Prove that : $ \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $

Question

Denoting $ \varphi=\frac{1+\sqrt{5}}{2}=\mathrm{Golden\ Ratio} $. How would you prove that : $$ \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $$

I came accross this strange identity while reading the following article : VALUES OF THE RIEMANN ZETA FUNCTION AND INTEGRALS INVOLVING $\log{\left(2\sinh{\frac{\theta}{2}}\right)}$ AND $\log{\left(2\sin{\frac{\theta}{2}}\right)}$. (No complete proof was given in the article). I'm still trying to prove it. For now I did try to set $ x=2\ln{\left(\sinh{\frac{\theta}{2}}\right)} $ to get rid of the golden ratio, but it doesn't seem to get any better. I know I'll have to expand the integrand to get to $ \zeta $ using, for example, the fact that : $ \ln{\left(\sinh{x}\right)}=x-\ln{2}-\sum\limits_{n=1}^{+\infty}{\frac{\mathrm{e}^{-2nx}}{n}} $, but I'm sure I cannot do it now, because of that golden ratio in the upper bound. Any ideas, hints will be appreciated.

Answer 1

First, we need a preliminary result:

First preliminary result :

$$\ln\left(2 \sinh\left(\frac{x}{2}\right)\right)=\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \tag{1}$$

Proof:

$$ \begin{aligned} \ln\left( \sinh(x)\right)&=\ln\left( \frac{1}{2} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+\ln\left( e^{x}-e^{-x} \right)\\ &=-\ln 2+\ln\left( \frac{e^{-x}}{e^{-x}} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+x+\ln\left( 1-e^{-2x} \right)\\ &=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n} \qquad \blacksquare \end{aligned} $$

Letting $x \to \frac{x}{2}$ completes the proof

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Than:

$$ \begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}&=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \\ &=-2\int_0^{2\ln(\phi)}x\left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \right)\,dx & \left( \text{by eq. (1)}\right)\\ &=-\int_0^{2\ln(\phi)}x^2\,dx+2\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{2\ln(\phi)} xe^{-kx}\,dx\\ &=-\frac{8}{3}\ln^3(\phi)+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{2\ln(\phi)\phi^{-2k}}{k}+\frac{1}{k}\int_0^{2\ln(\phi)} e^{-kx}\,dx \right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^2}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k^2}-\frac{(\phi^{-2})^k}{k^2}\right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})+2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^3}\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})-2\operatorname{Li}_3(\phi^{-2})+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\left( \frac{\pi^{2}}{15}-\ln ^{2} \phi\right)-2\left(\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \right)+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)+4\ln^3(\phi)-\frac{4}{3}\ln^3(\phi)-\frac85\zeta(3)+2\zeta(3)\\ &=\frac25\zeta(3) \qquad \blacksquare \end{aligned} $$

Which is the representation of $\zeta(3)$ that Apery used to prove the irrationality of $\zeta(3)$.

Note that we used

$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) =\frac{\pi^{2}}{15}-\ln ^{2} \phi$

A proof can be found here in my blog

And:

$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$

Proof of the second relation: Recall the Trilogarithm identity proved here

$ \operatorname{Li}_{3}(x)+\operatorname{Li}_{3}(1-x)+\operatorname{Li}_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}$

And the polylogarithm relation:

$\operatorname{Li}_{n}(x)+\operatorname{Li}_{n}(-x)=2^{1-n}\operatorname{Li}_{n}(x^2)$

Example, letting $x=\phi^{-1}$ in the last relation we obtain

$$\operatorname{Li}_{n}(\phi^{-1})+\operatorname{Li}_{n}(-\phi^{-1})=\frac14\operatorname{Li}_{n}(\phi^{-2})$$

Claim:

$$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$$

Proof:

