How to evaluate $\int_0^{\pi/2} x^2 \ln^2(2\cos{x}) \mathrm{d}x$

Question

In this post, It is mentioned that $$ \int_0^{\pi/2} x^2 \ln^2(2\cos{x}) \mathrm{d}x = \frac{11 \pi}{16} \zeta(4) $$ is easy to evaluate.

I thought it by doing integration by parts but if I assume $x^2$ as $1st$ function and $\ln^2(2\cos{x})$ as $2nd$ function but problem with this is that, We can't integrate $\ln^2(2\cos{x})$ directly.

I also tried to use Fourier series of $\ln(2\cos{x})$ $$ \ln(2 \cos(x)) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \cos(2 n x)$$ But it didn't work as well.

Please guide me how to solve it ? Thank you very much!!

Answer 1

We will prove

$$\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(2\cos\left(x\right)\right)dx = \frac{11\pi^{5}}{1440}.$$

Proof. Let the given integral be $I$. Then $$ \eqalign{ I&=\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(2\cos\left(x\right)\right)dx \cr &= \int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(\frac{e^{2ix}+1}{e^{ix}}\right)dx \cr &= \int_{0}^{\frac{\pi}{2}}x^{2}\left(\ln\left(e^{2ix}+1\right)-ix\right)^{2}dx \cr &= \int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(1+e^{2ix}\right)dx-2i\int_{0}^{\frac{\pi}{2}}x^{3}\ln\left(1+e^{2ix}\right)dx-\int_{0}^{\frac{\pi}{2}}x^{4}dx \cr &= \int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(1+e^{2ix}\right)dx-2i\int_{0}^{\frac{\pi}{2}}x^{3}\ln\left(1+e^{2ix}\right)dx-\frac{\pi^{5}}{160} \cr } $$

For the first integral, let $f(z) = z^2\log^2\left(1+e^{2iz}\right)$ and define a rectangular contour

$$C=\left\{z\in\mathbb{C} : \Re(z) \in \left[0,\frac{\pi}{2}\right] \wedge \Im(z) \in \left[0,R\right]\right\}$$

that we will traverse counterclockwise around. By Cauchy's Residue Theorem, we get

$$0 = \int_{0}^{\frac{\pi}{2}}f\left(z\right)dz+\int_{\frac{\pi}{2}}^{\frac{\pi}{2}+iR}f\left(z\right)dz+\int_{\frac{\pi}{2}+iR}^{iR}f\left(z\right)dz+\int_{iR}^{0}f\left(z\right)dz.$$

With some grunt work, the third integral will vanish as $R\to\infty$. The fourth integral will also vanish when we take $\Re$ on both sides. As $R \to \infty$, we get

$$ \eqalign{ \Re\int_{0}^{\frac{\pi}{2}}f\left(z\right)dz &= -\Re\int_{\frac{\pi}{2}}^{\frac{\pi}{2}+i\infty}f\left(z\right)dz \cr &= -\Re i\int_{0}^{R}f\left(\frac{\pi}{2}+ix\right)dx \cr &= -\Re i\int_{0}^{\infty}\left(\frac{\pi}{2}+ix\right)^{2}\log^{2}\left(1+e^{2i\left(\frac{\pi}{2}+ix\right)}\right)dx \cr &= \pi\int_{0}^{\infty}x\ln^{2}\left(1-e^{-2x}\right)dx \cr &= \pi\int_{0}^{\infty}x\left(-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n}\right)^{2}dx \cr &= \pi\int_{0}^{\infty}x\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n}\sum_{k=1}^{\infty}\frac{e^{-2kx}}{k}dx \cr &= \sum_{n,k=1}^{\infty}\frac{\pi}{nk}\int_{0}^{\infty}xe^{-2\left(n+k\right)x}dx \cr &= \sum_{n,k=1}^{\infty}\frac{\pi}{nk}\left(\frac{1}{4\left(n+k\right)^{2}}\right) \cr &= \frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{\left(n+k\right)^{2}-n^{2}-k^{2}}{n^{2}k^{2}\left(n+k\right)^{2}} \cr &= \frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{1}{n^{2}k^{2}}-\frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{1}{k^{2}\left(n+k\right)^{2}}-\frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{1}{n^{2}\left(n+k\right)^{2}} \cr &= \frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{1}{n^{2}k^{2}}-\frac{\pi}{8}\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{1}{k^{2}n^{2}}-\frac{\pi}{8}\sum_{n=1}^{\infty}\sum_{k=n+1}^{\infty}\frac{1}{n^{2}k^{2}} \cr &= \frac{\pi}{8}\sum_{n,k=1}^{\infty}\frac{1}{n^{2}k^{2}}-\frac{\pi}{8}\sum_{1\le k<n<\infty}^{ }\frac{1}{k^{2}n^{2}}-\frac{\pi}{8}\sum_{1\le n<k<\infty}^{ }\frac{1}{n^{2}k^{2}} \cr &= \frac{\pi}{8}\sum_{n=k=1}^{\infty}\frac{1}{n^{2}k^{2}} \cr &= \frac{\pi}{8}\left(\frac{\pi^{4}}{90}\right) \cr &= \frac{\pi^{5}}{720}. \cr } $$

