How to prove that...












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How to prove that $$int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx=frac{pi^{2}}{16}-frac{ln^{2}left(sqrt{2}+1right)}{4}?$$



I have encountered this integral recently, and I am aware that one can show how this is true with a substitution $u=ln{x}$ and then expanding it into a series of integrals of the form $int_{0}^{ln{(sqrt{2}+1)}}ue^{-nu}du$ where each integral is then calculated individually, but it feels pretty brute-forced. Is there any other way to do this? Thanks.






real-analysis calculus integration definite-integrals







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    The substitution $u=x^2$ turns it into the dilogarithm integral, right?
    $endgroup$
    – Ant
    Jan 10 at 2:29






  • 1




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    Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
    $endgroup$
    – Ant
    Jan 10 at 2:46








  • 1




    $begingroup$
    It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
    $endgroup$
    – Ant
    Jan 10 at 3:01






  • 3




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    A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
    $endgroup$
    – Kemono Chen
    Jan 10 at 3:43
















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$begingroup$


How to prove that $$int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx=frac{pi^{2}}{16}-frac{ln^{2}left(sqrt{2}+1right)}{4}?$$



I have encountered this integral recently, and I am aware that one can show how this is true with a substitution $u=ln{x}$ and then expanding it into a series of integrals of the form $int_{0}^{ln{(sqrt{2}+1)}}ue^{-nu}du$ where each integral is then calculated individually, but it feels pretty brute-forced. Is there any other way to do this? Thanks.






real-analysis calculus integration definite-integrals







share|cite|improve this question









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  • $begingroup$
    The substitution $u=x^2$ turns it into the dilogarithm integral, right?
    $endgroup$
    – Ant
    Jan 10 at 2:29






  • 1




    $begingroup$
    Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
    $endgroup$
    – Ant
    Jan 10 at 2:46








  • 1




    $begingroup$
    It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
    $endgroup$
    – Ant
    Jan 10 at 3:01






  • 3




    $begingroup$
    A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
    $endgroup$
    – Kemono Chen
    Jan 10 at 3:43














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$begingroup$


How to prove that $$int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx=frac{pi^{2}}{16}-frac{ln^{2}left(sqrt{2}+1right)}{4}?$$



I have encountered this integral recently, and I am aware that one can show how this is true with a substitution $u=ln{x}$ and then expanding it into a series of integrals of the form $int_{0}^{ln{(sqrt{2}+1)}}ue^{-nu}du$ where each integral is then calculated individually, but it feels pretty brute-forced. Is there any other way to do this? Thanks.






real-analysis calculus integration definite-integrals







share|cite|improve this question









$endgroup$




How to prove that $$int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx=frac{pi^{2}}{16}-frac{ln^{2}left(sqrt{2}+1right)}{4}?$$



I have encountered this integral recently, and I am aware that one can show how this is true with a substitution $u=ln{x}$ and then expanding it into a series of integrals of the form $int_{0}^{ln{(sqrt{2}+1)}}ue^{-nu}du$ where each integral is then calculated individually, but it feels pretty brute-forced. Is there any other way to do this? Thanks.






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real-analysis calculus integration definite-integrals






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asked Jan 10 at 2:13









Edward H.Edward H.

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1439












  • $begingroup$
    The substitution $u=x^2$ turns it into the dilogarithm integral, right?
    $endgroup$
    – Ant
    Jan 10 at 2:29






  • 1




    $begingroup$
    Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
    $endgroup$
    – Ant
    Jan 10 at 2:46








  • 1




    $begingroup$
    It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
    $endgroup$
    – Ant
    Jan 10 at 3:01






  • 3




    $begingroup$
    A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
    $endgroup$
    – Kemono Chen
    Jan 10 at 3:43


















  • $begingroup$
    The substitution $u=x^2$ turns it into the dilogarithm integral, right?
    $endgroup$
    – Ant
    Jan 10 at 2:29






  • 1




    $begingroup$
    Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
    $endgroup$
    – Ant
    Jan 10 at 2:46








  • 1




    $begingroup$
    It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
    $endgroup$
    – Ant
    Jan 10 at 3:01






  • 3




    $begingroup$
    A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
    $endgroup$
    – Kemono Chen
    Jan 10 at 3:43
















$begingroup$
The substitution $u=x^2$ turns it into the dilogarithm integral, right?
$endgroup$
– Ant
Jan 10 at 2:29




$begingroup$
The substitution $u=x^2$ turns it into the dilogarithm integral, right?
$endgroup$
– Ant
Jan 10 at 2:29




1




1




$begingroup$
Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
$endgroup$
– Ant
Jan 10 at 2:46






$begingroup$
Oops, my bad. The anti-derivative is $frac{1}{4}textrm{Li}_2(1-x^2)-textrm{Li}_2(1-x)$, though, so maybe there's another way...
$endgroup$
– Ant
Jan 10 at 2:46






1




1




$begingroup$
It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
$endgroup$
– Ant
Jan 10 at 3:01




$begingroup$
It's straightforward to see how that anti-derivative differentiates to your integrand; so reversing that says that you can re-write your integrand as the sum of two terms (and you need to use $u=x^2$ on one of them). But using the various dilogarithm identities to evaluate at your bounds is tricky.
$endgroup$
– Ant
Jan 10 at 3:01




3




3




$begingroup$
A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
$endgroup$
– Kemono Chen
Jan 10 at 3:43




$begingroup$
A possible approach $$I=int_{1}^{sqrt{2}+1}frac{ln{x}}{x^{2}-1}dx\ =int_{pi/4}^{3pi/8}frac{lntan x}{tan^2x-1}sec^2xdxtext{ (tangent substitution)}\ =int_{pi/4}^{3pi/8}-lntan xsec2xdxtext{ (trig identity)}\ =int_{pi/4}^{3pi/8}-csc2xleft(lncscleft(x-fracpi4right)+lnsinleft(fracpi4+xright)right)dx-frac12ln^2(1+sqrt2)text{ (IBP)}\ =int_{0}^{pi/8}sec2xlncot xdx-frac12ln^2(1+sqrt2)text{ (linear substitution)}\ =int_{0}^{pi/8}sum_{n=0}^inftyfrac{cos(4n+2)x}{2n+1}sec 2xdx-frac12ln^2(1+sqrt2)text{ (Fourier series)}$$
$endgroup$
– Kemono Chen
Jan 10 at 3:43










3 Answers
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Put
begin{equation*}
I=int_{1}^{sqrt{2}+1}dfrac{ln x}{x^2-1}, mathrm{d}x =[x=1/y] = int_{sqrt{2}-1}^{1}dfrac{ln y}{y^2-1}, mathrm{d}y.tag{1}
end{equation*}
After the substitution $ y=frac{1-z}{1+z}$ and integration by parts we have
begin{gather*}
I = int_{0}^{sqrt{2}-1}dfrac{lnleft(frac{1-z}{1+z}right)}{-2z}, mathrm{d}z = left[-dfrac{1}{2}ln(z)lnleft(frac{1-z}{1+z}right)right]_{0}^{sqrt{2}-1}+int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z =\[2ex]-dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z.tag{2}
end{gather*}
If we add (1) and (2) we get
begin{equation*}
2I = -dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{1}dfrac{ln z}{z^2-1}, mathrm{d}z.
end{equation*}
Consequently
begin{gather*}
I = -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}int_{0}^{1}left(sum_{k=0}^{infty}ln(z)z^{2k}right), mathrm{d}z =\[2ex] -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}sum_{k=0}^{infty}int_{0}^{1}ln(z)z^{2k}, mathrm{d}z =\[2ex]
-dfrac{1}{4}ln^2(sqrt{2}+1) +dfrac{1}{2}sum_{k=0}^{infty}dfrac{1}{(2k+1)^2} = dfrac{pi^2}{16} -dfrac{1}{4}ln^2(sqrt{2}+1).
end{gather*}






