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

        Today we will evaluate the beautiful integral found in this post


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


We have (see appendix bellow for the proof)


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

Then



\begin{aligned} \int_0^{\pi/2} x^2 \ln^2\left(2\cos x\right)\,dx&=\int_0^{\pi/2} x^4\,dx+2\sum_{k=1}^\infty \frac{(-1)^kH_{k-1}}{k}\int_0^{\pi/2} x^2 \cos(2 kx)\,dx\\ &=\frac{\pi^5}{160}+2\sum_{k=1}^\infty \frac{(-1)^kH_{k-1}}{k}\frac{\pi (-1)^k}{4 k^2}\\ &=\frac{\pi^5}{160}+\frac{\pi}{2}\sum_{k=1}^\infty \frac{H_{k-1}}{k^3}\\ &=\frac{\pi^5}{160}+\frac{\pi}{2}\sum_{k=1}^\infty \frac{H_{k}}{k^3}-\frac{\pi}{2}\sum_{k=1}^\infty \frac{1}{k^4} \qquad \left(H_k=H_{k-1}+\frac{1}{k} \right)\\ &=\frac{\pi^5}{160}+\frac{\pi}{2}\frac{\pi^4}{72}-\frac{\pi}{2}\zeta(4) \\ &=\frac{\pi^5}{160}+\frac{\pi^5}{144}-\frac{\pi^5}{180} \\ &=\frac{11 \pi^5}{1440} \qquad \blacksquare \end{aligned}

Appendix


Recall the generating function


\begin{aligned} &\sum_{k=1}^\infty H_{k-1}x^{k-1}=-\frac{\ln(1-x)}{1-x}\\ & \sum_{k=1}^\infty H_{k-1} \int_0^x w^{k-1}\,dw=-\int_0^x\frac{\ln(1-w)}{1-w}\,dw \qquad \text{integrating both sides from 0 to x}\\ & \sum_{k=1}^\infty \frac{H_{k-1} x^k}{k}=\int_1^{1-x}\frac{\ln(w)}{w}\,dw \qquad (1-w \to w)\\ \end{aligned}


Lets now focus only in the integral of the R.H.S.


\begin{aligned} &\int_1^{1-x}\frac{\ln(w)}{w}\,dw=\ln^2(w)\Big|_1^{1-x}-\int_1^{1-x}\frac{\ln(w)}{w}\,dw\\ &2\int_1^{1-x}\frac{\ln(w)}{w}\,dw=\ln^2(1-x)\\ &\int_1^{1-x}\frac{\ln(w)}{w}\,dw=\frac{\ln^2(1-x)}{2}\\ \end{aligned}

We then conclude that


\sum_{k=1}^\infty \frac{H_{k-1} x^k}{k}=\frac{\ln^2(1-x)}{2} \qquad \blacksquare(A.1)

Recall

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


\sum_{k=1}^\infty \frac{(-1)^k\sin(2 k x)}{k}=-x(A.3)


\sum_{k=1}^\infty \frac{(-1)^ke^{2 ik x}}{k}=\sum_{k=1}^\infty \frac{(-e^{2i x})^k}{k}=\ln(1+e^{2 i x})(A.4)


On the other hand


\begin{aligned} \sum_{k=1}^\infty \frac{(-1)^ke^{2 ik x}}{k}&=\sum_{k=1}^\infty \frac{(-1)^k\cos(2 k x)}{k}+i\sum_{k=1}^\infty \frac{(-1)^k\sin(2 k x)}{k}\\ &=-\ln\left(2 \cos x \right)-ix\\ \end{aligned}(A.5)

Therefore


\ln(1+e^{2 i x})=-\ln\left(2 \cos x \right)-ix(A.6)


Squaring both sides of (A.6) we obtain


\begin{aligned} \ln^2(1+e^{2 i x})&=\left(-\ln\left(2 \cos x \right)-ix \right)^2\\ &=\ln^2\left(2 \cos x \right)+i2x\ln\left(2 \cos x \right)-x^2\\ \end{aligned}(A.7)