If we let $x=\phi^{-1}$ in $(2)$ we obtain

$$ \begin{aligned} &\operatorname{Li}_{3}\left(\phi^{-1}\right)+\operatorname{Li}_{3}(1-\phi^{-1})+\operatorname{Li}_{3}\left(1-\frac{1}{\phi^{-1}}\right)=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}-\frac{\ln ^{2}(\phi) \ln (1-\phi^{-1})}{2}\\ &\operatorname{Li}_3\left(\phi^{-1} \right)+\operatorname{Li}_3\left(\phi^{-2} \right)+\operatorname{Li}_3\left(-\phi^{-1} \right)=\zeta(3)-\frac{\ln^3(\phi)}{6}-\frac{\pi^2\ln(\phi)}{6}-\frac{\ln^2(\phi)\ln(\phi^{-2})}{2}\\ &\frac14\operatorname{Li}_{3}\left(\phi^{-2}\right)+\operatorname{Li}_{3}(\phi^{-2})=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}+\ln ^{3}(\phi) \\ &\frac54\operatorname{Li}_{3}\left(\phi^{-2}\right)=\zeta(3)+\frac{5\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6} \\ &\operatorname{Li}_{3}\left(\phi^{-2}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2}\ln (\phi)}{15} \qquad \blacksquare \\ \end{aligned} $$

Answer 2

$$I=\int_0^{2\ln\varphi}\theta\ln\left(2\sinh\frac\theta2\right)\,d\theta$$ if we let $\alpha=\frac\theta2\Rightarrow d\theta=2d\alpha$ soL $$I=\int_0^{\ln\varphi}2\alpha\ln\left(2\sinh\alpha\right).2\,d\alpha=4\int_0^{\ln\varphi}\alpha[\ln(2)+\ln(\sinh\alpha)]d\alpha$$ now you can break it up: $I=4(I_1+I_2)$ so: $$I_1=\int_0^{\ln\varphi}\alpha\ln(2)\,d\alpha=\frac{\ln2}{2}\left[\alpha^2\right]_0^{\ln\varphi}=\frac12\ln(2)\ln^2(\varphi)$$ now: $$I_2=\int_0^{\ln\varphi}\alpha\ln(\sinh(\alpha))\,d\alpha=\int_0^{\ln\varphi}\left(\alpha^2-\alpha\ln(2)-\alpha\sum_{n=1}^\infty\frac{e^{-2n\alpha}}{n}\right)d\alpha$$ now notice the middle term is just $I_1$ so that will cancel nicely so we may as well redefine: $I=I_3-I_4$ $$I_3=\int_0^{\ln\varphi}\alpha^2\,d\alpha$$ $$I_4=\sum_{n=1}^\infty\frac1n\int_0^{\ln\varphi}\alpha e^{-2n\alpha}\,d\alpha$$ now the first integral is easy, and with the second integral you can just use IBP, you should get a series involving powers of $\varphi$ for which relations do exist, hope this helps :)

The bottom integral would evaluate to: $$\frac{1}{4n^2}-\frac{\varphi^{-2n}\left(2n\ln\varphi+1\right)}{4n^2}$$ which does look tricky to link to $1/n^3$

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Remember that the $n$th term of the Fibonacci sequence is given by: $$F(n)=\frac{\varphi^n-(-\bar{\varphi})^n}{\sqrt{5}}$$ so maybe there is a weird link here somewhere

Answer 3

A CAS gives the antiderivative (have a look here) $$I(\theta)=\int\theta \log \Bigg[2 \sinh \left(\frac{\theta }{2}\right)\Bigg]\,d\theta$$ $$I(\theta)=\frac{1}{12} \left(\theta ^3+6 \theta ^2 \left(-\log \left(1-e^{\theta }\right)+\log \left(\sinh \left(\frac{\theta }{2}\right)\right)+\log (2)\right)-12 \theta \text{Li}_2\left(e^{\theta }\right)+12 \text{Li}_3\left(e^{\theta }\right)-i \pi ^3\right)$$ $$\lim_{\theta \to 0} \, I(\theta )=\zeta (3)-\frac{i \pi ^3}{12}$$ and the surprise is that, after simplifications $$I(2\log(\varphi))=\frac{4 }{5}\zeta (3)-\frac{i \pi ^3}{12}$$

