What is the value of this expression involving the dilogarithm $operatorname{Li}_2$ and $sqrt{2}$?
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
add a comment |
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
1
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
2
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.
– Mason
Nov 25 '18 at 7:23
1
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
1
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48
add a comment |
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
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
edited Nov 25 '18 at 7:25
Lord Shark the Unknown
101k958132
101k958132
asked Nov 25 '18 at 7:18
alpha
233
233
1
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
2
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.
– Mason
Nov 25 '18 at 7:23
1
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
1
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48
add a comment |
1
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
2
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.
– Mason
Nov 25 '18 at 7:23
1
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
1
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48
1
1
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
2
2
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.
– Mason
Nov 25 '18 at 7:23
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.
– Mason
Nov 25 '18 at 7:23
1
1
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
1
1
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48
add a comment |
1 Answer
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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 $frac{pi^2}{12}-frac12ln^22$.
add a comment |
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1 Answer
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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 $frac{pi^2}{12}-frac12ln^22$.
add a comment |
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 $frac{pi^2}{12}-frac12ln^22$.
add a comment |
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 $frac{pi^2}{12}-frac12ln^22$.
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 $frac{pi^2}{12}-frac12ln^22$.
answered Nov 26 '18 at 5:45
Mikalai Parshutsich
463315
463315
add a comment |
add a comment |
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1
Where we can presume: $operatorname{Li}_2(z)=sum_{k=1}^infty frac{z^k}{k^2}$ the dilogarithm function.
– Mason
Nov 25 '18 at 7:20
2
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.
– Mason
Nov 25 '18 at 7:23
1
the closed from follows from integrating by parts $Li_2(z) = -int frac{log(1-z)}{z}dz$ @Mason
– reuns
Nov 25 '18 at 7:33
1
Also I think one may be able to arrive there by exploiting the functional equations that appear in the first link I offered.
– Mason
Nov 25 '18 at 7:34
@Mason Yes that's the point of integrating by parts $int_1^z frac{log(1-s)}{s}ds$
– reuns
Nov 25 '18 at 7:48