Second Moment of Logsine squared

Today´s post We will compute the following logsine integral


\int_0^{\pi/2}x^2\ln^2\left(2 \sin(x) \right)dx=-\frac{3 \pi^{5}}{320}+\pi \operatorname{Li}_{4}\left(\frac{1}{2}\right)+\frac{7}{8} \ln (2) \zeta(3) \pi+\frac{\pi}{24} \ln ^{4}(2)-\frac{\pi^{3}}{24} \ln ^{2}(2)


Analogous to what we did previously in this post with regards to the pair


\int_0^{\pi/2}x \ln\left(\cos(x)\right)\,dx-\frac{\pi^2\ln(2)}{8}-\frac{7\zeta(3)}{16}

\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16}


We will find a series representation to \ln^2\left(2 \sin(x) \right) and then use it to the evaluation.


Recall the following relations:

\sum_{k=1}^\infty\frac{\sin\left(2kx \right)}{k}=\frac{\pi}2-x(1)

and

\sum_{k=1}^\infty\frac{\cos\left(2kx \right)}{k}=-\ln\left(2 \sin(x) \right)(2)


and by Euler´s formula we have


\begin{aligned} \sum_{k=1}^\infty\frac{e^{2ikx} }{k}&=\sum_{k=1}^\infty\frac{\cos\left(2kx \right)}{k}+i\sum_{k=1}^\infty\frac{\sin\left(2kx \right)}{k}\\ &=-\ln\left(2 \sin(x) \right)+i\left(\frac{\pi}2-x\right) \end{aligned}(3)


But the left hand of (3) is equal to


\sum_{k=1}^\infty\frac{e^{2ikx} }{k}=-\ln(1-{e^{2ix})(4)

And therefore we conclude that


\ln(1-{e^{2ix})=\ln\left(2 \sin(x) \right)-i\left(\frac{\pi}2-x \right)(5)


If we square both sides of (5) we obtain


\begin{aligned} \ln^2(1-{e^{2ix})&=\left(\ln\left(2 \sin(x) \right)-i\left(\frac{\pi}2-x \right)\right)^2\\ &=\ln^2\left(2 \sin(x) \right)-\left(\frac{\pi}2-x\right)^2-2i\ln\left(2 \sin(x) \right)\left(\frac{\pi}2-x \right) \end{aligned}(6)

Which gives us

\Re\left\{\ln^2(1-{e^{2ix}) \right\}=\ln^2\left(2 \sin(x) \right)-\left(\frac{\pi}2-x\right)^2(7)

\Im\left\{\ln^2(1-{e^{2ix}) \right\}=-2\ln\left(2 \sin(x) \right)\left(\frac{\pi}2-x \right)(8)


Now recall the generating function of Harmonic numbers (see here)


\sum_{n=1}^\infty \text{H}_{n-1}x^{n-1}=-\frac{\ln(1-x)}{1-x}(9)


integrating both sides of (9) w.r. to x


\begin{aligned} \frac{\ln^2(1-x)}{2}&=\sum_{n=1}^\infty \frac{\text{H}_{n-1}x^{n}}{n}\\ &=\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n}-\sum_{n=1}^\infty \frac{x^{n}}{n^2} \qquad \left( \text{H}_{n-1}=\text{H}_{n}-\frac1{n}\right)\\ &=\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n}-\text{Li}_2(x) \end{aligned}(10)

\ln^2(1-x)=2\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n}-2\text{Li}_2(x)(11)


Now if we let x=e^{2ix} in (11) we obtain


\begin{aligned} \ln^2(1-e^{2ix})&=2\sum_{n=1}^\infty \frac{\text{H}_{n}e^{2inx}}{n}-2\text{Li}_2\left(e^{2ix}\right)\\ &=2\sum_{n=1}^\infty \frac{\text{H}_{n}e^{2inx}}{n}-2\sum_{n=1}^\infty \frac{e^{2inx}}{n^2}\\ &=2\left( \sum_{n=1}^\infty \frac{\text{H}_{n}\cos(2nx)}{n}-\sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}\right)+2i\left( \sum_{n=1}^\infty \frac{\text{H}_{n}\sin(2nx)}{n}-\sum_{n=1}^\infty \frac{\sin(2nx)}{n^2}\right) \end{aligned}(12)

Which gives us

\Re\left\{\ln^2(1-{e^{2ix}) \right\}=2 \sum_{n=1}^\infty \frac{\text{H}_{n}\cos(2nx)}{n}-2\sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}(13)