What is also interesting is to see how fast converges the integral after a series expansion $$\theta \log \Bigg[2 \sinh \left(\frac{\theta }{2}\right)\Bigg]=\theta\log(\theta)+\sum_{n=1}^k a_k\,\theta^{2k+1}$$

$$\left( \begin{array}{cc} k & \text{from series expansion}\\ 1 & -0.240365891808504 \\ 2 & -0.240411880810115 \\ 3 & -0.240411373694167 \\ 4 & -0.240411380739980 \\ 5 & -0.240411380630111 \\ 6 & -0.240411380631951 \\ 7 & -0.240411380631918 \\ 8 & -0.240411380631919 \end{array} \right)$$

Answer 4

This may not be the exact answer you're looking for since it requires two polylogarithmic identities that must be calculated beforehand. And given how most of the terms in the penultimate step seem to cancel each other out, it makes me believe that there is a better solution. Nevertheless, I will present this solution in hopes of possibly inspiring someone else to come up with a more concise and efficient answer!

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Calling the integral $\mathfrak{I}$, enforce the transformation $x\mapsto\tfrac 12x$ to get

$$\mathfrak{I}=4\int\limits_0^{\log\varphi}\mathrm dx\, x\log(2\sinh x)$$

Now, recall that $\sinh x$ can be rewritten in terms of $e$ as such

$$\log(2\sinh x)=\log\left(e^{x}-e^{-x}\right)=x+\log\left(1-e^{-2x}\right)$$

After splitting up the integral and evaluating each component, one arrives at

$$\mathfrak{I}=\frac 43\log^3\varphi+4\int\limits_0^{\log\varphi}\mathrm dx\, x\log\left(1-e^{-2x}\right)$$

For the natural logarithm, use its Taylor Expansion to rewrite the integral with an infinite sum.

$$\mathfrak{I}=\frac 43\log^3\varphi-4\sum\limits_{n\geq1}\frac 1n\int\limits_0^{\log\varphi}\mathrm dx\, xe^{-2nx}$$

Integrate the resulting expression using integration by parts on $u=x$ and $\mathrm dv=e^{-2nx}\mathrm dx$ to see that

$$\int\limits_0^{\log\varphi}\mathrm dx\, xe^{-2nx}=-\frac 1{2n}e^{-2n\log\varphi}\log\varphi-\frac 1{4n^2}e^{-2n\log\varphi}+\frac 1{4n^2}$$

Hence

$$\begin{align*}\mathfrak{I} & =\frac 43\log^3\varphi-4\sum\limits_{n\geq1}\frac 1{n}\left(\frac 1{4n^2}-\frac 1{4n^2}e^{-2n\log\varphi}-\frac 1{2n}e^{-2n\log\varphi}\log\varphi\right)\\ & =\frac 43\log^3\varphi-\zeta(3)+\sum\limits_{n\geq1}\frac 1{n^3}\varphi^{-2n}+2\sum\limits_{n\geq1}\frac 1{n^2}\varphi^{-2n}\log\varphi\\ & =\frac 43\log^3\varphi-\zeta(3)+\operatorname{Li}_3(\varphi^{-2})+2\operatorname{Li}_2(\varphi^{-2})\log\varphi\end{align*}$$

The complex polylogarithmic results can be simplified with the following two identities

$$\begin{align*}\operatorname{Li}_2\left(\varphi^{-2}\right)& =\frac {\pi^2}{15}-\log^2\varphi\\\operatorname{Li}_3\left(\varphi^{-2}\right) & =\frac 45\zeta(3)-\frac {2\pi^2}{15}\log\varphi+\frac 23\log^3\varphi\end{align*}$$

To give

$$\int\limits_0^{2\log\varphi}\mathrm dx\, x\log\left(2\sinh\frac x2\right)\color{blue}{=-\frac 15\zeta(3)}$$