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  • 2




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    Now that is a solution! Very nice (+1)
    $endgroup$
    – omegadot
    Jan 10 at 22:06






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    @ omegadot$ Thank you.
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    – JanG
    Jan 11 at 5:56



















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A Complete Solution Now



Consider
$$F(s)=int_1^sfrac{log x}{x^2-1}mathrm dx$$
Like before,
$$F(s)=frac12int_1^{s}frac{log x}{x-1}mathrm dx-frac12int_1^{s}frac{log x}{x+1}mathrm dx$$
$$F(s)=-frac12mathrm{Li}_2(1-s)-frac12J(s)$$
For $J(s)$, we integrate by parts with $mathrm dv=frac{mathrm dx}{1+x}$ to get
$$J(s)=log(s)log(s+1)-int_1^sfrac{log(1+x)}{x}mathrm dx$$
$$J(s)=log(s)log(s+1)+int_2^{1+s}frac{log x}{1-x}mathrm dx$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(1-x)bigg|_2^{1+s}$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)-mathrm{Li}_2(-1)$$
Using $mathrm{Li}_k(-1)=(2^{1-k}-1)zeta(k)$,
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)+frac{pi^2}{12}$$
Plugging in $s=1+sqrt2$,
$$J(1+sqrt2)=log(1+sqrt2)log(2+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+sqrt2)=log^2(1+sqrt2)+frac12log(2)log(1+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
And again as the OP noted,
$$F(1+sqrt2)=-frac{pi^2}{48}-frac14log^2(1+sqrt2)+frac14log2log(1+sqrt2)+frac1{16}log^22+frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)$$
And since
$$frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)=frac{pi^2}{12}-frac14log2log(1+sqrt2)-frac1{16}log^22$$
We have
$$F(1+sqrt2)=frac{pi^2}{16}-frac14log^2(1+sqrt2)$$






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    Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
    $endgroup$
    – Ant
    Jan 10 at 3:07






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    @Ant Right... I'm on it
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    – clathratus
    Jan 10 at 3:08






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    I think the WolframAlpha result is using equality (6~7) from this MathWorld page
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    – Edward H.
    Jan 10 at 5:24






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    Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:14






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    begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:22



















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Here is one possible approach.



Enforcing a substitution of $x mapsto 1/x$ to begin with gives
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -int_{sqrt{2}-1}^1 frac{ln x}{1 - x^2} , dx = -sum_{n = 0}^infty int_{sqrt{2} - 1}^1 x^{2n} ln x , dx,
end{align}
where we have taken advantage of the geometric sum for $1/(1 - x^2)$.
Integrating by parts then leads to
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= - ln (1 + sqrt{2}) sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} + sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&qquad - sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2}. tag1
end{align}



Observing that
$$tanh^{-1} z = sum_{n = 0}^infty frac{z^{2n + 1}}{2n + 1}, qquad |z| < 1, qquad (*)$$
the first of the sums appearing in (1) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} = tanh^{-1} (sqrt{2} - 1) = frac{1}{2} ln (1 + sqrt{2}).$$



To find the second and third sums in (1), divide ($*$) by $x$ before integrating up from $0$ to $x$. Thus
begin{align}
sum_{n = 0}^infty frac{1}{2n + 1} int_0^x t^{2n} , dt &= int_0^x frac{tanh^{-1} t}{t} , dt\
sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} &= frac{1}{2} int_0^x ln left (frac{1 + t}{1 - t} right ) frac{dt}{t}\
&= frac{1}{2} int_0^{-x} frac{ln (1 - t)}{t} , dt - frac{1}{2} int_0^x frac{ln (1 - t)}{t} , dt
end{align}
or
$$sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (x) - operatorname{Li}_2 (-x) right ], qquad (**)$$
where $operatorname{Li}_2 (x)$ is the dilogarithm function.



The second and third sums appearing in (1) can now be found using ($**$). For the second sum, setting $x = 1$ gives
$$sum_{n = 1}^infty frac{1}{(2n + 1)^2} = frac{1}{2} operatorname{Li}_2 (1) - frac{1}{2} operatorname{Li}_2 (-1) = frac{1}{2} left (frac{pi^2}{6} + frac{pi^2}{12} right ) = frac{pi^2}{8},$$
where the well-known special values for the dilogarithm function at 1 and $-1$ have been used.



For the third sum, setting $x = sqrt{2} - 1$ gives
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) right ]. qquad (dagger)$$
It now remains to express the difference between the two dilogarithms is terms of elementary constants.



To do this, we will make use of the following identities for the dilogarithm:
begin{align}
operatorname{Li}_2 (1-x) + operatorname{Li}_2 left (1 - frac{1}{x} right ) &= -frac{1}{2} ln^2 x tag2\
operatorname{Li}_2 (x) + operatorname{Li}_2 (1-x) &= frac{pi^2}{6} - ln x ln (1 - x) tag3\
operatorname{Li}_2 (x) + operatorname{Li}_2 (-x) &= frac{1}{2} operatorname{Li}_2 (x^2) tag4
end{align}
So here we go.
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) &= operatorname{Li}_2 [1 - (2 - sqrt{2})]\
&= -operatorname{Li}_2 left (1 - frac{1}{2 - sqrt{2}} right ) - frac{1}{2} ln^2 (2 - sqrt{2}) qquad text{(by (2))}\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 left (frac{sqrt{2}}{1 + sqrt{2}} right )\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) -frac{1}{2} left [frac{1}{2} ln 2 - ln (1 + sqrt{2}) right ]^2\
&= -operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 + frac{1}{2} ln 2 ln (1 + sqrt{2}) - frac{1}{2} ln^2 (1 + sqrt{2})
end{align}
And
begin{align}
operatorname{Li}_2 (1 - sqrt{2}) &= -operatorname{Li}_2 left (1 - frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 sqrt{2} qquad text{(by (2))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + ln left (frac{1}{sqrt{2}} right ) ln left (1 - frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 qquad text{(by (3))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} -frac{1}{2} ln 2 ln left (frac{1}{sqrt{2}(1 + sqrt{2})} right ) - frac{1}{8} ln^2 2\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 left [frac{1}{2} ln 2 + ln (1 + sqrt{2}) right ] - frac{1}{8} ln^2 2\
&=operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 ln (1 + sqrt{2}) + frac{1}{8} ln^2 2.
end{align}
Thus
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) &= - left [operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) + operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) right ] +frac{pi^2}{6} - frac{1}{2} ln^2 (1 + sqrt{2}) - frac{1}{4} ln^2 2\
&= -frac{1}{2} operatorname{Li}_2 left (frac{1}{2} right ) + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) - frac{1}{4} ln^2 2 qquad text{(by (4))} \
&= -frac{1}{2} left [frac{pi^2}{12} - frac{1}{2} ln^2 2 right ] + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) -frac{1}{4} ln^2 2\
&= frac{pi^2}{8} - frac{1}{2} ln^2 (1 + sqrt{2}).
end{align}
Here the well-known value for the dilogarithm function at $x = 1/2$ has been used.



