Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $

Solution 1:

Using the third of these equations we can express your integrals in terms of the Clausen function: \begin{align} I &\equiv \int \limits_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \log (\cot(t))\, \mathrm{d} t + \int \limits_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \log (\cot(t))\, \mathrm{d} t \\ &= \frac{1}{2} \left[- \operatorname{Cl}_2 \left(\vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)}\frac{\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{3\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{5\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{7\pi}{18}\right) \right. \\ &\phantom{= \frac{1}{2} \left[ \vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)} \right.} \left.+ \operatorname{Cl}_2 \left(\frac{11\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{13\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{15\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{17\pi}{18}\right)\right] \\ &= \frac{1}{2} \left[\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) - \sum \limits_{k=0}^{8} (-1)^k \operatorname{Cl}_2 \left(\frac{(2k+1)\pi}{18}\right) \right] \, . \end{align} The Fourier series of $\operatorname{Cl}_2$ and its special value $\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) = \mathrm{G}$ allow us to write this result as $$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{n=1}^\infty \frac{\alpha(n)}{n^2}\right]$$ with $\alpha \colon \mathbb{N} \to \mathbb{R}$ given by $$\alpha (n) = \sum \limits_{k=0}^8 (-1)^k \sin \left(\frac{(2k+1)n\pi}{18}\right) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \sum \limits_{k=0}^{8} \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right) k} \right] \, .$$ Clearly, we have $$\alpha (9 + 18 m) = \operatorname{Im} \left[\mathrm{i} (-1)^m \sum \limits_{k=0}^{8} 1 \right] = 9 (-1)^m $$ for $m \in \mathbb{N}_0$, while for all other $n \in \mathbb{N}$ we find $$ \alpha (n) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \frac{1 - \mathrm{e}^{\mathrm{i} \pi (9+n)}}{1 - \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right)}} \right] = \operatorname{Im} \left[2 \frac{1 + (-1)^n}{\left \lvert 1 + \mathrm{e}^{\mathrm{i} \pi n/9} \right \rvert^2} \cos \left(\frac{n \pi}{18}\right)\right] = 0 \, .$$ Therefore, $$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{m=0}^\infty \frac{9 (-1)^m}{(9 + 18 m)^2}\right] = \frac{1}{2} \left[\mathrm{G} - \frac{1}{9}\sum \limits_{m=0}^\infty \frac{(-1)^m}{(2m+1)^2}\right] = \frac{4}{9} \mathrm{G} \, .$$

The same method leads to the more general result \begin{align} \sum \limits_{\mu = 0}^{\nu - 1} (-1)^{\nu - 1 - \mu} \int \limits_0^{\frac{2\mu +1}{2\nu+1} \frac{\pi}{4}} \log(\cot(t)) \, \mathrm{d} t &= \frac{1}{2} \left[\mathrm{G} + (-1)^{\nu - 1} \sum \limits_{k=0}^{2\nu} (-1)^k \operatorname{Cl}_2 \left(\frac{2k+1}{2\nu+1} \frac{\pi}{2}\right)\right] \\ &= \frac{\nu + 1_{2\mathbb{N} - 1} (\nu)}{2\nu+1} \mathrm{G} \end{align} for $\nu \in \mathbb{N}$ ($1_{2\mathbb{N}-1}$ is the indicator function of the odd positive integers). The first three cases are \begin{align} \int \limits_0^{\frac{\pi}{12}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{3} \mathrm{G} \, , \\ \int \limits_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{5} \mathrm{G} \, , \\ \int \limits_0^{\frac{\pi}{28}} \log(\cot(t)) \, \mathrm{d} t + \int \limits_{\frac{3\pi}{28}}^{\frac{5\pi}{28}} \log(\cot(t)) \, \mathrm{d} t &= \frac{4}{7} \mathrm{G} \end{align} and $\nu = 4$ corresponds to the original question.

Solution 2:

Rewriting

Use the trigonometric identity $\csc^2(x) = \cot^2+1$. Rewrite the integral as $$\int\limits_{\alpha}^{\beta}\csc^2(x)\left(-\frac{\log\cot x}{\cot^2+1}\right)\,\mathrm{d}x$$ After substituting $u=\cot x$ we get the integral $$-\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u^2+1}\,\mathrm{d}u=-\left[\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u+i}\,\mathrm{d}u-\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u-i}\,\mathrm{d}u\right].$$

First integral

Sub $v=u+i$. You should get $$\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log v-i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log iv+1}{v}\,\mathrm{d}v+\log(-i)\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{1}{v}\,\mathrm{d}v.$$ Integral 1

Sub $w=-iv$. This will get us to $$-\int\limits_{1-i\cot(\alpha)}^{1-i\cot(\beta)}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(1-i\cot(\beta))-\operatorname{Li}_2(1-i\cot(\alpha))\Big]$$

Integral 2

Note that $\log(-i)=-i\pi/2$. We get: $$-i\pi/2\Bigg[\log(\cot(\beta)+i)-\log(\cot(\alpha)+i)\Bigg].$$

Second Integral

Sub $v=u-i$. You should get $$\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log v+i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log 1-iv}{v}\,\mathrm{d}v+\log(i)\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{1}{v}\,\mathrm{d}v.$$ Integral 1

Sub $w=iv$. This will get us to $$-\int\limits_{i\cot(\alpha)-1}^{i\cot(\beta)-1}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(i\cot(\beta)-1)-\operatorname{Li}_2(i\cot(\alpha)-1)\Big]$$

