{384} ln ^42+frac{11}{768} pi ^2 ln ^22+i left(-frac{1}{64} pi ln ^32-frac{7}{256} pi ^3 ln 2right)$$
      It seems like there is a method that Mathematica knows to simplify $L_n$. What is the method?






      calculus polylogarithm




      calculus polylogarithm






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      asked Jan 7 at 9:12









      Kemono ChenKemono Chen

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          Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${rm{Li}}_n(1-i)=(-1)^{n-1}{rm{Li}}_nleft(frac{1+i}2right)-frac{(2pi i)^n}{n!}B_nleft(frac{ln(i-1)}{2pi i}+frac12right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-frac{(2pi i)^n}{n!}B_nleft(frac{frac{ln2}2+frac{3ipi}4}{2pi i}+frac12right)=-frac{(2pi i)^n}{n!}B_nleft(frac78-frac{ln2}{4pi}iright)$$ where $B_n(cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.






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            $begingroup$

            Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${rm{Li}}_n(1-i)=(-1)^{n-1}{rm{Li}}_nleft(frac{1+i}2right)-frac{(2pi i)^n}{n!}B_nleft(frac{ln(i-1)}{2pi i}+frac12right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-frac{(2pi i)^n}{n!}B_nleft(frac{frac{ln2}2+frac{3ipi}4}{2pi i}+frac12right)=-frac{(2pi i)^n}{n!}B_nleft(frac78-frac{ln2}{4pi}iright)$$ where $B_n(cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.






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              $begingroup$

              Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${rm{Li}}_n(1-i)=(-1)^{n-1}{rm{Li}}_nleft(frac{1+i}2right)-frac{(2pi i)^n}{n!}B_nleft(frac{ln(i-1)}{2pi i}+frac12right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-frac{(2pi i)^n}{n!}B_nleft(frac{frac{ln2}2+frac{3ipi}4}{2pi i}+frac12right)=-frac{(2pi i)^n}{n!}B_nleft(frac78-frac{ln2}{4pi}iright)$$ where $B_n(cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.






              share|cite|improve this answer









              $endgroup$
















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                4





                $begingroup$

                Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${rm{Li}}_n(1-i)=(-1)^{n-1}{rm{Li}}_nleft(frac{1+i}2right)-frac{(2pi i)^n}{n!}B_nleft(frac{ln(i-1)}{2pi i}+frac12right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-frac{(2pi i)^n}{n!}B_nleft(frac{frac{ln2}2+frac{3ipi}4}{2pi i}+frac12right)=-frac{(2pi i)^n}{n!}B_nleft(frac78-frac{ln2}{4pi}iright)$$ where $B_n(cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.






                share|cite|improve this answer









                $endgroup$



                Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${rm{Li}}_n(1-i)=(-1)^{n-1}{rm{Li}}_nleft(frac{1+i}2right)-frac{(2pi i)^n}{n!}B_nleft(frac{ln(i-1)}{2pi i}+frac12right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-frac{(2pi i)^n}{n!}B_nleft(frac{frac{ln2}2+frac{3ipi}4}{2pi i}+frac12right)=-frac{(2pi i)^n}{n!}B_nleft(frac78-frac{ln2}{4pi}iright)$$ where $B_n(cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Jan 7 at 9:49









                TheSimpliFireTheSimpliFire

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