So the sum in ($dagger$) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}).$$



Putting the final pieces of the puzzle together, (1) becomes
begin{align}
int_{1}^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -frac{1}{2} ln^2 (1 + sqrt{2}) + frac{pi^2}{8} - left [frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}) right ]\
&= frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}),
end{align}
as required.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
    $endgroup$
    – Kemono Chen
    Jan 10 at 8:12












  • $begingroup$
    That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
    $endgroup$
    – omegadot
    Jan 10 at 9:52








  • 1




    $begingroup$
    Really neat! (+1) Good work
    $endgroup$
    – clathratus
    Jan 10 at 15:19











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3 Answers
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3 Answers
3






active

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active

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9












$begingroup$

Put
begin{equation*}
I=int_{1}^{sqrt{2}+1}dfrac{ln x}{x^2-1}, mathrm{d}x =[x=1/y] = int_{sqrt{2}-1}^{1}dfrac{ln y}{y^2-1}, mathrm{d}y.tag{1}
end{equation*}
After the substitution $ y=frac{1-z}{1+z}$ and integration by parts we have
begin{gather*}
I = int_{0}^{sqrt{2}-1}dfrac{lnleft(frac{1-z}{1+z}right)}{-2z}, mathrm{d}z = left[-dfrac{1}{2}ln(z)lnleft(frac{1-z}{1+z}right)right]_{0}^{sqrt{2}-1}+int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z =\[2ex]-dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z.tag{2}
end{gather*}
If we add (1) and (2) we get
begin{equation*}
2I = -dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{1}dfrac{ln z}{z^2-1}, mathrm{d}z.
end{equation*}
Consequently
begin{gather*}
I = -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}int_{0}^{1}left(sum_{k=0}^{infty}ln(z)z^{2k}right), mathrm{d}z =\[2ex] -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}sum_{k=0}^{infty}int_{0}^{1}ln(z)z^{2k}, mathrm{d}z =\[2ex]
-dfrac{1}{4}ln^2(sqrt{2}+1) +dfrac{1}{2}sum_{k=0}^{infty}dfrac{1}{(2k+1)^2} = dfrac{pi^2}{16} -dfrac{1}{4}ln^2(sqrt{2}+1).
end{gather*}






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Now that is a solution! Very nice (+1)
    $endgroup$
    – omegadot
    Jan 10 at 22:06






  • 1




    $begingroup$
    @ omegadot$ Thank you.
    $endgroup$
    – JanG
    Jan 11 at 5:56
















9












$begingroup$

Put
begin{equation*}
I=int_{1}^{sqrt{2}+1}dfrac{ln x}{x^2-1}, mathrm{d}x =[x=1/y] = int_{sqrt{2}-1}^{1}dfrac{ln y}{y^2-1}, mathrm{d}y.tag{1}
end{equation*}
After the substitution $ y=frac{1-z}{1+z}$ and integration by parts we have
begin{gather*}
I = int_{0}^{sqrt{2}-1}dfrac{lnleft(frac{1-z}{1+z}right)}{-2z}, mathrm{d}z = left[-dfrac{1}{2}ln(z)lnleft(frac{1-z}{1+z}right)right]_{0}^{sqrt{2}-1}+int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z =\[2ex]-dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z.tag{2}
end{gather*}
If we add (1) and (2) we get
begin{equation*}
2I = -dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{1}dfrac{ln z}{z^2-1}, mathrm{d}z.
end{equation*}
Consequently
begin{gather*}
I = -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}int_{0}^{1}left(sum_{k=0}^{infty}ln(z)z^{2k}right), mathrm{d}z =\[2ex] -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}sum_{k=0}^{infty}int_{0}^{1}ln(z)z^{2k}, mathrm{d}z =\[2ex]
-dfrac{1}{4}ln^2(sqrt{2}+1) +dfrac{1}{2}sum_{k=0}^{infty}dfrac{1}{(2k+1)^2} = dfrac{pi^2}{16} -dfrac{1}{4}ln^2(sqrt{2}+1).
end{gather*}






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Now that is a solution! Very nice (+1)
    $endgroup$
    – omegadot
    Jan 10 at 22:06






  • 1




    $begingroup$
    @ omegadot$ Thank you.
    $endgroup$
    – JanG
    Jan 11 at 5:56














9












9








9





$begingroup$

Put
begin{equation*}
I=int_{1}^{sqrt{2}+1}dfrac{ln x}{x^2-1}, mathrm{d}x =[x=1/y] = int_{sqrt{2}-1}^{1}dfrac{ln y}{y^2-1}, mathrm{d}y.tag{1}
end{equation*}
After the substitution $ y=frac{1-z}{1+z}$ and integration by parts we have
begin{gather*}
I = int_{0}^{sqrt{2}-1}dfrac{lnleft(frac{1-z}{1+z}right)}{-2z}, mathrm{d}z = left[-dfrac{1}{2}ln(z)lnleft(frac{1-z}{1+z}right)right]_{0}^{sqrt{2}-1}+int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z =\[2ex]-dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z.tag{2}
end{gather*}
If we add (1) and (2) we get
begin{equation*}
2I = -dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{1}dfrac{ln z}{z^2-1}, mathrm{d}z.
end{equation*}
Consequently
begin{gather*}
I = -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}int_{0}^{1}left(sum_{k=0}^{infty}ln(z)z^{2k}right), mathrm{d}z =\[2ex] -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}sum_{k=0}^{infty}int_{0}^{1}ln(z)z^{2k}, mathrm{d}z =\[2ex]
-dfrac{1}{4}ln^2(sqrt{2}+1) +dfrac{1}{2}sum_{k=0}^{infty}dfrac{1}{(2k+1)^2} = dfrac{pi^2}{16} -dfrac{1}{4}ln^2(sqrt{2}+1).
end{gather*}






share|cite|improve this answer









$endgroup$



Put
begin{equation*}
I=int_{1}^{sqrt{2}+1}dfrac{ln x}{x^2-1}, mathrm{d}x =[x=1/y] = int_{sqrt{2}-1}^{1}dfrac{ln y}{y^2-1}, mathrm{d}y.tag{1}
end{equation*}
After the substitution $ y=frac{1-z}{1+z}$ and integration by parts we have
begin{gather*}
I = int_{0}^{sqrt{2}-1}dfrac{lnleft(frac{1-z}{1+z}right)}{-2z}, mathrm{d}z = left[-dfrac{1}{2}ln(z)lnleft(frac{1-z}{1+z}right)right]_{0}^{sqrt{2}-1}+int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z =\[2ex]-dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{sqrt{2}-1}dfrac{ln z}{z^2-1}, mathrm{d}z.tag{2}
end{gather*}
If we add (1) and (2) we get
begin{equation*}
2I = -dfrac{1}{2}ln^2(sqrt{2}+1) + int_{0}^{1}dfrac{ln z}{z^2-1}, mathrm{d}z.
end{equation*}
Consequently
begin{gather*}
I = -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}int_{0}^{1}left(sum_{k=0}^{infty}ln(z)z^{2k}right), mathrm{d}z =\[2ex] -dfrac{1}{4}ln^2(sqrt{2}+1) -dfrac{1}{2}sum_{k=0}^{infty}int_{0}^{1}ln(z)z^{2k}, mathrm{d}z =\[2ex]
-dfrac{1}{4}ln^2(sqrt{2}+1) +dfrac{1}{2}sum_{k=0}^{infty}dfrac{1}{(2k+1)^2} = dfrac{pi^2}{16} -dfrac{1}{4}ln^2(sqrt{2}+1).
end{gather*}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 21:58