Integral 2

Note that $\log(i)=i\pi/2$. We get: $$i\pi/2\Bigg[\log(\cot(\beta)-i)-\log(\cot(\alpha)-i)\Bigg].$$

Values for $\alpha$ and $\beta$

For the first orignal integral we have

$$-\frac{{\pi}\left(\log\left(\tan^2\left(\frac{7{\pi}}{36}\right)+1\right)-\log\left(\tan^2\left(\frac{5{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{7{\pi}}{36}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{7{\pi}}{36}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{5{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{5{\pi}}{36}\right)\right)\right)}{4}$$

For the second orignal integral we have

$$-\dfrac{{\pi}\left(\log\left(\tan^2\left(\frac{{\pi}}{12}\right)+1\right)-\log\left(\tan^2\left(\frac{{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{12}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{12}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{36}\right)\right)\right)}{4}$$

A few last words

I added these values numerically and I got the right result. The integral is solved but it still needs some long (algebraic) operations to prove the end result. But the Polylogarithm $\operatorname{Li}_2$ looks promising. Maybe I will add the final derivation later on, but first I have to take a break of this monster challenge. Also, if someone finds a faster way and all of this is useless I would be twice as happy. Nonetheless I hope it has helped you at least a bit.

Solution 3:

Too long for comments.

Let $$f(a)=\int_{5a}^{7a}\log (\cot (t))\,dt+\int_{a}^{3a}\log (\cot (t))\,dt$$ leads to $$f(a)=a \Big[-\log (\cot (a))+3 \log (\cot (3 a))-5 \log (\cot (5 a))+7 \log (\cot (7 a))\Big]+$$ $$\frac{\pi}{2} \Big[ \log (\cot (a))-\log (\cot (3 a))+\log (\cot (5 a))-\log (\cot (7 a))\Big]+$$ $$\frac{i}{2} (\text{Li}_2(i \cot (a))-\text{Li}_2(-i \cot (a)))-$$ $$\frac{i}{2} (\text{Li}_2(i \cot (3a))-\text{Li}_2(-i \cot (3a)))+$$ $$\frac{i}{2} (\text{Li}_2(i \cot (5a))-\text{Li}_2(-i \cot (5a)))-$$ $$\frac{i}{2} (\text{Li}_2(i \cot (7a))-\text{Li}_2(-i \cot (7a)))$$ At this point, I am stuck.

Expanding $f(a)$ as a (long) Taylor series around $a=0$ and making $a=\frac \pi {36}$ leads to $$f\left(\frac{\pi}{36}\right)= 0.407095819634319562246490451081059604788510834$$ which is recognized as $\frac{4 }{9}C$ by the inverse symbolic calculator (notice that all digits are exact).

One point on the side : $\frac{4 }{9}C$ is the maximum value of $f(a)$.

Solution 4:

Here is an attempt at an elementary solution. The identity is readily seen to be the same as $$\int_{\pi/4}^{3\pi/4}\log\cot\frac v9\,dv+\int_{5\pi/4}^{7\pi/4}\log\cot\frac v9\,dv=4G.$$ We can combine both integrals into one through the linear substitution $2u=v-3\pi/4\pm\pi/2$ to get $$\int_0^{\pi/4}\log\cot\left(\frac{2u}9+\frac\pi{36}\right)+\log\cot\left(\frac{2u}9+\frac{5\pi}{36}\right)\,du=2G.$$ Taking advantage of symmetry over $t=2u/9+\pi/12$, the integral can be written as \begin{align}\frac49G&=\int_{\pi/12}^{5\pi/36}\log\left(\cot\left(t-\frac\pi{18}\right)\cot\left(t+\frac\pi{18}\right)\right)\,dt\\&=\int_{\pi/12}^{5\pi/36}\log\frac{-\cos2t-\cos\pi/9}{\cos2t-\cos\pi/9}\,dt\end{align} or through a Taylor series expansion,$$\frac89G=\int_{\pi/6}^{5\pi/18}\log\frac{\cos\pi/9+\cos x}{\cos\pi/9-\cos x}\,dx=\int_{\pi/6}^{5\pi/18}\sum_{k=1}^\infty\frac2{2k-1}\left(\frac{\cos x}{\cos\pi/9}\right)^{2k-1}\,dx.$$ We can exchange operators to get \begin{align}\frac49G&=\sum_{k=1}^\infty\frac{(\sec\pi/9)^{2k-1}}{2k-1}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx\end{align} and since \begin{align}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx&=\int_{\pi/6}^{5\pi/18}\frac1{2^{2k-2}}\sum_{m=0}^{k-1}\binom{2k-1}m\cos(2k-2m-1)x\,dx\\&=\frac4{4^k}\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{5(2k-2m-1)\pi}{18}-\sin\frac{(2k-2m-1)\pi}6}{2k-2m-1}\end{align} the identity is equivalent to $$\frac1{18}G=\sum_{k=1}^\infty\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{(2k-2m-1)\pi}9\cos\frac{4(2k-2m-1)\pi}9}{4^k(2k-1)(2k-2m-1)(\cos\pi/9)^{2k-1}}.$$ The trig expressions should reduce to something simpler (and not involving trig functions) — need help!

Remark: if $\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx=a+b\sin\pi/18+c\cos\pi/9+d\cos2\pi/9$ for some rational $a,b,c,d$ then $$\frac49G=\sum_{k=1}^\infty\frac{a+(b-d)\sin\pi/18+(c+d)\cos\pi/9}{(2k-1)(\cos\pi/9)^{2k-1}}$$