HARD INTEGRAL - PART II

Today We will compute the following integral following the same ideas of the previous post:

$\begin{aligned} \int_0^{\pi/3}\ln^2\left( \sin(x) \right)\,dx7&=\frac{\pi \ln^2{2}}{3}-\frac{\pi \ln^2(3)}{12}+\frac{\pi^3}{81}-\Im \ {\rm Li}_3(1-e^{i2\pi/3})\\ \\ & \qquad+\frac{\ln(2)}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \end{aligned}$

Recall (see here)

$\ln^2\left( \sin(x) \right)=-\ln^2(2)-2\ln(2)\ln(\sin(x))+\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\cos(2nx)}{n}-2 \sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}$ (1)

Integrating both sides of (1) from 0 to $\pi/3$

$\begin{aligned} \int_0^{\pi/3}\ln^2\left( \sin(x) \right)\,dx&=-\ln^2(2)\int_0^{\pi/3}\,dx-2\ln(2)\int_0^{\pi/3}\ln(\sin(x))\,dx+\frac{\pi^2}4\int_0^{\pi/3}\,dx-\pi\int_0^{\pi/3} x\,dx+\int_0^{\pi/3}x^2\,dx+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n}\int_0^{\pi/3}\cos(2nx)\,dx-2 \sum_{n=1}^\infty \frac{1}{n^2}\int_0^{\pi/3}\cos(2nx)\,dx\\ &=-\frac{\pi \ln^2(2)}{3}-2\ln(2)\left(-\ln(2)\int_0^{\pi/3}\,dx-\sum_{k=1}^\infty\frac{1}{n}\int_0^{\pi/3}\cos(2 n \pi)\,dx \right)+\frac{\pi^3}{12}-\frac{\pi^3}{18}+\frac{\pi^3}{81}+ \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left(\frac{2 \pi n}{3} \right)}{n^2}- \sum_{n=1}^\infty \frac{\sin\left(\frac{2 \pi n}{3} \right)}{n^3}\\ &=-\frac{\pi \ln^2(2)}{3}+\frac{2\pi\ln^2(2)}{3}+\ln(2)\operatorname{Cl}_2\left( \frac{2 \pi }{3}\right)+\frac{13\pi^3}{324}+ \operatorname{Gl}_3\left( \frac{2 \pi }{3}\right)- \Im \operatorname{Li}_3\left(1-e^{i 2\pi/3}\right)-\frac{\pi^3}{36}-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)-\frac{ \pi \ln^2(3)}{12} -\operatorname{Gl}_3\left( \frac{2 \pi }{3}\right)\\ &=\frac{\pi \ln^2{2}}{3}-\frac{\pi \ln^2(3)}{12}+\frac{\pi^3}{81}-\Im \ {\rm Li}_3(1-e^{i2\pi/3})+\frac{\ln(2)}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \qquad \blacksquare \end{aligned}$

Computing $\sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{2 \pi n}{3}\right)}{n^2}$

Recall (see here)

$\sum_{n=1}^\infty \frac{\operatorname{H}_{n}x^n}{n^2}=\operatorname{Li}_3\left( x\right)-\operatorname{Li}_3\left(1-x \right)+\ln\left( 1-x\right)\operatorname{Li}_2\left(1-x \right)+\frac12\ln\left(x\right)\ln^2\left(1-x\right)+\zeta(3)$

Letting $x=e^{i 2\pi/3}$ we obtain

$\begin{aligned} \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\cos\left( \frac{ 2\pi n}{3}\right)}{n^2}+i\sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{2 \pi n}{3}\right)}{n^2}&=\operatorname{Li}_3\left( \frac{2 \pi}{3}\right)-\operatorname{Li}_3\left(1-e^{i 2\pi/3}\right)+\ln\left( 1-e^{i 2\pi/3}\right)\operatorname{Li}_2\left(1-e^{i 2\pi/3} \right)+\frac12\ln\left(e^{i 2\pi/3}}\right)\ln^2\left(1-e^{i 2\pi/3}\right)+\zeta(3)\\ \end{aligned}$

We are looking for the Imaginary part of the equation above:

$\begin{aligned} \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{2 \pi n}{3}\right)}{n^2}&=\operatorname{Gl}_3\left( \frac{2 \pi }{3}\right)- \Im \operatorname{Li}_3\left(1-e^{i 2\pi/3}\right)-\frac{\pi\ln^2(3)}{6}-\frac{\pi^3}{54}-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)+\frac{ \pi \ln^2(3)}{12}-\frac{\pi^3 }{108}\\ &=\operatorname{Gl}_3\left( \frac{2 \pi }{3}\right)- \Im \operatorname{Li}_3\left(1-e^{i 2\pi/3}\right)-\frac{\pi^3}{36}-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)-\frac{ \pi \ln^2(3)}{12}\\ \end{aligned}$

Computing the quantities:

$\begin{aligned} \ln\left(1-e^{i 2\pi/3} \right)&=\ln\left(1-\cos\left( \frac{2\pi}{3}\right)-i\sin\left( \frac{2\pi}{3} \right)\right)\\ &=\ln\left(1 +\frac{1}{2}-i \frac{\sqrt{3}}{2} \right)\\ &=\ln\left(\frac{3}{2}-i \frac{\sqrt{3}}{2} \right)\\ &=\frac12\ln\left(\frac{9}{4}+ \frac{3}{4} \right)-i\arctan\left( \frac{\sqrt{3}}{3}\right)\\ &=\frac12\ln\left(3 \right)-\frac{i \pi}{6} \qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} \ln\left^2(1-e^{i 2\pi/3} \right) &=\left(\frac12\ln\left(3 \right)-\frac{i \pi}{6} \right)^2\\ &=\frac{\ln^2(3)}{4}+i\frac{\pi\ln(3)}{6}-\frac{\pi^2}{36}\qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} {\rm Re} \operatorname{Li}_2\left( e^{i 2\pi/3}\right)&=\operatorname{Gl}_2\left(\frac{2 \pi}{3} \right)\\ &=\sum_{n=1}^\infty\frac{\cos\left(\frac{2 n \pi}{3} \right)}{n^2}\\ &=\zeta(2)-\frac{\pi}{2}\left(\frac{2 \pi}{3} \right)+\frac14 \left(\frac{2 \pi}{3} \right)^2\\ &=\frac{\pi^2}{6}-\frac{\pi^2}{3}+\frac{\pi^2}{9}\\ &=-\frac{\pi^2}{18} \qquad \blacksquare \end{aligned}$

$\begin{aligned} {\rm Im} \operatorname{Li}_2\left( e^{i 2\pi/3}\right)&=\operatorname{Cl}_2\left(\frac{2 \pi}{3} \right)\\ &=\sum_{n=1}^\infty\frac{\sin\left(\frac{2 n \pi}{3} \right)}{n^2}\\ &=\frac{\sin\left(\frac{2 \pi}{3} \right)}{1^2}+\frac{\sin\left(\frac{4 \pi}{3} \right)}{2^2}+\frac{\sin\left(\frac{6 \pi}{3} \right)}{3^2}+\frac{\sin\left(\frac{8 \pi}{3} \right)}{4^2}+\frac{\sin\left(\frac{10\pi}{3} \right)}{5^2}+\frac{\sin\left(\frac{12 \pi}{3} \right)}{6^2}+\cdots\\ &=\frac{\left(\frac{\sqrt{3}}{2}\right)}{1}+\frac{\left(-\frac{\sqrt{3}}{2}\right)}{4}+\frac{0}{9}+\frac{\left(\frac{\sqrt{3}}{2}\right)}{16}+\frac{\left(-\frac{\sqrt{3}}{2}\right)}{25}+\frac{0}{36}+\cdots\\ &=\frac{\sqrt{3}}{2}\sum_{k=0}^\infty \frac{1}{(3k+1)^2}-\frac{\sqrt{3}}{2}\sum_{k=0}^\infty \frac{1}{(3k+2)^2}\\ &=\frac{\sqrt{3}}{18}\sum_{k=0}^\infty \frac{1}{\left(k+\frac13\right)^2}-\frac{\sqrt{3}}{18}\sum_{k=0}^\infty \frac{1}{\left(k+\frac{2}{3}\right)^2}\\ &=\frac{1}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \qquad \blacksquare\\ \end{aligned}$

$\begin{aligned}{\rm Li}_2(1-e^{i2\pi/3}) &=\frac{\pi^2}{6}-\ln(e^{i2\pi/3})\ln(1-e^{i2\pi/3})-{\rm Li}_2(e^{i2\pi/3})\\ &=\frac{\pi^2}{6}-\left(\frac{2 \pi i}{3} \right)\left( \frac12\ln\left(3 \right)-\frac{i \pi}{6}\right)+\frac{\pi^2}{18}-\frac{i}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \\ &=\frac{\pi^2}{6}-\frac{\pi^2}{9}-\frac{i \pi \ln(3)}{3}+\frac{\pi^2}{18}-\frac{i}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \\ &=\frac{\pi^2}{9}-i\left(\frac{ \pi \ln(3)}{3}-\frac{1}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)\right) \qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} {\rm Im}\ln\left( 1-e^{i 2\pi/3}\right)\operatorname{Li}_2\left(1-e^{i 2\pi/3} \right)&={\rm Im}\left(\frac12\ln\left(3 \right)-\frac{i \pi}{6} \right)\left( \frac{\pi^2}{9}-i\frac{ \pi \ln(3)}{3}-i\frac{1}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)\right)\\ &=-\frac{\pi\ln^2(3)}{6}-\frac{\pi^3}{54}-\frac{\ln(3)}{12\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)\qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} {\rm Im}\ln\left(e^{i 2\pi/3}}\right)\ln^2\left(1-e^{i 2\pi/3}\right)&={\rm Im}\left(\frac{2 \pi i}{3} \right)\left( \frac{\ln^2(3)}{4}+i\frac{\pi\ln(3)}{6}-\frac{\pi^2}{36}\right)\\ &=\frac{ \pi \ln^2(3)}{6}-\frac{\pi^3 }{54}\qquad \blacksquare\\ \end{aligned}$

The Glaisher function

We know that (see here):

$\operatorname{Gl}_1\left(x \right)&=\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi}{2}-\frac{x}{2}$

If we integrate $\operatorname{Gl}_1\left(x \right)$ from 0 to x we obtain

$\operatorname{Gl}_2\left(x \right)&=\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}=\zeta(2)-\frac{\pi}{2}x-\frac{x^2}{4}$

Integrating $\operatorname{Gl}_2\left(x \right)$ from 0 to x we obtain

$\operatorname{Gl}_3\left(x \right)&=\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=\frac{\pi^2}{6}x-\frac{\pi}{4}x^2-\frac{x^3}{12}$

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