JanGJanG

2,802514




2,802514








  • 2




    $begingroup$
    Now that is a solution! Very nice (+1)
    $endgroup$
    – omegadot
    Jan 10 at 22:06






  • 1




    $begingroup$
    @ omegadot$ Thank you.
    $endgroup$
    – JanG
    Jan 11 at 5:56














  • 2




    $begingroup$
    Now that is a solution! Very nice (+1)
    $endgroup$
    – omegadot
    Jan 10 at 22:06






  • 1




    $begingroup$
    @ omegadot$ Thank you.
    $endgroup$
    – JanG
    Jan 11 at 5:56








2




2




$begingroup$
Now that is a solution! Very nice (+1)
$endgroup$
– omegadot
Jan 10 at 22:06




$begingroup$
Now that is a solution! Very nice (+1)
$endgroup$
– omegadot
Jan 10 at 22:06




1




1




$begingroup$
@ omegadot$ Thank you.
$endgroup$
– JanG
Jan 11 at 5:56




$begingroup$
@ omegadot$ Thank you.
$endgroup$
– JanG
Jan 11 at 5:56











5












$begingroup$

A Complete Solution Now



Consider
$$F(s)=int_1^sfrac{log x}{x^2-1}mathrm dx$$
Like before,
$$F(s)=frac12int_1^{s}frac{log x}{x-1}mathrm dx-frac12int_1^{s}frac{log x}{x+1}mathrm dx$$
$$F(s)=-frac12mathrm{Li}_2(1-s)-frac12J(s)$$
For $J(s)$, we integrate by parts with $mathrm dv=frac{mathrm dx}{1+x}$ to get
$$J(s)=log(s)log(s+1)-int_1^sfrac{log(1+x)}{x}mathrm dx$$
$$J(s)=log(s)log(s+1)+int_2^{1+s}frac{log x}{1-x}mathrm dx$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(1-x)bigg|_2^{1+s}$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)-mathrm{Li}_2(-1)$$
Using $mathrm{Li}_k(-1)=(2^{1-k}-1)zeta(k)$,
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)+frac{pi^2}{12}$$
Plugging in $s=1+sqrt2$,
$$J(1+sqrt2)=log(1+sqrt2)log(2+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+sqrt2)=log^2(1+sqrt2)+frac12log(2)log(1+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
And again as the OP noted,
$$F(1+sqrt2)=-frac{pi^2}{48}-frac14log^2(1+sqrt2)+frac14log2log(1+sqrt2)+frac1{16}log^22+frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)$$
And since
$$frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)=frac{pi^2}{12}-frac14log2log(1+sqrt2)-frac1{16}log^22$$
We have
$$F(1+sqrt2)=frac{pi^2}{16}-frac14log^2(1+sqrt2)$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
    $endgroup$
    – Ant
    Jan 10 at 3:07






  • 1




    $begingroup$
    @Ant Right... I'm on it
    $endgroup$
    – clathratus
    Jan 10 at 3:08






  • 1




    $begingroup$
    I think the WolframAlpha result is using equality (6~7) from this MathWorld page
    $endgroup$
    – Edward H.
    Jan 10 at 5:24






  • 1




    $begingroup$
    Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:14






  • 1




    $begingroup$
    begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:22
















5












$begingroup$

A Complete Solution Now



Consider
$$F(s)=int_1^sfrac{log x}{x^2-1}mathrm dx$$
Like before,
$$F(s)=frac12int_1^{s}frac{log x}{x-1}mathrm dx-frac12int_1^{s}frac{log x}{x+1}mathrm dx$$
$$F(s)=-frac12mathrm{Li}_2(1-s)-frac12J(s)$$
For $J(s)$, we integrate by parts with $mathrm dv=frac{mathrm dx}{1+x}$ to get
$$J(s)=log(s)log(s+1)-int_1^sfrac{log(1+x)}{x}mathrm dx$$
$$J(s)=log(s)log(s+1)+int_2^{1+s}frac{log x}{1-x}mathrm dx$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(1-x)bigg|_2^{1+s}$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)-mathrm{Li}_2(-1)$$
Using $mathrm{Li}_k(-1)=(2^{1-k}-1)zeta(k)$,
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)+frac{pi^2}{12}$$
Plugging in $s=1+sqrt2$,
$$J(1+sqrt2)=log(1+sqrt2)log(2+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+sqrt2)=log^2(1+sqrt2)+frac12log(2)log(1+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
And again as the OP noted,
$$F(1+sqrt2)=-frac{pi^2}{48}-frac14log^2(1+sqrt2)+frac14log2log(1+sqrt2)+frac1{16}log^22+frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)$$
And since
$$frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)=frac{pi^2}{12}-frac14log2log(1+sqrt2)-frac1{16}log^22$$
We have
$$F(1+sqrt2)=frac{pi^2}{16}-frac14log^2(1+sqrt2)$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
    $endgroup$
    – Ant
    Jan 10 at 3:07






  • 1




    $begingroup$
    @Ant Right... I'm on it
    $endgroup$
    – clathratus
    Jan 10 at 3:08






  • 1




    $begingroup$
    I think the WolframAlpha result is using equality (6~7) from this MathWorld page
    $endgroup$
    – Edward H.
    Jan 10 at 5:24






  • 1




    $begingroup$
    Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:14






  • 1




    $begingroup$
    begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:22














5












5








5





$begingroup$

A Complete Solution Now



Consider
$$F(s)=int_1^sfrac{log x}{x^2-1}mathrm dx$$
Like before,
$$F(s)=frac12int_1^{s}frac{log x}{x-1}mathrm dx-frac12int_1^{s}frac{log x}{x+1}mathrm dx$$
$$F(s)=-frac12mathrm{Li}_2(1-s)-frac12J(s)$$
For $J(s)$, we integrate by parts with $mathrm dv=frac{mathrm dx}{1+x}$ to get
$$J(s)=log(s)log(s+1)-int_1^sfrac{log(1+x)}{x}mathrm dx$$
$$J(s)=log(s)log(s+1)+int_2^{1+s}frac{log x}{1-x}mathrm dx$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(1-x)bigg|_2^{1+s}$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)-mathrm{Li}_2(-1)$$
Using $mathrm{Li}_k(-1)=(2^{1-k}-1)zeta(k)$,
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)+frac{pi^2}{12}$$
Plugging in $s=1+sqrt2$,
$$J(1+sqrt2)=log(1+sqrt2)log(2+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+sqrt2)=log^2(1+sqrt2)+frac12log(2)log(1+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
And again as the OP noted,
$$F(1+sqrt2)=-frac{pi^2}{48}-frac14log^2(1+sqrt2)+frac14log2log(1+sqrt2)+frac1{16}log^22+frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)$$
And since
$$frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)=frac{pi^2}{12}-frac14log2log(1+sqrt2)-frac1{16}log^22$$
We have
$$F(1+sqrt2)=frac{pi^2}{16}-frac14log^2(1+sqrt2)$$