Additionally, we have

$$ \eqalign{ \int_{0}^{\frac{\pi}{2}}x^{3}\ln\left(1+e^{2ix}\right)dx &= \int_{0}^{\frac{\pi}{2}}x^{3}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}e^{2nix}}{n}dx \cr &= \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\int_{0}^{\frac{\pi}{2}}x^{3}e^{2nix}dx \cr &= \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\left(\frac{6+e^{i\pi n}\left(-6+\pi n\left(\pi n\left(3-i\pi n\right)+6i\right)\right)}{16n^{4}}\right) \cr &= \frac{3}{8}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n^{5}}+\frac{3}{8}\sum_{n=1}^{\infty}\frac{1}{n^{5}}-\frac{3\pi i}{8}\sum_{n=1}^{\infty}\frac{1}{n^{4}}-\frac{3\pi^{2}}{8}\sum_{n=1}^{\infty}\frac{1}{n^{3}}+\frac{i\pi^{3}}{16}\sum_{n=1}^{\infty}\frac{1}{n^{2}} \cr &= \frac{3}{8}\left(\frac{15}{16}\zeta{(5)}\right)+\frac{3}{8}\zeta{(5)}-\frac{3\pi i}{8}\left(\frac{\pi^{4}}{90}\right)-\frac{3\pi^{2}}{16}\zeta{(3)}+\frac{i\pi^{3}}{16}\left(\frac{\pi^{2}}{6}\right) \cr &= \frac{i\pi^{5}}{160}+\frac{93\zeta{(5)}}{128}-\frac{3\pi^{2}\zeta{(3)}}{16}. \cr } $$

Gathering everything together and taking the real part of $I$, we get

$$ \Re I = \frac{\pi^{5}}{720}-\Re 2i\left(\frac{i\pi^{5}}{160}+\frac{93\zeta{(5)}}{128}-\frac{3\pi^{2}\zeta{(3)}}{16}\right)-\frac{\pi^{5}}{160} = \frac{11\pi^{5}}{1440}. $$

In conclusion,

$$\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}\left(2\cos\left(x\right)\right)dx = \frac{11\pi^{5}}{1440}.$$

Q.E.D.

Answer 2

Using the Fourier expansion does work, though I have yet to completely evaluate every sum that makes up the overall integral. Recall the log-sine expansion,

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

Expand the integrand as

$$x^2 \ln^2(2\cos(x)) = x^2 \bigg(\ln^2(2) + 2 \ln(2) \ln(\cos(x)) + \ln^2(\cos(x))\bigg)$$

Let

$$I_{m,n} = \int_0^{\frac\pi2} x^m \ln^n(\sin(x)) \, dx$$

so that

$$\begin{align*} \mathcal I &= \int_0^{\frac\pi2} x^2 \ln^2(2\cos(x)) \, dx \\[1ex] &= \ln^2(2) I_{2,0} + 2\ln(2) I_{2,1} + I_{2,2} \end{align*}$$

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$$\begin{align*} I_{2,0} &= \int_0^{\frac\pi2} x^2 \, dx \\[1ex] &= \frac{\pi^3}{24} \end{align*}$$

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$$\begin{align*} I_{2,1} &= - \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \left(\ln(2) + \sum_{k=1}^\infty \frac{\cos(2kx)}k\right) \, dx \\[1ex] &= -\frac{\pi^3}{24}\ln(2) - \sum_{k=1}^\infty \frac1k \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \cos(2kx) \, dx \\[1ex] &= -\frac{\pi^3}{24}\ln(2) - \sum_{k=1}^\infty \frac1{2k} \int_0^\pi \left(\frac\pi2-\frac x2\right)^2 \cos(kx) \, dx \\[1ex] &= -\frac{\pi^3}{24}\ln(2) - \sum_{k=1}^\infty \frac\pi{4k} \left(\frac{(-1)^k}{k^2} - \frac{-1+(-1)^k}{k^2}\right) \\[1ex] &= -\frac{\pi^3}{24}\ln(2) - \frac\pi4 \zeta(3) \end{align*}$$