Now, let x=-e^{2 i x} in (A.1), we obtain


\sum_{k=1}^\infty \frac{(-1)^kH_{k-1} e^{2 i x}}{k}=\frac{\ln^2(1+e^{2 i x})}{2}(A.8)


Plugging the R.H.S. of (A.7) in the R.H.S. of (A.8) we get


2\sum_{k=1}^\infty \frac{(-1)^kH_{k-1} e^{2 i x}}{k}=\ln^2\left(2 \cos x \right)+i2x\ln\left(2 \cos x \right)-x^2(A.9)


Equating Real and Imaginary parts in (A.9) we obtain


2\sum_{k=1}^\infty \frac{(-1)^kH_{k-1} \cos(2 x)}{k}=\ln^2\left(2 \cos x \right)-x^2(A.10)

2\sum_{k=1}^\infty \frac{(-1)^kH_{k-1} \sin(2 x)}{k}=2x\ln\left(2 \cos x \right)(A.11)

\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}(A.11)

Recall that     H_{n}=\int_{0}^{1} \frac{1-t^{n}}{1-t} d t,     then


\begin{aligned} &\sum_{n=0}^{\infty} \frac{H_{n}}{n^{3}}=\sum_{n=0}^{\infty} \frac{1}{n^{3}} \int_{0}^{1} \frac{1-t^{n}}{1-t} d t \\ &=\int_{0}^{1} \frac{1}{1-t}\left\{\sum_{n=0}^{\infty} \frac{1-t^{n}}{n^{3}}\right\} d t \\ &=\int_{0}^{1} \frac{1}{1-t}\left\{\sum_{n=0}^{\infty} \frac{1}{n^{3}}-\sum_{n=0}^{\infty} \frac{t^{n}}{n^{3}}\right\} d t \\ &=\int_{0}^{1} \frac{1}{1-t}\left\{\zeta(3)-L i_{3}(t)\right\} d t \\ &=\zeta(3) \int_{0}^{1} \frac{d t}{1-t}-\int_{0}^{1} \frac{L i_{3}(t)}{1-t} d t \\ &=-\left.\zeta(3) \ln (1-x)\right|_{0} ^{1}-\int_{0}^{1} \frac{L i_{3}(t)}{1-t} d t \end{aligned}

Lets integrate by part the remaining integral, considering that     L i_{3}(t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n^{3}}



\begin{aligned} \sum_{n=0}^{\infty} \frac{H_{n}}{n^{3}}&=-\left.\zeta(3) \ln (1-x)\right|_{0} ^{1}-\left\{-\left.L i_{3}(t) \ln (1-x)\right|_{0} ^{1}+\int_{0}^{1} \frac{L i_{2}(t) \ln (1-t)}{t} d t\right\} \\ &=\left.\left(L i_{3}(t)-\zeta(3)\right) \ln (1-x)\right|_{0} ^{1}-\int_{0}^{1} \frac{L i_{2}(t) \ln (1-t)}{t} \,d t\\ &=-\underbrace{\int_{0}^{1} \frac{L i_{2}(t) \ln (1-t)}{t} d t}_{J} \\ \end{aligned}


Integrating by parts    u=L i_{2}(t)$ and $v=\int \frac{\ln (1-t)}{t} d t=-L i_{2}(t)


\begin{aligned} J&=-L i_{2}^{2}(t)\Big|\right|_{0} ^{1}+\underbrace{\int_{0}^{1} \frac{L i_{2}(t) \ln (1-t)}{t} d t}_{-J} \\ &=\frac12\left(-\sum_{n=1}^\infty\frac{t^n}{n^2}\right)^2\Bigg|_{0} ^{1}\\ &=\frac{-\zeta(2)^2}{2} \end{aligned}

Therefore

\sum_{n=0}^{\infty} \frac{H_{n}}{n^{3}}=\frac{\zeta(2)^2}{2}=\frac{\pi^4}{72} \qquad \blacksquare(A.12)

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