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How to prove

$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$

This problem is proposed by Cornel which can be found here where he suggested that the problem can be solved with and without harmonic series.

Here is my approach but I got stuck at the blue integral:

Using the common identity

$$ \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$$

Set $p=-\cos(x)$ we get

$$ \sum_{n=1}^{\infty}(-1)^n \cos^n(x) \cos(nx)=-\frac{2\cos^2(x)}{1+3\cos^2(x)}=-\frac23+\frac23\frac1{1+3\cos^2(x)}$$

Multiply both sides by $-x^2$ then integrate from $x=0$ to $\pi/2$ we get

$$\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac23\int_0^{\pi/2} x^2dx-\frac23\color{blue}{\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx}\\=\frac{\pi^3}{36}-\frac23\left(\color{blue}{\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)}\right)=\frac{\pi^3}{72}-\frac{\pi}{6}\operatorname{Li}_2\left(\frac13\right)$$


I have two Questions:

1) Can we evaluate $I$ in a different way?

2) How to finish the blue integral?


My try to the blue integral is using integration by parts

$$\int\frac{dx}{1+3\cos^2(x)}=\frac12\tan^{-1}\left(\frac{\tan(x)}{2}\right)=-\frac12\tan^{-1}\left(2\cot(x)\right)$$

which gives us

$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\frac{\pi^3}{16}-\int_0^{\pi/2}x\tan^{-1}\left(\frac{\tan(x)}{2}\right)dx$$

Or

$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\int_0^{\pi/2}x\tan^{-1}\left(2\cot(x)\right)dx$$

I also tried the trick $x\to \pi/2-x$ but got complicated


Proof of the identity:

\begin{align} \sum_{n=0}^\infty p^ne^{inx}&=\sum_{n=0}^\infty\left(p e^{ix}\right)^n=\frac{1}{1-pe^{ix}},\quad |p|<1\\&=\frac{1}{1-p\cos(x)-ip\sin(x)}=\frac{1-p\cos(x)+ip\sin(x)}{1-2p\cos(x)+p^2}\\ &=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}+i\frac{p\sin(x)}{1-2p\cos(x)+p^2} \end{align}

By comparing the real and imaginary parts, we get

$$\sum_{n=\color{blue}{0}}^\infty p^n \cos(nx)=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{blue}{1}}^\infty p^{n-1} \cos(nx)=\frac{\cos(x)-p}{1-2p\cos(x)+p^2}$$

and

$$\sum_{n=\color{red}{0}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{red}{1}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}$$

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We can use the following fourier series:$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$ Plugging $a=5, b=3$ and $t=2x$ we get: $$\frac{1}{1+3\cos^2 x}=\frac{2}{5+3\cos(2x)}=\frac{1}{2}+\sum_{n=1}^\infty (-1)^n\left(\frac{1}{3}\right)^n\cos(2nx)$$ $$\Rightarrow \int_0^\frac{\pi}{2}\frac{x^2}{1+3\cos^2 x}dx=\frac12\int_0^\frac{\pi}{2} x^2dx+\sum_{n=1}^\infty(-1)^n \left(\frac13\right)^n\int_0^\frac{\pi}{2}x^2 \cos(2nx)dx$$ $$=\frac{\pi^3}{48}+\frac{\pi}4\sum_{n=1}^\infty \left(\frac13\right)^n\frac{1}{n^2}=\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)$$


Using the series obtain above, we can also conclude that: $$\sum_{n=1}^{\infty}(-1)^n \cos^n(x) \cos(nx)=-\frac13+\frac23\sum_{n=1}^\infty \left(-\frac{1}{3}\right)^n\cos(2nx)$$

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    $\begingroup$ Impressive work thank you (+1). may I ask how you got this idea? $\endgroup$ – Ali Shadhar Nov 9 '19 at 23:49
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    $\begingroup$ Some time ago I stumbled upon an integral of the type $f(x) \frac{1}{a+b\cos x}$ (posted here) and after that I always used to solve integrals of that type by expanding the denominator into Fourier series and most of the time it turned out to be a good approach. $\endgroup$ – Zacky Nov 10 '19 at 0:03
  • $\begingroup$ However it wasn't really my first try here, as I tried to continue your approach to integrate by parts and afterwards to consider the integral $I(t)=\int_0^\frac{\pi}{2}x\arctan(t\cot x)dx$ and differentiate under the integral sign. I still think it can work, since $I(1)=\int_0^\frac{\pi}{2}x(\pi/2-x)dx=\frac{\pi^3}{48}$, but it appears to be much harder. $\endgroup$ – Zacky Nov 10 '19 at 0:05
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    $\begingroup$ Yes It got complicated when i used $x\to \pi/2-x$. The path you took is much easier. $\endgroup$ – Ali Shadhar Nov 10 '19 at 0:10
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Evaluating the blue integral:

First we write

$$\frac1{1+3\cos^2(x)}=\frac{1}{5+3\cos(2x)}$$

Using the same identity in the post body

$$\sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1\tag1$$

But lets manipulate the denominator to have it in the form of $\frac1{5+3\cos(x)}$:

$$\frac1{1-2p\cos(x)+p^2}=\frac{-\frac{3}{2p}}{-\frac{3(1+p^2)}{2p}+3\cos(x)}$$

Now set $$-\frac{3(1+p^2)}{2p}=5\Longrightarrow p=-3,-\frac13$$

and since $|p|<1$, so we take $p=-\frac13$. Plug this value in (1) and replace $x$ by $2x$ we get

$$\frac{1}{5+3\cos(2x)}=\frac{1}{4}+\frac12\sum_{n=1}^\infty (-1)^n\left(\frac{1}{3}\right)^n\cos(2nx)\tag2$$

Multiply both sides of (2) by $x^2$ and integrate between $0$ and $\pi/2$ we get

$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)$$


Addendum:

The identity used by @Zacky above:

$$\frac{1}{a+b\cos(x)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nx)},\ a>b\tag{3}$$

can be derived the same way:

$$\frac1{1-2p\cos(x)+p^2}=\frac{-\frac{b}{2p}}{-\frac{b(1+p^2)}{2p}+b\cos(x)}$$

If we set $$-\frac{b(1+p^2)}{2p}=a\tag{4}$$

we can write

$$\frac1{1-2p\cos(x)+p^2}=\frac{\frac{a}{1+p^2}}{a+b\cos x}$$

We proved above that

$$\sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}=-\frac12-\frac{p^2-1}{1-2p\cos(x)+p^2}$$

$$=-\frac12-\frac12 \color{red}{\frac{p^2-1}{p^2+1}}\frac{\color{red}{a}}{1-2p\cos(x)+p^2}\tag5$$

From $(4)$ we find $p=\frac{\sqrt{a^2-b^2}-a}{b}$. Note that we ignored $p=\frac{\sqrt{a^2-b^2}+a}{b}$ as $|p|<1$.

Substitute this root in $(5)$ we get

$$\sum_{n=1}^\infty\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos(nx)=-\frac12-\frac12\cdot\frac{\color{red}{-\sqrt{a^2-b^2}}}{a+b\cos(x)}$$

or

$$\frac{1}{a+b\cos(x)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nx)}$$

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  • $\begingroup$ I like more your approach for that identity, as using this we keep everything away using complex analysis stuff. $\endgroup$ – Zacky Feb 8 '20 at 11:41
  • $\begingroup$ @Zacky thank you glad you like it. $\endgroup$ – Ali Shadhar Feb 8 '20 at 15:43

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