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$$\begin{align*} I_{2,2} &= \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \left(-\ln(2)-\sum_{k=1}^\infty \frac{\cos(2kx)}k\right)^2 \, dx \\[1ex] &= \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \left(\ln^2(2) + 2 \ln(2) \sum_{k=1}^\infty \frac{\cos(2kx)}k + \left(\sum_{k=1}^\infty \frac{\cos(2kx)}k\right)^2\right) \, dx \\[1ex] &= \frac{\pi^3}{24}\ln^2(2) + \frac\pi4 \zeta(3) + \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \left(\sum_{k=1}^\infty \frac{\cos(2kx)}k\right)^2 \, dx \\[1ex] &= \frac{\pi^3}{24}\ln^2(2) + \frac\pi4 \zeta(3) \\ &\qquad + \int_0^{\frac\pi2} \left(\frac\pi2-x\right)^2 \left(\sum_{k=1}^\infty \frac{\cos^2(2kx)}{k^2} + 2 \sum_{(k_1,k_2)\in\Bbb N^2} \frac{\cos(2k_1x)\cos(2k_2)x}{k_1k_2}\right) \, dx \\[1ex] &= \frac{\pi^3}{24}\ln^2(2) + \frac\pi4 \zeta(3) + \sum_{k=1}^\infty \frac1{4k^2} \left(\frac{\pi^3}{12} + \frac\pi{8k^2}\right) \\ &\qquad + \frac\pi4 \sum_{(k_1,k_2)\in\Bbb N^2} \frac1{k_1k_2} \left(\frac1{(k_1+k_2)^2} + \frac1{(k_1-k_2)^2}\right) \\[1ex] &= \frac{\pi^3}{24}\ln^2(2) + \frac\pi4 \zeta(3) + \frac{11\pi^5}{2880} \\&\qquad + \frac\pi4 \sum_{(k_1,k_2)\in\Bbb N^2} \frac1{k_1k_2} \left(\frac1{(k_1+k_2)^2} + \frac1{(k_1-k_2)^2}\right) \\[1ex] &= \frac{\pi^3}{24}\ln^2(2) + \frac\pi8 \zeta(3) + \frac{3\pi^5}{320} \\&\qquad + \frac\pi4 \sum_{(k_1,k_2)\in\Bbb N^2} \frac1{k_1k_2} \frac1{(k_1-k_2)^2} \end{align*}$$

where in the double sum, $k_1\neq k_2$.

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To evaluate the first sum in $I_{2,2}$, I've used the same approach as shown here. Let

$$g(x) = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab(a+b)^2}$$

Note that this includes the case of $a=b$. Differentiate and multiply by $x$ to recover $f$ (as in linked answer), integrate by parts to solve for $g$, and let $x\to1$ from below.

$$\begin{align*} g'(x) &= \frac1x \int_0^x \frac{\ln^2(1-y)}y\,dy\\ &= \frac{\ln(x)\ln^2(1-x)+2\ln(1-x)\operatorname{Li}_2(1-x)-2\operatorname{Li}_3(1-x)}x \end{align*}$$

Integrate by parts again to get

$$\begin{align*} \sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)^2} &= \int_0^1 \frac{\ln(x)\ln^2(1-x)+2\ln(1-x)\operatorname{Li}_2(1-x)-2\operatorname{Li}_3(1-x)}x\,dx \\ &= -\int_0^1 \frac{\ln(x)\ln^2(1-x)}x \, dx \end{align*}$$

where IBP shows the last two integrals are absorbed into the first. Recalling the generating function for $H_n$ the harmonic numbers, $-\frac{\ln(1-x)}{1-x}$, it follows that

$$\sum_{(k_1,k_2)\in\Bbb N^2} \frac1{k_1k_3(k_1+k_2)^2} = \sum_{n=1}^\infty \frac{2H_n}{(n+1)^3} - \sum_{n=1}^\infty \frac1{2n^3} = \frac{\pi^4}{180} - \frac{\zeta(3)}2$$

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in-progress

The second sum of $I_{2,2}$ still remains,

$$\sum_{(k_1,k_2)\in\Bbb N^2} \frac1{k_1k_2} \frac1{(k_1-k_2)^2}$$

Answer 3

Using $$\ln^2(2\cos x)=x^2+2\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}\cos(2nx),\quad |x|<\frac{\pi}{2}$$ we have

$$\int_0^{\pi/2} x^2 \ln^2(2\cos{x}) \mathrm{d}x=\int_0^{\pi/2} x^4 \mathrm{d}x+2\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}\int_0^{\pi/2} x^2\cos(2nx) \mathrm{d}x$$

$$=\frac{\pi^5}{160}+2\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}\left(\frac{\pi\cos(n\pi)}{4n^2}-\frac{\sin(n\pi)}{4n^3}+\frac{\pi^2\sin(n\pi)}{8n}\right)$$

$$=\frac{\pi^5}{160}+2\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}\left(\frac{\pi(-1)^n}{4n^2}\right)$$ $$=\frac{\pi^5}{160}+\frac{\pi}{2}\sum_{n=1}^\infty\frac{H_{n}}{n^3}-\frac{\pi}{2}\zeta(4)=\frac{11\pi^{5}}{1440}.$$