share|cite|improve this answer











$endgroup$



A Complete Solution Now



Consider
$$F(s)=int_1^sfrac{log x}{x^2-1}mathrm dx$$
Like before,
$$F(s)=frac12int_1^{s}frac{log x}{x-1}mathrm dx-frac12int_1^{s}frac{log x}{x+1}mathrm dx$$
$$F(s)=-frac12mathrm{Li}_2(1-s)-frac12J(s)$$
For $J(s)$, we integrate by parts with $mathrm dv=frac{mathrm dx}{1+x}$ to get
$$J(s)=log(s)log(s+1)-int_1^sfrac{log(1+x)}{x}mathrm dx$$
$$J(s)=log(s)log(s+1)+int_2^{1+s}frac{log x}{1-x}mathrm dx$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(1-x)bigg|_2^{1+s}$$
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)-mathrm{Li}_2(-1)$$
Using $mathrm{Li}_k(-1)=(2^{1-k}-1)zeta(k)$,
$$J(s)=log(s)log(s+1)+mathrm{Li}_2(-s)+frac{pi^2}{12}$$
Plugging in $s=1+sqrt2$,
$$J(1+sqrt2)=log(1+sqrt2)log(2+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+sqrt2)=log^2(1+sqrt2)+frac12log(2)log(1+sqrt2)+mathrm{Li}_2(-1-sqrt2)+frac{pi^2}{12}$$
And again as the OP noted,
$$F(1+sqrt2)=-frac{pi^2}{48}-frac14log^2(1+sqrt2)+frac14log2log(1+sqrt2)+frac1{16}log^22+frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)$$
And since
$$frac12mathrm{Li}_2bigg(frac1{sqrt2}bigg)-frac12mathrm{Li}_2(1-sqrt2)=frac{pi^2}{12}-frac14log2log(1+sqrt2)-frac1{16}log^22$$
We have
$$F(1+sqrt2)=frac{pi^2}{16}-frac14log^2(1+sqrt2)$$







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share|cite|improve this answer








edited Jan 10 at 19:32

























answered Jan 10 at 3:02









clathratusclathratus

4,234336




4,234336








  • 1




    $begingroup$
    Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
    $endgroup$
    – Ant
    Jan 10 at 3:07






  • 1




    $begingroup$
    @Ant Right... I'm on it
    $endgroup$
    – clathratus
    Jan 10 at 3:08






  • 1




    $begingroup$
    I think the WolframAlpha result is using equality (6~7) from this MathWorld page
    $endgroup$
    – Edward H.
    Jan 10 at 5:24






  • 1




    $begingroup$
    Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:14






  • 1




    $begingroup$
    begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:22














  • 1




    $begingroup$
    Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
    $endgroup$
    – Ant
    Jan 10 at 3:07






  • 1




    $begingroup$
    @Ant Right... I'm on it
    $endgroup$
    – clathratus
    Jan 10 at 3:08






  • 1




    $begingroup$
    I think the WolframAlpha result is using equality (6~7) from this MathWorld page
    $endgroup$
    – Edward H.
    Jan 10 at 5:24






  • 1




    $begingroup$
    Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:14






  • 1




    $begingroup$
    begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
    $endgroup$
    – Edward H.
    Jan 10 at 18:22








1




1




$begingroup$
Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
$endgroup$
– Ant
Jan 10 at 3:07




$begingroup$
Evaluating $frac14mathrm{Li}_2(1-x^2)-mathrm{Li}_2(1-x)$ at $x=1$ gives $0$, right?
$endgroup$
– Ant
Jan 10 at 3:07




1




1




$begingroup$
@Ant Right... I'm on it
$endgroup$
– clathratus
Jan 10 at 3:08




$begingroup$
@Ant Right... I'm on it
$endgroup$
– clathratus
Jan 10 at 3:08




1




1




$begingroup$
I think the WolframAlpha result is using equality (6~7) from this MathWorld page
$endgroup$
– Edward H.
Jan 10 at 5:24




$begingroup$
I think the WolframAlpha result is using equality (6~7) from this MathWorld page
$endgroup$
– Edward H.
Jan 10 at 5:24




1




1




$begingroup$
Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
$endgroup$
– Edward H.
Jan 10 at 18:14




$begingroup$
Seeing @omegadot 's answer I'm thinking that yours can also work, like this: begin{align*} &quadfrac{1}{4}text{Li}_2(-2-2sqrt{2})-text{Li}_2(-sqrt{2})\ &=frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2}+1)ln(sqrt{2}+2)-text{Li}_2(-sqrt{2}-1)+text{Li}_2(-sqrt{2})right)-text{Li}_2(-sqrt{2})\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}text{Li}_2(-sqrt{2}-1)-frac{1}{2}text{Li}_2(-sqrt{2})\ end{align*} (tbc)
$endgroup$
– Edward H.
Jan 10 at 18:14




1




1




$begingroup$
begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
$endgroup$
– Edward H.
Jan 10 at 18:22




$begingroup$
begin{align*} &quaddots\ &=-frac{pi^2}{24}-frac{1}{2}ln^2(sqrt{2}+1)-frac{1}{4}ln{2}ln(sqrt{2}+1)-frac{1}{2}left(-frac{1}{2}ln^2(sqrt{2}+2)-text{Li}_2(frac{1}{sqrt{2}})right)\ &quad-frac{1}{2}left(-frac{pi^2}{12}-ln(sqrt{2})ln(sqrt{2}+1)+text{Li}_2(1-sqrt{2})-frac{1}{2}text{Li}_2(-1)right)\ &=-frac{pi^2}{48}-frac{1}{4}ln^2(sqrt{2}+1)+frac{1}{4}ln{2}ln(sqrt{2}+1)+frac{1}{16}ln^2{2}+frac{1}{2}text{Li}_2(frac{1}{sqrt{2}})-frac{1}{2}text{Li}_2(1-sqrt{2}) end{align*} (tbc)
$endgroup$
– Edward H.
Jan 10 at 18:22











4












$begingroup$

Here is one possible approach.



Enforcing a substitution of $x mapsto 1/x$ to begin with gives
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -int_{sqrt{2}-1}^1 frac{ln x}{1 - x^2} , dx = -sum_{n = 0}^infty int_{sqrt{2} - 1}^1 x^{2n} ln x , dx,
end{align}
where we have taken advantage of the geometric sum for $1/(1 - x^2)$.
Integrating by parts then leads to
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= - ln (1 + sqrt{2}) sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} + sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&qquad - sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2}. tag1
end{align}



Observing that
$$tanh^{-1} z = sum_{n = 0}^infty frac{z^{2n + 1}}{2n + 1}, qquad |z| < 1, qquad (*)$$
the first of the sums appearing in (1) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} = tanh^{-1} (sqrt{2} - 1) = frac{1}{2} ln (1 + sqrt{2}).$$



To find the second and third sums in (1), divide ($*$) by $x$ before integrating up from $0$ to $x$. Thus
begin{align}
sum_{n = 0}^infty frac{1}{2n + 1} int_0^x t^{2n} , dt &= int_0^x frac{tanh^{-1} t}{t} , dt\
sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} &= frac{1}{2} int_0^x ln left (frac{1 + t}{1 - t} right ) frac{dt}{t}\
&= frac{1}{2} int_0^{-x} frac{ln (1 - t)}{t} , dt - frac{1}{2} int_0^x frac{ln (1 - t)}{t} , dt
end{align}
or
$$sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (x) - operatorname{Li}_2 (-x) right ], qquad (**)$$
where $operatorname{Li}_2 (x)$ is the dilogarithm function.



The second and third sums appearing in (1) can now be found using ($**$). For the second sum, setting $x = 1$ gives
$$sum_{n = 1}^infty frac{1}{(2n + 1)^2} = frac{1}{2} operatorname{Li}_2 (1) - frac{1}{2} operatorname{Li}_2 (-1) = frac{1}{2} left (frac{pi^2}{6} + frac{pi^2}{12} right ) = frac{pi^2}{8},$$
where the well-known special values for the dilogarithm function at 1 and $-1$ have been used.



