What is the value of this expression involving the dilogarithm $operatorname{Li}_2$ and $sqrt{2}$?












1












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I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







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$endgroup$








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48


















1












$begingroup$


I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48
















1












1








1


0



$begingroup$


I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm







share|cite|improve this question











$endgroup$




I'm carrying out a calculation, and the end result is $$operatorname{Li}_2(1/sqrt2) - operatorname{Li}_2(1 - 1/sqrt2) +
operatorname{Li}_2(2 - sqrt2) - operatorname{Li}_2(sqrt{2} - 1).$$
The value appears to be $pi^2/12$. How does one prove this?






definite-integrals special-functions polylogarithm




definite-integrals special-functions polylogarithm






share|cite|improve this question















share|cite|improve this question













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edited Nov 25 '18 at 7:25









Lord Shark the Unknown

102k1059132




102k1059132










asked Nov 25 '18 at 7:18









alphaalpha

233




233








  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48
















  • 1




    $begingroup$
    Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:20








  • 2




    $begingroup$
    Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:23






  • 1




    $begingroup$
    the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
    $endgroup$
    – reuns
    Nov 25 '18 at 7:33








  • 1




    $begingroup$
    Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:34












  • $begingroup$
    @Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
    $endgroup$
    – reuns
    Nov 25 '18 at 7:48










1




1




$begingroup$
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
$endgroup$
– Mason
Nov 25 '18 at 7:20






$begingroup$
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
$endgroup$
– Mason
Nov 25 '18 at 7:20






2




2




$begingroup$
Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
$endgroup$
– Mason
Nov 25 '18 at 7:23




$begingroup$
Note that $pi^2/12 =zeta(2)/2 =frac{1}{2} sum_{k=1}^infty frac{1}{k^2}$. So this can give you a direction in which to manipulate this expression.
$endgroup$
– Mason
Nov 25 '18 at 7:23




1




1




$begingroup$
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
$endgroup$
– reuns
Nov 25 '18 at 7:33






$begingroup$
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
$endgroup$
– reuns
Nov 25 '18 at 7:33






1




1




$begingroup$
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
$endgroup$
– Mason
Nov 25 '18 at 7:34






$begingroup$
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
$endgroup$
– Mason
Nov 25 '18 at 7:34














$begingroup$
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
$endgroup$
– reuns
Nov 25 '18 at 7:48






$begingroup$
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
$endgroup$
– reuns
Nov 25 '18 at 7:48












1 Answer
1






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oldest

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2












$begingroup$

The following identity could be found in Russion version of article about dilogarithm:



$$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
&-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
Therefore
$$begin{aligned}
operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
&+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
end{aligned}$$
Now we can simplify the original expression (let's name it $E$):



$$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
&=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
&+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
&=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
&=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
end{aligned}$$



In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






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    2












    $begingroup$

    The following identity could be found in Russion version of article about dilogarithm:



    $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
    &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
    Therefore
    $$begin{aligned}
    operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
    &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
    end{aligned}$$
    Now we can simplify the original expression (let's name it $E$):



    $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
    &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
    &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
    &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
    &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
    end{aligned}$$



    In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The following identity could be found in Russion version of article about dilogarithm:



      $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
      &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
      Therefore
      $$begin{aligned}
      operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
      &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
      end{aligned}$$
      Now we can simplify the original expression (let's name it $E$):



      $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
      &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
      &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
      &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
      &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
      end{aligned}$$



      In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The following identity could be found in Russion version of article about dilogarithm:



        $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
        &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
        Therefore
        $$begin{aligned}
        operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
        &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
        end{aligned}$$
        Now we can simplify the original expression (let's name it $E$):



        $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
        &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
        &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
        &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
        &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
        end{aligned}$$



        In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.






        share|cite|improve this answer









        $endgroup$



        The following identity could be found in Russion version of article about dilogarithm:



        $$begin{aligned}operatorname{Li}_2(xy)&=operatorname{Li}_2(x)+operatorname{Li}_2(y)-operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)-operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)-\
        &-lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)end{aligned}$$
        Therefore
        $$begin{aligned}
        operatorname{Li}_2(x)+operatorname{Li}_2(y)&=operatorname{Li}_2(xy)+operatorname{Li}_2left(frac{x(1-y)}{1-xy}right)+operatorname{Li}_2left(frac{y(1-x)}{1-xy}right)+\
        &+lnleft(frac{1-x}{1-xy}right)lnleft(frac{1-y}{1-xy}right)
        end{aligned}$$
        Now we can simplify the original expression (let's name it $E$):



        $$begin{aligned}E&=left(operatorname{Li}_2frac1{sqrt2}+operatorname{Li}_2left(2-sqrt2right)right)-operatorname{Li}_2left(1-frac1{sqrt2}right)-operatorname{Li}_2(sqrt2-1)=\
        &=color{red}{operatorname{Li}_2(sqrt2-1)}+operatorname{Li}_2left(frac{frac1{sqrt2}(1-(2-sqrt2))}{1-(sqrt2-1)}right)+operatorname{Li}_2left(frac{(2-sqrt2)(1-frac1{sqrt2})}{1-(sqrt2-1)}right)+\
        &+lnleft(frac{1-(2-sqrt2)}{1-(sqrt2-1)}right)lnleft(frac{1-frac1{sqrt2}}{1-(sqrt2-1)}right)-operatorname{Li}_2left(1-frac1{sqrt2}right)color{red}{-operatorname{Li}_2(sqrt2-1)}=\
        &=operatorname{Li}_2left(frac12right)+color{red}{operatorname{Li}_2left(1-frac1{sqrt2}right)}+lnleft(frac1{sqrt2}right)lnleft(frac12right)color{red}{-operatorname{Li}_2left(1-frac1{sqrt2}right)}=\
        &=frac{pi^2}{12}-frac12ln^22+frac12ln^22=frac{pi^2}{12}
        end{aligned}$$



        In the last line special value of $operatorname{Li}_2left(frac12right)$ was used which equals to $frac{pi^2}{12}-frac12ln^22$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 5:45









        Mikalai ParshutsichMikalai Parshutsich

        473315




        473315






























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            Is it possible to collect Nectar points via Trainline?

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            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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