\Im\left\{\ln^2(1-{e^{2ix}) \right\}=2\sum_{n=1}^\infty \frac{\text{H}_{n}\sin(2nx)}{n}-2\sum_{n=1}^\infty \frac{\sin(2nx)}{n^2}(14)


Comparing equation (7) with (13) we conclude that


\ln^2\left(2 \sin(x) \right)+\left(\frac{\pi}2-x\right)^2=2 \sum_{n=1}^\infty \frac{\text{H}_{n}\cos(2nx)}{n}-2\sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}(15)

Or

\ln^2\left(2 \sin(x) \right)=\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\text{H}_{n}\cos(2nx)}{n}-2\sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}(16)

Now, lets evaluate the integral


\begin{aligned} \int_0^{\pi/2}x^2\ln^2\left(2 \sin(x) \right)\,dx&=\frac{\pi^2}4\int_0^{\pi/2}x^2\,dx-\pi\int_0^{\pi/2}x^3\,dx+\int_0^{\pi/2}x^4\,dx+2\left( \sum_{n=1}^\infty \frac{\text{H}_{n}}{n}-\sum_{n=1}^\infty \frac{1}{n^2}\right)\int_0^{\pi/2}x^2\cos(2nx)\,dx\\ &=\frac{\pi^5}{96}-\frac{\pi^4}{64}+\frac{\pi^5}{160}+2\left( \sum_{n=1}^\infty \frac{\text{H}_{n}}{n}-\sum_{n=1}^\infty \frac{1}{n^2}\right)\left( \frac{\pi}{4}\frac{(-1)^n}{n^2}\right)\\ &=\frac{\pi^5}{96}-\frac{\pi^4}{64}+\frac{\pi^5}{160}+\frac{\pi}{2}\left( \sum_{n=1}^\infty \frac{(-1)^n\text{H}_{n}}{n^3}-\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\right)\\ &=\frac{\pi^5}{96}-\frac{\pi^4}{64}+\frac{\pi^5}{160}+\frac{\pi}{2}\left( \sum_{n=1}^\infty \frac{(-1)^n\text{H}_{n}}{n^3}+\eta(4)\right)\\ &=\frac{\pi^5}{96}-\frac{\pi^4}{64}+\frac{\pi^5}{160}+\frac{7 \pi^5}{1440}+\frac{\pi}{2}\left( \sum_{n=1}^\infty \frac{(-1)^n\text{H}_{n}}{n^3}\right)\\ &=\frac{\pi^5}{96}-\frac{\pi^5}{64}+\frac{\pi^5}{160}+\frac{7 \pi^5}{1440}+\frac{\pi}{2}\left( -\frac{11 \pi^4}{360}+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12}\right)\\ &=-\frac{3 \pi^{5}}{320}+\pi \operatorname{Li}_{4}\left(\frac{1}{2}\right)+\frac{7}{8} \ln (2) \zeta(3) \pi+\frac{\pi}{24} \ln ^{4}(2)-\frac{\pi^{3}}{24} \ln ^{2}(2) \qquad \blacksquare \end{aligned}

Where we used the results

\sum_{n=1}^\infty\frac{(-1)^{n}\text{H}_n}{n^3}=-\frac{11 \pi^4}{360}+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12}

proved here.

Also for the integral:

\begin{aligned} \int_0^{\pi/2} x^2 \cos(2 kx)\,dx&=\frac{1}{(2 k)^3}\int_0^{\pi k} x^2 \cos(x)\,dx \qquad (2 kx \to x)\\ &=\frac{1}{(2 k)^3}\left(x^2\sin(x)\Big|_0^{ \pi k} -2\int_0^{ \pi k}x\sin(x)\,dx\right)\\ &=\frac{-2}{(2 k)^3}\left(\int_0^{ \pi k}x \sin x \,dx\right)\\ &=\frac{-2}{(2 k)^3}\left(-x\cos x\Big|_0^{ \pi k}+\int_0^{ \pi k}\cos x \,dx\right)\\ &=\frac{2}{(2 k)^3}\left(\pi k (-1)^k \right)\\ &= \frac{\pi (-1)^k}{4 k^2}\qquad \blacksquare \end{aligned}

Check here the entry where we computed the sister of today´s integral:

\int_0^{\pi/2} x^2 \ln^2\left(2\cos x\right)\,dx=\frac{11 \pi^5}{1440}

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