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






active

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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    active

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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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            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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