For the third sum, setting $x = sqrt{2} - 1$ gives
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) right ]. qquad (dagger)$$
It now remains to express the difference between the two dilogarithms is terms of elementary constants.



To do this, we will make use of the following identities for the dilogarithm:
begin{align}
operatorname{Li}_2 (1-x) + operatorname{Li}_2 left (1 - frac{1}{x} right ) &= -frac{1}{2} ln^2 x tag2\
operatorname{Li}_2 (x) + operatorname{Li}_2 (1-x) &= frac{pi^2}{6} - ln x ln (1 - x) tag3\
operatorname{Li}_2 (x) + operatorname{Li}_2 (-x) &= frac{1}{2} operatorname{Li}_2 (x^2) tag4
end{align}
So here we go.
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) &= operatorname{Li}_2 [1 - (2 - sqrt{2})]\
&= -operatorname{Li}_2 left (1 - frac{1}{2 - sqrt{2}} right ) - frac{1}{2} ln^2 (2 - sqrt{2}) qquad text{(by (2))}\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 left (frac{sqrt{2}}{1 + sqrt{2}} right )\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) -frac{1}{2} left [frac{1}{2} ln 2 - ln (1 + sqrt{2}) right ]^2\
&= -operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 + frac{1}{2} ln 2 ln (1 + sqrt{2}) - frac{1}{2} ln^2 (1 + sqrt{2})
end{align}
And
begin{align}
operatorname{Li}_2 (1 - sqrt{2}) &= -operatorname{Li}_2 left (1 - frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 sqrt{2} qquad text{(by (2))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + ln left (frac{1}{sqrt{2}} right ) ln left (1 - frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 qquad text{(by (3))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} -frac{1}{2} ln 2 ln left (frac{1}{sqrt{2}(1 + sqrt{2})} right ) - frac{1}{8} ln^2 2\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 left [frac{1}{2} ln 2 + ln (1 + sqrt{2}) right ] - frac{1}{8} ln^2 2\
&=operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 ln (1 + sqrt{2}) + frac{1}{8} ln^2 2.
end{align}
Thus
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) &= - left [operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) + operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) right ] +frac{pi^2}{6} - frac{1}{2} ln^2 (1 + sqrt{2}) - frac{1}{4} ln^2 2\
&= -frac{1}{2} operatorname{Li}_2 left (frac{1}{2} right ) + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) - frac{1}{4} ln^2 2 qquad text{(by (4))} \
&= -frac{1}{2} left [frac{pi^2}{12} - frac{1}{2} ln^2 2 right ] + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) -frac{1}{4} ln^2 2\
&= frac{pi^2}{8} - frac{1}{2} ln^2 (1 + sqrt{2}).
end{align}
Here the well-known value for the dilogarithm function at $x = 1/2$ has been used.



So the sum in ($dagger$) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}).$$



Putting the final pieces of the puzzle together, (1) becomes
begin{align}
int_{1}^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -frac{1}{2} ln^2 (1 + sqrt{2}) + frac{pi^2}{8} - left [frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}) right ]\
&= frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}),
end{align}
as required.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
    $endgroup$
    – Kemono Chen
    Jan 10 at 8:12












  • $begingroup$
    That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
    $endgroup$
    – omegadot
    Jan 10 at 9:52








  • 1




    $begingroup$
    Really neat! (+1) Good work
    $endgroup$
    – clathratus
    Jan 10 at 15:19
















4












$begingroup$

Here is one possible approach.



Enforcing a substitution of $x mapsto 1/x$ to begin with gives
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -int_{sqrt{2}-1}^1 frac{ln x}{1 - x^2} , dx = -sum_{n = 0}^infty int_{sqrt{2} - 1}^1 x^{2n} ln x , dx,
end{align}
where we have taken advantage of the geometric sum for $1/(1 - x^2)$.
Integrating by parts then leads to
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= - ln (1 + sqrt{2}) sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} + sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&qquad - sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2}. tag1
end{align}



Observing that
$$tanh^{-1} z = sum_{n = 0}^infty frac{z^{2n + 1}}{2n + 1}, qquad |z| < 1, qquad (*)$$
the first of the sums appearing in (1) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} = tanh^{-1} (sqrt{2} - 1) = frac{1}{2} ln (1 + sqrt{2}).$$



To find the second and third sums in (1), divide ($*$) by $x$ before integrating up from $0$ to $x$. Thus
begin{align}
sum_{n = 0}^infty frac{1}{2n + 1} int_0^x t^{2n} , dt &= int_0^x frac{tanh^{-1} t}{t} , dt\
sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} &= frac{1}{2} int_0^x ln left (frac{1 + t}{1 - t} right ) frac{dt}{t}\
&= frac{1}{2} int_0^{-x} frac{ln (1 - t)}{t} , dt - frac{1}{2} int_0^x frac{ln (1 - t)}{t} , dt
end{align}
or
$$sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (x) - operatorname{Li}_2 (-x) right ], qquad (**)$$
where $operatorname{Li}_2 (x)$ is the dilogarithm function.



The second and third sums appearing in (1) can now be found using ($**$). For the second sum, setting $x = 1$ gives
$$sum_{n = 1}^infty frac{1}{(2n + 1)^2} = frac{1}{2} operatorname{Li}_2 (1) - frac{1}{2} operatorname{Li}_2 (-1) = frac{1}{2} left (frac{pi^2}{6} + frac{pi^2}{12} right ) = frac{pi^2}{8},$$
where the well-known special values for the dilogarithm function at 1 and $-1$ have been used.



For the third sum, setting $x = sqrt{2} - 1$ gives
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) right ]. qquad (dagger)$$
It now remains to express the difference between the two dilogarithms is terms of elementary constants.



To do this, we will make use of the following identities for the dilogarithm:
begin{align}
operatorname{Li}_2 (1-x) + operatorname{Li}_2 left (1 - frac{1}{x} right ) &= -frac{1}{2} ln^2 x tag2\
operatorname{Li}_2 (x) + operatorname{Li}_2 (1-x) &= frac{pi^2}{6} - ln x ln (1 - x) tag3\
operatorname{Li}_2 (x) + operatorname{Li}_2 (-x) &= frac{1}{2} operatorname{Li}_2 (x^2) tag4
end{align}
So here we go.
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) &= operatorname{Li}_2 [1 - (2 - sqrt{2})]\
&= -operatorname{Li}_2 left (1 - frac{1}{2 - sqrt{2}} right ) - frac{1}{2} ln^2 (2 - sqrt{2}) qquad text{(by (2))}\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 left (frac{sqrt{2}}{1 + sqrt{2}} right )\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) -frac{1}{2} left [frac{1}{2} ln 2 - ln (1 + sqrt{2}) right ]^2\
&= -operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 + frac{1}{2} ln 2 ln (1 + sqrt{2}) - frac{1}{2} ln^2 (1 + sqrt{2})
end{align}
And
begin{align}
operatorname{Li}_2 (1 - sqrt{2}) &= -operatorname{Li}_2 left (1 - frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 sqrt{2} qquad text{(by (2))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + ln left (frac{1}{sqrt{2}} right ) ln left (1 - frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 qquad text{(by (3))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} -frac{1}{2} ln 2 ln left (frac{1}{sqrt{2}(1 + sqrt{2})} right ) - frac{1}{8} ln^2 2\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 left [frac{1}{2} ln 2 + ln (1 + sqrt{2}) right ] - frac{1}{8} ln^2 2\
&=operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 ln (1 + sqrt{2}) + frac{1}{8} ln^2 2.
end{align}
Thus
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) &= - left [operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) + operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) right ] +frac{pi^2}{6} - frac{1}{2} ln^2 (1 + sqrt{2}) - frac{1}{4} ln^2 2\
&= -frac{1}{2} operatorname{Li}_2 left (frac{1}{2} right ) + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) - frac{1}{4} ln^2 2 qquad text{(by (4))} \
&= -frac{1}{2} left [frac{pi^2}{12} - frac{1}{2} ln^2 2 right ] + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) -frac{1}{4} ln^2 2\
&= frac{pi^2}{8} - frac{1}{2} ln^2 (1 + sqrt{2}).
end{align}
Here the well-known value for the dilogarithm function at $x = 1/2$ has been used.



So the sum in ($dagger$) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}).$$



Putting the final pieces of the puzzle together, (1) becomes
begin{align}
int_{1}^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -frac{1}{2} ln^2 (1 + sqrt{2}) + frac{pi^2}{8} - left [frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}) right ]\
&= frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}),
end{align}
as required.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
    $endgroup$
    – Kemono Chen
    Jan 10 at 8:12












  • $begingroup$
    That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
    $endgroup$
    – omegadot
    Jan 10 at 9:52








  • 1




    $begingroup$
    Really neat! (+1) Good work
    $endgroup$
    – clathratus
    Jan 10 at 15:19














4












4








4





$begingroup$

Here is one possible approach.



Enforcing a substitution of $x mapsto 1/x$ to begin with gives
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -int_{sqrt{2}-1}^1 frac{ln x}{1 - x^2} , dx = -sum_{n = 0}^infty int_{sqrt{2} - 1}^1 x^{2n} ln x , dx,
end{align}
where we have taken advantage of the geometric sum for $1/(1 - x^2)$.
Integrating by parts then leads to
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= - ln (1 + sqrt{2}) sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} + sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&qquad - sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2}. tag1
end{align}



Observing that
$$tanh^{-1} z = sum_{n = 0}^infty frac{z^{2n + 1}}{2n + 1}, qquad |z| < 1, qquad (*)$$
the first of the sums appearing in (1) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} = tanh^{-1} (sqrt{2} - 1) = frac{1}{2} ln (1 + sqrt{2}).$$



To find the second and third sums in (1), divide ($*$) by $x$ before integrating up from $0$ to $x$. Thus
begin{align}
sum_{n = 0}^infty frac{1}{2n + 1} int_0^x t^{2n} , dt &= int_0^x frac{tanh^{-1} t}{t} , dt\
sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} &= frac{1}{2} int_0^x ln left (frac{1 + t}{1 - t} right ) frac{dt}{t}\
&= frac{1}{2} int_0^{-x} frac{ln (1 - t)}{t} , dt - frac{1}{2} int_0^x frac{ln (1 - t)}{t} , dt
end{align}
or
$$sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (x) - operatorname{Li}_2 (-x) right ], qquad (**)$$
where $operatorname{Li}_2 (x)$ is the dilogarithm function.



The second and third sums appearing in (1) can now be found using ($**$). For the second sum, setting $x = 1$ gives
$$sum_{n = 1}^infty frac{1}{(2n + 1)^2} = frac{1}{2} operatorname{Li}_2 (1) - frac{1}{2} operatorname{Li}_2 (-1) = frac{1}{2} left (frac{pi^2}{6} + frac{pi^2}{12} right ) = frac{pi^2}{8},$$
where the well-known special values for the dilogarithm function at 1 and $-1$ have been used.



For the third sum, setting $x = sqrt{2} - 1$ gives
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) right ]. qquad (dagger)$$
It now remains to express the difference between the two dilogarithms is terms of elementary constants.



To do this, we will make use of the following identities for the dilogarithm:
begin{align}
operatorname{Li}_2 (1-x) + operatorname{Li}_2 left (1 - frac{1}{x} right ) &= -frac{1}{2} ln^2 x tag2\
operatorname{Li}_2 (x) + operatorname{Li}_2 (1-x) &= frac{pi^2}{6} - ln x ln (1 - x) tag3\
operatorname{Li}_2 (x) + operatorname{Li}_2 (-x) &= frac{1}{2} operatorname{Li}_2 (x^2) tag4
end{align}
So here we go.
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) &= operatorname{Li}_2 [1 - (2 - sqrt{2})]\
&= -operatorname{Li}_2 left (1 - frac{1}{2 - sqrt{2}} right ) - frac{1}{2} ln^2 (2 - sqrt{2}) qquad text{(by (2))}\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 left (frac{sqrt{2}}{1 + sqrt{2}} right )\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) -frac{1}{2} left [frac{1}{2} ln 2 - ln (1 + sqrt{2}) right ]^2\
&= -operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 + frac{1}{2} ln 2 ln (1 + sqrt{2}) - frac{1}{2} ln^2 (1 + sqrt{2})
end{align}
And
begin{align}
operatorname{Li}_2 (1 - sqrt{2}) &= -operatorname{Li}_2 left (1 - frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 sqrt{2} qquad text{(by (2))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + ln left (frac{1}{sqrt{2}} right ) ln left (1 - frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 qquad text{(by (3))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} -frac{1}{2} ln 2 ln left (frac{1}{sqrt{2}(1 + sqrt{2})} right ) - frac{1}{8} ln^2 2\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 left [frac{1}{2} ln 2 + ln (1 + sqrt{2}) right ] - frac{1}{8} ln^2 2\
&=operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 ln (1 + sqrt{2}) + frac{1}{8} ln^2 2.
end{align}
Thus
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) &= - left [operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) + operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) right ] +frac{pi^2}{6} - frac{1}{2} ln^2 (1 + sqrt{2}) - frac{1}{4} ln^2 2\
&= -frac{1}{2} operatorname{Li}_2 left (frac{1}{2} right ) + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) - frac{1}{4} ln^2 2 qquad text{(by (4))} \
&= -frac{1}{2} left [frac{pi^2}{12} - frac{1}{2} ln^2 2 right ] + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) -frac{1}{4} ln^2 2\
&= frac{pi^2}{8} - frac{1}{2} ln^2 (1 + sqrt{2}).
end{align}
Here the well-known value for the dilogarithm function at $x = 1/2$ has been used.



So the sum in ($dagger$) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}).$$



Putting the final pieces of the puzzle together, (1) becomes
begin{align}
int_{1}^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -frac{1}{2} ln^2 (1 + sqrt{2}) + frac{pi^2}{8} - left [frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}) right ]\
&= frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}),
end{align}
as required.






share|cite|improve this answer











$endgroup$



Here is one possible approach.



Enforcing a substitution of $x mapsto 1/x$ to begin with gives
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -int_{sqrt{2}-1}^1 frac{ln x}{1 - x^2} , dx = -sum_{n = 0}^infty int_{sqrt{2} - 1}^1 x^{2n} ln x , dx,
end{align}
where we have taken advantage of the geometric sum for $1/(1 - x^2)$.
Integrating by parts then leads to
begin{align}
int_1^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= - ln (1 + sqrt{2}) sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} + sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&qquad - sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2}. tag1
end{align}



Observing that
$$tanh^{-1} z = sum_{n = 0}^infty frac{z^{2n + 1}}{2n + 1}, qquad |z| < 1, qquad (*)$$
the first of the sums appearing in (1) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{2n+1} = tanh^{-1} (sqrt{2} - 1) = frac{1}{2} ln (1 + sqrt{2}).$$



To find the second and third sums in (1), divide ($*$) by $x$ before integrating up from $0$ to $x$. Thus
begin{align}
sum_{n = 0}^infty frac{1}{2n + 1} int_0^x t^{2n} , dt &= int_0^x frac{tanh^{-1} t}{t} , dt\
sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} &= frac{1}{2} int_0^x ln left (frac{1 + t}{1 - t} right ) frac{dt}{t}\
&= frac{1}{2} int_0^{-x} frac{ln (1 - t)}{t} , dt - frac{1}{2} int_0^x frac{ln (1 - t)}{t} , dt
end{align}
or
$$sum_{n = 0}^infty frac{x^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (x) - operatorname{Li}_2 (-x) right ], qquad (**)$$
where $operatorname{Li}_2 (x)$ is the dilogarithm function.



The second and third sums appearing in (1) can now be found using ($**$). For the second sum, setting $x = 1$ gives
$$sum_{n = 1}^infty frac{1}{(2n + 1)^2} = frac{1}{2} operatorname{Li}_2 (1) - frac{1}{2} operatorname{Li}_2 (-1) = frac{1}{2} left (frac{pi^2}{6} + frac{pi^2}{12} right ) = frac{pi^2}{8},$$
where the well-known special values for the dilogarithm function at 1 and $-1$ have been used.



For the third sum, setting $x = sqrt{2} - 1$ gives
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{1}{2} left [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) right ]. qquad (dagger)$$
It now remains to express the difference between the two dilogarithms is terms of elementary constants.



To do this, we will make use of the following identities for the dilogarithm:
begin{align}
operatorname{Li}_2 (1-x) + operatorname{Li}_2 left (1 - frac{1}{x} right ) &= -frac{1}{2} ln^2 x tag2\
operatorname{Li}_2 (x) + operatorname{Li}_2 (1-x) &= frac{pi^2}{6} - ln x ln (1 - x) tag3\
operatorname{Li}_2 (x) + operatorname{Li}_2 (-x) &= frac{1}{2} operatorname{Li}_2 (x^2) tag4
end{align}
So here we go.
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) &= operatorname{Li}_2 [1 - (2 - sqrt{2})]\
&= -operatorname{Li}_2 left (1 - frac{1}{2 - sqrt{2}} right ) - frac{1}{2} ln^2 (2 - sqrt{2}) qquad text{(by (2))}\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 left (frac{sqrt{2}}{1 + sqrt{2}} right )\
&= - operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) -frac{1}{2} left [frac{1}{2} ln 2 - ln (1 + sqrt{2}) right ]^2\
&= -operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 + frac{1}{2} ln 2 ln (1 + sqrt{2}) - frac{1}{2} ln^2 (1 + sqrt{2})
end{align}
And
begin{align}
operatorname{Li}_2 (1 - sqrt{2}) &= -operatorname{Li}_2 left (1 - frac{1}{sqrt{2}} right ) - frac{1}{2} ln^2 sqrt{2} qquad text{(by (2))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + ln left (frac{1}{sqrt{2}} right ) ln left (1 - frac{1}{sqrt{2}} right ) - frac{1}{8} ln^2 2 qquad text{(by (3))}\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} -frac{1}{2} ln 2 ln left (frac{1}{sqrt{2}(1 + sqrt{2})} right ) - frac{1}{8} ln^2 2\
&= operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 left [frac{1}{2} ln 2 + ln (1 + sqrt{2}) right ] - frac{1}{8} ln^2 2\
&=operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) - frac{pi^2}{6} + frac{1}{2} ln 2 ln (1 + sqrt{2}) + frac{1}{8} ln^2 2.
end{align}
Thus
begin{align}
operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2}) &= - left [operatorname{Li}_2 left (frac{1}{sqrt{2}} right ) + operatorname{Li}_2 left (-frac{1}{sqrt{2}} right ) right ] +frac{pi^2}{6} - frac{1}{2} ln^2 (1 + sqrt{2}) - frac{1}{4} ln^2 2\
&= -frac{1}{2} operatorname{Li}_2 left (frac{1}{2} right ) + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) - frac{1}{4} ln^2 2 qquad text{(by (4))} \
&= -frac{1}{2} left [frac{pi^2}{12} - frac{1}{2} ln^2 2 right ] + frac{pi^2}{6} - frac{1}{2} ln^2(1 + sqrt{2}) -frac{1}{4} ln^2 2\
&= frac{pi^2}{8} - frac{1}{2} ln^2 (1 + sqrt{2}).
end{align}
Here the well-known value for the dilogarithm function at $x = 1/2$ has been used.



So the sum in ($dagger$) can be expressed as
$$sum_{n = 0}^infty frac{(sqrt{2} - 1)^{2n + 1}}{(2n + 1)^2} = frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}).$$



Putting the final pieces of the puzzle together, (1) becomes
begin{align}
int_{1}^{1 + sqrt{2}} frac{ln x}{x^2 - 1} , dx &= -frac{1}{2} ln^2 (1 + sqrt{2}) + frac{pi^2}{8} - left [frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}) right ]\
&= frac{pi^2}{16} - frac{1}{4} ln^2 (1 + sqrt{2}),
end{align}
as required.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 12:54

























answered Jan 10 at 7:39









omegadotomegadot

5,5402728




5,5402728












  • $begingroup$
    Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
    $endgroup$
    – Kemono Chen
    Jan 10 at 8:12












  • $begingroup$
    That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
    $endgroup$
    – omegadot
    Jan 10 at 9:52








  • 1




    $begingroup$
    Really neat! (+1) Good work
    $endgroup$
    – clathratus
    Jan 10 at 15:19


















  • $begingroup$
    Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
    $endgroup$
    – Kemono Chen
    Jan 10 at 8:12












  • $begingroup$
    That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
    $endgroup$
    – omegadot
    Jan 10 at 9:52








  • 1




    $begingroup$
    Really neat! (+1) Good work
    $endgroup$
    – clathratus
    Jan 10 at 15:19
















$begingroup$
Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
$endgroup$
– Kemono Chen
Jan 10 at 8:12






$begingroup$
Actually, the solution can be simplified by using the chi function's known value which is re-proved by you.
$endgroup$
– Kemono Chen
Jan 10 at 8:12














$begingroup$
That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
$endgroup$
– omegadot
Jan 10 at 9:52






$begingroup$
That is correct, but I thought perhaps the dilogarithm function and some of its special values would be more familiar to work with rather than $chi_2 (x)$, the Legendre chi function of order 2. Some special values for this function are also known and can be found here, one of which is the value I found for ($dagger$), namely $chi_2 (sqrt{2} - 1) = frac{1}{2} [operatorname{Li}_2 (sqrt{2} - 1) - operatorname{Li}_2 (1 - sqrt{2})] = frac{pi^2}{16} - frac{1}{4} ln^2(1 + sqrt{2})$.
$endgroup$
– omegadot
Jan 10 at 9:52






1




1




$begingroup$
Really neat! (+1) Good work
$endgroup$
– clathratus
Jan 10 at 15:19




$begingroup$
Really neat! (+1) Good work
$endgroup$
– clathratus
Jan 10 at 15:19


















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